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PROFESSOR: So if you want to
hand in your problem sets, we
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00:00:03,870 --> 00:00:05,610
have three handouts.
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00:00:05,610 --> 00:00:10,305
The old problem set solutions,
the new problem set, and we're
4
00:00:10,305 --> 00:00:12,536
also handing out chapter
eight today.
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Of course, these are
all on the web.
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Reminders of previously
announced events.
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Next week I'm going
to be away.
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Ralf Koetter will be a
much-improved substitute.
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00:00:25,540 --> 00:00:28,790
He'll be talking about
Reed-Solomon codes and
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Reed-Solomon decoding.
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He's one of the world experts
on these things.
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He's a great lecturer.
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So you're really lucky,
I believe.
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I will, of course, look at the
video afterwards and see how
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he did, but I'm sure he'll
do it better than I
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would have done it.
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And midterm is in two weeks.
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That's just a reminder.
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When I come back, we'll have
a Monday class to go over
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anything that Ralf may have
missed, or really anything in
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00:01:02,570 --> 00:01:05,300
the whole course up through
chapter eight.
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You can bring questions
in to that.
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I don't know how much I feel
I'll need to embellish and
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00:01:13,330 --> 00:01:15,800
cover, so I don't know
how much time
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there will be for questions.
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But that can be a fairly
free form lecture.
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And then Ashish is going to run
a review session on the
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00:01:22,620 --> 00:01:27,560
Tuesday with exactly
the same purpose.
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So hopefully that'll put you in
good shape for the midterm.
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00:01:31,431 --> 00:01:34,440
We have to make up
the midterm.
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No idea what's going to
be on the midterm yet.
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OK.
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00:01:39,460 --> 00:01:43,350
Any questions about
these things?
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00:01:43,350 --> 00:01:46,280
Homework processes going OK?
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Office hours?
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Solutions?
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00:01:50,060 --> 00:01:52,010
Everything seems to be
all right so far?
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Good.
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00:01:53,260 --> 00:01:55,800
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We're in chapter seven.
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I certainly plan to
complete it today.
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I'm not going to cover all of
chapter seven, but the key
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00:02:04,380 --> 00:02:06,260
elements that I want
you to know.
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You're responsible only
for what's covered in
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00:02:08,070 --> 00:02:10,930
class, by the way.
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00:02:10,930 --> 00:02:15,205
We've talked about a lot
of algebraic objects.
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00:02:15,205 --> 00:02:18,190
50
00:02:18,190 --> 00:02:22,590
Our main emphasis has been to
get the finite field, and at
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00:02:22,590 --> 00:02:25,120
this point, we have
prime fields.
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00:02:25,120 --> 00:02:27,970
Fields with a prime number
p of elements.
53
00:02:27,970 --> 00:02:33,410
And we find that we can
construct such a field by
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00:02:33,410 --> 00:02:36,500
taking the integers mod p.
55
00:02:36,500 --> 00:02:38,830
You can write that
as z mod p z.
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00:02:38,830 --> 00:02:42,830
You can equally well consider
this as the equivalence
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00:02:42,830 --> 00:02:52,040
classes of integers of the
cosets of pz nz which are each
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00:02:52,040 --> 00:02:57,050
identified by a remainder
or residue r between
59
00:02:57,050 --> 00:03:02,430
0 and p minus 1.
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00:03:02,430 --> 00:03:02,840
OK.
61
00:03:02,840 --> 00:03:06,630
So basically, the elements of
the field are these remainders
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00:03:06,630 --> 00:03:09,560
or these cosets.
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00:03:09,560 --> 00:03:10,950
There are p of them.
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00:03:10,950 --> 00:03:16,080
We do arithmetic by simply
doing mod p addition and
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00:03:16,080 --> 00:03:17,510
multiplication.
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00:03:17,510 --> 00:03:22,190
So if you understand arithmetic
mod p, you
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00:03:22,190 --> 00:03:24,710
understand this field.
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00:03:24,710 --> 00:03:27,475
Yes?
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00:03:27,475 --> 00:03:31,100
AUDIENCE: Is Fp and
Zp isomorphic?
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00:03:31,100 --> 00:03:34,020
PROFESSOR: As an additive group,
it's isomorphic to Zp.
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00:03:34,020 --> 00:03:37,960
Zp we use that notation
for the group.
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00:03:37,960 --> 00:03:40,650
We use this notation
for the field.
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00:03:40,650 --> 00:03:45,170
And I write Zp is isomorphic
to Z mod pZ
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00:03:45,170 --> 00:03:46,580
as a quotient group.
75
00:03:46,580 --> 00:03:49,920
Here I've added in the
multiplication operation mod p
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00:03:49,920 --> 00:03:51,970
to make this a field.
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00:03:51,970 --> 00:03:57,290
So this is somewhat more than
just a quotient group.
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00:03:57,290 --> 00:04:00,050
Sorry.
79
00:04:00,050 --> 00:04:04,210
The notation is supposed to be
more suggestive than precise.
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00:04:04,210 --> 00:04:06,560
This is not a math class.
81
00:04:06,560 --> 00:04:08,485
I hope it's helpful rather
than otherwise.
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00:04:08,485 --> 00:04:12,170
83
00:04:12,170 --> 00:04:12,570
OK.
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00:04:12,570 --> 00:04:16,100
So today, we're going to
construct all the rest of the
85
00:04:16,100 --> 00:04:19,060
finite fields.
86
00:04:19,060 --> 00:04:23,330
By the way, we showed that these
are the only fields with
87
00:04:23,330 --> 00:04:25,090
a prime number of elements.
88
00:04:25,090 --> 00:04:31,240
Today we're going to construct
fields with a prime power
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00:04:31,240 --> 00:04:36,240
number of elements in a very
analogous way, and it will
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00:04:36,240 --> 00:04:38,460
turn out -- although I'm not
going to prove this -- that
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00:04:38,460 --> 00:04:43,790
these are the only
finite fields.
92
00:04:43,790 --> 00:04:47,240
Well, these are generalization
of this, so all finite fields
93
00:04:47,240 --> 00:04:50,090
have a prime power number of
elements and are basically
94
00:04:50,090 --> 00:04:53,040
isomorphic to one
of these fields.
95
00:04:53,040 --> 00:04:56,990
96
00:04:56,990 --> 00:04:59,090
How do we construct
this field?
97
00:04:59,090 --> 00:05:03,660
To give you a preview, very
analogously to the way we
98
00:05:03,660 --> 00:05:05,450
constructed this field.
99
00:05:05,450 --> 00:05:09,130
You'll see that the character
of the arguments is
100
00:05:09,130 --> 00:05:10,670
very much the same.
101
00:05:10,670 --> 00:05:14,770
And that's because there is a
great algebraic similarity
102
00:05:14,770 --> 00:05:20,030
between the integers and the
polynomials over a field.
103
00:05:20,030 --> 00:05:22,220
And we'll talk about
that in a second.
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00:05:22,220 --> 00:05:27,520
Basically, they're both
countably infinite rings, or
105
00:05:27,520 --> 00:05:29,060
infinite rings.
106
00:05:29,060 --> 00:05:33,250
Countably infinite if it's
over a finite field.
107
00:05:33,250 --> 00:05:36,960
And they both have analogous
factorization properties.
108
00:05:36,960 --> 00:05:39,900
They're both unique
factorization domains, would
109
00:05:39,900 --> 00:05:45,690
be one characterization,
algebraically.
110
00:05:45,690 --> 00:05:46,230
All right.
111
00:05:46,230 --> 00:05:47,850
How do we construct
this field?
112
00:05:47,850 --> 00:05:50,540
We're basically going to
construct it by taking the set
113
00:05:50,540 --> 00:05:56,320
of all polynomials over Fp which
is denoted by Fp square
114
00:05:56,320 --> 00:05:58,510
brackets x.
115
00:05:58,510 --> 00:06:03,690
And we're going to take that mod
the set of all polynomials
116
00:06:03,690 --> 00:06:08,080
that are divisible by G of x, or
G of x times the set of all
117
00:06:08,080 --> 00:06:08,820
polynomials.
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00:06:08,820 --> 00:06:12,500
The set of all multiples
of G of x.
119
00:06:12,500 --> 00:06:15,110
Or more simply, just
mod G of x.
120
00:06:15,110 --> 00:06:19,910
Where we're going to take G of
x to be a prime polynomial.
121
00:06:19,910 --> 00:06:26,060
And I guess I have to add that
the degree of G of x is going
122
00:06:26,060 --> 00:06:28,320
to be equal to m.
123
00:06:28,320 --> 00:06:32,080
So we basically need to find a
prime polynomial degree m.
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00:06:32,080 --> 00:06:36,620
Then we take the set of all
polynomials modulo this prime
125
00:06:36,620 --> 00:06:37,800
polynomial.
126
00:06:37,800 --> 00:06:41,970
They're going to be exactly p
to the m residue classes, or
127
00:06:41,970 --> 00:06:47,070
equivalence classes, or
remainders modulo g of x.
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00:06:47,070 --> 00:06:51,290
And we'll use the arithmetic
operations from mod G of x
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00:06:51,290 --> 00:06:55,680
arithmetic, addition mod G of x,
multiplication mod G of x,
130
00:06:55,680 --> 00:06:59,910
and we'll find that the
resulting object has satisfied
131
00:06:59,910 --> 00:07:02,170
the axioms of the field.
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00:07:02,170 --> 00:07:02,550
OK?
133
00:07:02,550 --> 00:07:05,026
So that's where we're going.
134
00:07:05,026 --> 00:07:08,210
You with me?
135
00:07:08,210 --> 00:07:08,630
OK.
136
00:07:08,630 --> 00:07:10,710
So let's talk about
factorization.
137
00:07:10,710 --> 00:07:18,330
And I think it makes it easy to
understand polynomials if
138
00:07:18,330 --> 00:07:22,530
we understand the analogies
to the integers.
139
00:07:22,530 --> 00:07:25,960
In particular where I'm headed
is unique factorization.
140
00:07:25,960 --> 00:07:28,740
141
00:07:28,740 --> 00:07:32,450
Both the integers and the
polynomials have unique
142
00:07:32,450 --> 00:07:33,230
factorizations.
143
00:07:33,230 --> 00:07:38,210
The integers into product of
integers, and any polynomial
144
00:07:38,210 --> 00:07:42,010
is uniquely factorizable into
a product of polynomials.
145
00:07:42,010 --> 00:07:45,300
Now I wasn't quite precise
when I said that.
146
00:07:45,300 --> 00:07:52,560
We have to be a little bit
more precise when we talk
147
00:07:52,560 --> 00:07:53,810
about factorization.
148
00:07:53,810 --> 00:07:55,810
149
00:07:55,810 --> 00:07:59,600
In particular, we have a
certain kind of trivial
150
00:07:59,600 --> 00:08:02,870
factorization that
involves units.
151
00:08:02,870 --> 00:08:06,470
Units, if you remember, are
the invertible elements.
152
00:08:06,470 --> 00:08:09,870
We're in a ring, so not every
element has an inverse, but
153
00:08:09,870 --> 00:08:11,120
some of them do.
154
00:08:11,120 --> 00:08:13,170
155
00:08:13,170 --> 00:08:17,930
And in the integers, what were
the invertible integers under
156
00:08:17,930 --> 00:08:19,030
multiplication?
157
00:08:19,030 --> 00:08:22,910
Everything here is about
multiplication, you know?
158
00:08:22,910 --> 00:08:28,220
A ring is something that
satisfies all of the
159
00:08:28,220 --> 00:08:32,710
properties of the field,
except that some of the
160
00:08:32,710 --> 00:08:35,039
elements don't have inverses,
so that division is not
161
00:08:35,039 --> 00:08:39,850
necessarily well-defined, even
for non-zero elements.
162
00:08:39,850 --> 00:08:40,250
OK?
163
00:08:40,250 --> 00:08:41,440
Sorry.
164
00:08:41,440 --> 00:08:42,919
I think I said that before.
165
00:08:42,919 --> 00:08:46,430
But we're not emphasizing
that these are rings,
166
00:08:46,430 --> 00:08:47,515
although they are.
167
00:08:47,515 --> 00:08:48,775
That's an informal definition.
168
00:08:48,775 --> 00:08:51,650
169
00:08:51,650 --> 00:08:55,260
So in the integers, of course
12 doesn't have a
170
00:08:55,260 --> 00:08:58,480
multiplicative inverse,
but it's an integer.
171
00:08:58,480 --> 00:09:00,805
But which integers do have
multiplicative inverses?
172
00:09:00,805 --> 00:09:03,680
173
00:09:03,680 --> 00:09:04,930
Plus or minus 1.
174
00:09:04,930 --> 00:09:07,980
175
00:09:07,980 --> 00:09:11,950
So if an integer is divisible
by n, it's also
176
00:09:11,950 --> 00:09:14,880
divisible by minus n.
177
00:09:14,880 --> 00:09:16,350
All right?
178
00:09:16,350 --> 00:09:18,870
Trivially.
179
00:09:18,870 --> 00:09:21,370
These are the only ones
in the integers.
180
00:09:21,370 --> 00:09:24,210
And last time we actually
talked, if I remember
181
00:09:24,210 --> 00:09:27,320
correctly, about the units
and the polynomials.
182
00:09:27,320 --> 00:09:29,800
Which polynomials
have inverses?
183
00:09:29,800 --> 00:09:33,970
184
00:09:33,970 --> 00:09:34,056
Excuse me.
185
00:09:34,056 --> 00:09:34,930
The degree 0.
186
00:09:34,930 --> 00:09:36,180
Thank you.
187
00:09:36,180 --> 00:09:40,480
188
00:09:40,480 --> 00:09:45,760
The degree 0 polynomials have
inverses because they are
189
00:09:45,760 --> 00:09:48,390
basically the non-zero elements
of the field.
190
00:09:48,390 --> 00:09:51,070
191
00:09:51,070 --> 00:09:51,320
OK?
192
00:09:51,320 --> 00:09:55,440
It's slight abuse of notation.
193
00:09:55,440 --> 00:09:58,500
We identify the field elements
with the degree zero
194
00:09:58,500 --> 00:09:59,250
polynomials.
195
00:09:59,250 --> 00:10:03,710
These are polynomials just of
the form F of x equals F_0 a
196
00:10:03,710 --> 00:10:05,940
constant, where the constant
is non-zero.
197
00:10:05,940 --> 00:10:08,570
198
00:10:08,570 --> 00:10:08,990
OK.
199
00:10:08,990 --> 00:10:17,960
So similarly, we will have to
have representatives of
200
00:10:17,960 --> 00:10:20,370
equivalence classes with
respect to the units.
201
00:10:20,370 --> 00:10:23,620
202
00:10:23,620 --> 00:10:29,330
So if both plus or minus n
divide something, what we take
203
00:10:29,330 --> 00:10:34,704
as the representative is we take
the positive integers.
204
00:10:34,704 --> 00:10:37,690
When we talk about divisibility,
we talk only
205
00:10:37,690 --> 00:10:42,700
about divisibility by positive
integers or factorization by
206
00:10:42,700 --> 00:10:43,720
positive integers.
207
00:10:43,720 --> 00:10:47,650
It's trivial that if something
is divisible by n, it's also
208
00:10:47,650 --> 00:10:49,680
divisible by minus n.
209
00:10:49,680 --> 00:10:56,400
Similarly for polynomials, the
representatives are taken as
210
00:10:56,400 --> 00:10:57,650
mnemonic polynomials.
211
00:10:57,650 --> 00:11:03,880
212
00:11:03,880 --> 00:11:09,790
And this means that the highest
order term, Fm is
213
00:11:09,790 --> 00:11:11,040
equal to 1.
214
00:11:11,040 --> 00:11:17,620
215
00:11:17,620 --> 00:11:23,600
So we can take any polynomial,
and by multiplying it by one
216
00:11:23,600 --> 00:11:29,050
of these invertible elements,
basically by a non-zero
217
00:11:29,050 --> 00:11:33,420
constant in the ground field, we
can make the highest order
218
00:11:33,420 --> 00:11:34,500
term equal to 1.
219
00:11:34,500 --> 00:11:35,400
Right?
220
00:11:35,400 --> 00:11:41,990
So if a polynomial is divisible
by F of x, it's also
221
00:11:41,990 --> 00:11:46,750
divisible by alpha F of x, where
alpha is any non-zero
222
00:11:46,750 --> 00:11:49,490
field element, right?
223
00:11:49,490 --> 00:11:54,230
And so we may as well fix the
highest order coefficient
224
00:11:54,230 --> 00:11:55,610
equal to 1.
225
00:11:55,610 --> 00:11:59,390
Actually, for some purposes in
the literature, like you've
226
00:11:59,390 --> 00:12:04,560
probably seen this in filter
design, or we always make the
227
00:12:04,560 --> 00:12:06,490
lowest order coefficient
equal to one.
228
00:12:06,490 --> 00:12:08,060
F0 equal to 1.
229
00:12:08,060 --> 00:12:11,990
You could also adopt that
convention, and say that
230
00:12:11,990 --> 00:12:16,530
that's going to be a mnemonic
polynomial.
231
00:12:16,530 --> 00:12:19,120
Here we'll focus on the high
order coefficient, but you
232
00:12:19,120 --> 00:12:21,720
could do it either way.
233
00:12:21,720 --> 00:12:25,380
And these both have the nice
property that the product of
234
00:12:25,380 --> 00:12:28,080
positive integers is
a positive integer.
235
00:12:28,080 --> 00:12:30,640
The product of monic polynomials
is a monic
236
00:12:30,640 --> 00:12:33,450
polynomial, right?
237
00:12:33,450 --> 00:12:36,440
The highest order term of the
product is going to have a
238
00:12:36,440 --> 00:12:38,880
highest order term equal to 1.
239
00:12:38,880 --> 00:12:40,520
Or if you chose the
lowest order one,
240
00:12:40,520 --> 00:12:41,560
that would work, too.
241
00:12:41,560 --> 00:12:46,920
The lowest order term of the
product would be equal to 1.
242
00:12:46,920 --> 00:12:47,390
All right.
243
00:12:47,390 --> 00:12:52,340
So having recognized this,
when we talk about unique
244
00:12:52,340 --> 00:12:55,930
factorization, just as with the
integers, what we really
245
00:12:55,930 --> 00:12:58,820
mean is factorization of a
positive integer into a
246
00:12:58,820 --> 00:13:00,720
product of positive integers.
247
00:13:00,720 --> 00:13:03,280
It's unique up to units.
248
00:13:03,280 --> 00:13:09,040
We can always put units either
on the integer that we're
249
00:13:09,040 --> 00:13:13,730
factoring or on any of the
factors, and we can freely
250
00:13:13,730 --> 00:13:16,180
multiply any of these things by
units, and it won't affect
251
00:13:16,180 --> 00:13:18,670
the factorization.
252
00:13:18,670 --> 00:13:21,920
Similarly over here, when
we talk about unique
253
00:13:21,920 --> 00:13:25,090
factorization, we mean
unique up to units.
254
00:13:25,090 --> 00:13:27,730
We're basically going to talk
about the factorization of
255
00:13:27,730 --> 00:13:35,250
monic polynomials into a product
of monic polynomials.
256
00:13:35,250 --> 00:13:41,100
257
00:13:41,100 --> 00:13:43,560
Not getting a real positive
feeling that everybody's
258
00:13:43,560 --> 00:13:43,980
following me.
259
00:13:43,980 --> 00:13:46,020
Would it help if I wrote
down more things,
260
00:13:46,020 --> 00:13:47,270
or wrote some examples?
261
00:13:47,270 --> 00:13:53,670
262
00:13:53,670 --> 00:13:57,770
In F2 of x, for instance,
we have --
263
00:13:57,770 --> 00:13:59,940
well, this isn't a very
good example.
264
00:13:59,940 --> 00:14:05,070
Let's write R of x.
265
00:14:05,070 --> 00:14:06,320
OK?
266
00:14:06,320 --> 00:14:08,080
267
00:14:08,080 --> 00:14:10,910
We're going to talk about
factorizations like this. x
268
00:14:10,910 --> 00:14:17,900
squared minus 1 equals x
minus 1 times x plus 1.
269
00:14:17,900 --> 00:14:21,730
That's a factorization of a
monic polynomial into a
270
00:14:21,730 --> 00:14:27,990
product of monic polynomials
of a lower degree.
271
00:14:27,990 --> 00:14:29,160
Happens in F2 of x.
272
00:14:29,160 --> 00:14:32,560
Since there is only one non-zero
field element, then
273
00:14:32,560 --> 00:14:36,164
all polynomials are monic,
except for the 0 polynomial.
274
00:14:36,164 --> 00:14:39,070
275
00:14:39,070 --> 00:14:41,540
The only non-zero term we
have play with is one.
276
00:14:41,540 --> 00:14:45,082
So the highest order non-zero
term is always 1.
277
00:14:45,082 --> 00:14:48,000
All right.
278
00:14:48,000 --> 00:14:48,095
OK.
279
00:14:48,095 --> 00:14:51,440
So that's what we're going to
mean by unique factorization.
280
00:14:51,440 --> 00:14:55,200
Now, there's one other
qualifier.
281
00:14:55,200 --> 00:14:56,640
There's some trivial factors.
282
00:14:56,640 --> 00:15:04,760
283
00:15:04,760 --> 00:15:10,190
We took some care to use the
standard mathematical
284
00:15:10,190 --> 00:15:13,580
terminology in the notes, so if
it says trivial devisors,
285
00:15:13,580 --> 00:15:14,330
then that's what I mean.
286
00:15:14,330 --> 00:15:18,280
What would the trivial divisors
of integer n be?
287
00:15:18,280 --> 00:15:23,450
288
00:15:23,450 --> 00:15:27,160
1 and n are always going
to divide in.
289
00:15:27,160 --> 00:15:29,000
And we're really not interested
in those when we
290
00:15:29,000 --> 00:15:30,250
talk about factorization.
291
00:15:30,250 --> 00:15:34,210
292
00:15:34,210 --> 00:15:36,820
Similarly, over here, the
trivial divisor of a
293
00:15:36,820 --> 00:15:41,740
polynomial F of x are
1 and F of x.
294
00:15:41,740 --> 00:15:43,060
We're not interested in those.
295
00:15:43,060 --> 00:15:46,570
296
00:15:46,570 --> 00:15:53,340
So when we talk about unique
factorization, we mean up to
297
00:15:53,340 --> 00:15:59,010
units and nontrivial factors.
298
00:15:59,010 --> 00:16:04,100
299
00:16:04,100 --> 00:16:07,400
And for the integers, that means
that we're going to talk
300
00:16:07,400 --> 00:16:13,980
about divisors d, let's
say, such that d is
301
00:16:13,980 --> 00:16:15,360
between 1 and n.
302
00:16:15,360 --> 00:16:21,080
303
00:16:21,080 --> 00:16:26,440
And for polynomials, what this
means is we're not interested
304
00:16:26,440 --> 00:16:29,200
in degree 0 factors.
305
00:16:29,200 --> 00:16:34,630
The only factor of degree the
same as F of x is going to be
306
00:16:34,630 --> 00:16:37,070
F of x up to units.
307
00:16:37,070 --> 00:16:44,010
We're interested in divisors d
of x such that the degree that
308
00:16:44,010 --> 00:16:50,200
0 is less than the degree of
the divisor less than the
309
00:16:50,200 --> 00:16:53,685
degree of what it's
dividing into.
310
00:16:53,685 --> 00:16:54,090
Sorry.
311
00:16:54,090 --> 00:16:57,290
I'm not defining everything,
but I hope that's clear.
312
00:16:57,290 --> 00:17:02,480
We're just interested in factors
that have degree less
313
00:17:02,480 --> 00:17:05,910
than the polynomial that we're
factoring, but we're not
314
00:17:05,910 --> 00:17:07,310
interested in degree
0 factors.
315
00:17:07,310 --> 00:17:10,349
316
00:17:10,349 --> 00:17:14,109
So that's what's meant by
unique factorization.
317
00:17:14,109 --> 00:17:17,585
It also shows you the
analogy in general.
318
00:17:17,585 --> 00:17:20,280
319
00:17:20,280 --> 00:17:27,269
In the case of integers, the key
thing in a divisor is that
320
00:17:27,269 --> 00:17:31,670
it have magnitude
between 1 and n.
321
00:17:31,670 --> 00:17:37,370
The key thing in a polynomial
is it have degree between 0
322
00:17:37,370 --> 00:17:41,370
and the degree of F of x.
323
00:17:41,370 --> 00:17:45,790
Basically, we want to factor
something into
324
00:17:45,790 --> 00:17:47,240
smaller things here.
325
00:17:47,240 --> 00:17:49,850
And when we say smaller, we
talk about magnitude.
326
00:17:49,850 --> 00:17:52,680
Here when we say smaller, we're
talking about degrees.
327
00:17:52,680 --> 00:17:54,650
In general, we go between
these two things.
328
00:17:54,650 --> 00:17:57,780
The concept of magnitude is
replaced by the concept of
329
00:17:57,780 --> 00:18:01,350
degree, to say how
big something is.
330
00:18:01,350 --> 00:18:05,120
331
00:18:05,120 --> 00:18:11,600
In both of these, the key to
all proofs is the Euclidean
332
00:18:11,600 --> 00:18:12,850
division algorithm.
333
00:18:12,850 --> 00:18:24,510
334
00:18:24,510 --> 00:18:33,810
Suppose we want to see if
n is a divisor of m.
335
00:18:33,810 --> 00:18:36,965
I've forgotten what I
put in the notes.
336
00:18:36,965 --> 00:18:41,380
Then we go through division, and
we find that m is equal to
337
00:18:41,380 --> 00:18:46,440
q, some quotient times n,
plus a remainder r.
338
00:18:46,440 --> 00:18:49,190
This is standard grade
school division.
339
00:18:49,190 --> 00:18:52,520
But it's really the key in the
universe in talking about
340
00:18:52,520 --> 00:18:55,680
these two domains.
341
00:18:55,680 --> 00:18:58,440
And this is what we've used,
really, to prove everything
342
00:18:58,440 --> 00:19:02,270
about the factorization
properties of integers.
343
00:19:02,270 --> 00:19:07,400
And how would you actually prove
that everything can be
344
00:19:07,400 --> 00:19:08,050
written this way?
345
00:19:08,050 --> 00:19:09,510
There's an important caveat.
346
00:19:09,510 --> 00:19:13,730
That the remainder we can always
choose to have to be in
347
00:19:13,730 --> 00:19:16,180
the range from 0 less
than r less than or
348
00:19:16,180 --> 00:19:19,380
equal to n minus 1.
349
00:19:19,380 --> 00:19:23,920
And the remainder is what
we call m mod n.
350
00:19:23,920 --> 00:19:28,340
351
00:19:28,340 --> 00:19:31,910
We divide and we get a remainder
that's one of these
352
00:19:31,910 --> 00:19:37,030
n things, and that's the main
thing we get out of this
353
00:19:37,030 --> 00:19:44,080
division, is m mod n, which is
equal to the remainder r.
354
00:19:44,080 --> 00:19:47,070
And there are precisely
n remainders.
355
00:19:47,070 --> 00:19:49,340
Now, how do you actually
prove this?
356
00:19:49,340 --> 00:19:51,600
You prove this, if
you want, very
357
00:19:51,600 --> 00:19:55,310
easily, just by recursion.
358
00:19:55,310 --> 00:20:03,630
You take m, and you ask, is
it already in this range?
359
00:20:03,630 --> 00:20:05,740
If it is, you're done.
360
00:20:05,740 --> 00:20:08,850
m is the remainder
and q is zero.
361
00:20:08,850 --> 00:20:13,510
If not, then you subtract n from
m, thereby reducing the
362
00:20:13,510 --> 00:20:14,410
magnitude of m.
363
00:20:14,410 --> 00:20:17,850
You can use the magnitude as an
indicator of how far you've
364
00:20:17,850 --> 00:20:19,720
gotten in this process.
365
00:20:19,720 --> 00:20:25,220
You reduce the magnitude, and
then you ask, is the result in
366
00:20:25,220 --> 00:20:28,070
this range?
367
00:20:28,070 --> 00:20:28,180
OK.
368
00:20:28,180 --> 00:20:32,140
If it's in that range, fine,
you finish this and q is 1.
369
00:20:32,140 --> 00:20:34,170
Otherwise, you continue.
370
00:20:34,170 --> 00:20:36,980
And in the recursion, you're
continually reducing the
371
00:20:36,980 --> 00:20:40,870
magnitude, and it's easy to
show that eventually, the
372
00:20:40,870 --> 00:20:44,990
magnitude has to fall into this
range and be one of these
373
00:20:44,990 --> 00:20:50,060
n remainder numbers, and
then you're done.
374
00:20:50,060 --> 00:20:50,550
OK?
375
00:20:50,550 --> 00:20:53,290
So it's a descending chain
where the chain
376
00:20:53,290 --> 00:20:54,880
has a bottom at 0.
377
00:20:54,880 --> 00:20:58,020
If you start with a positive
integer, you can't go below 0.
378
00:20:58,020 --> 00:20:59,790
And this is the only way
it can come out.
379
00:20:59,790 --> 00:21:01,040
Very easy to prove that.
380
00:21:01,040 --> 00:21:03,420
381
00:21:03,420 --> 00:21:03,850
All right.
382
00:21:03,850 --> 00:21:09,950
Similarly in polynomials, we get
an analogous expression.
383
00:21:09,950 --> 00:21:16,340
If we want to take F of x and
see if G of x is a divisor, we
384
00:21:16,340 --> 00:21:21,220
take F of x, and we can always
write this as some quotient
385
00:21:21,220 --> 00:21:25,900
times G of x plus
some remainder.
386
00:21:25,900 --> 00:21:28,770
Where the important thing here
about the remainder is that
387
00:21:28,770 --> 00:21:33,220
the degree of the remainder
is less than the
388
00:21:33,220 --> 00:21:36,460
degree of G of x.
389
00:21:36,460 --> 00:21:42,245
And here the remainder is called
F of x mod G of x.
390
00:21:42,245 --> 00:21:45,310
391
00:21:45,310 --> 00:21:51,750
Just as the remainder here
is called m mod n.
392
00:21:51,750 --> 00:21:54,790
393
00:21:54,790 --> 00:22:00,190
And there's a unique
remainder.
394
00:22:00,190 --> 00:22:05,900
And again, how would
you prove this?
395
00:22:05,900 --> 00:22:08,580
You could just take any long
division algorithm that you
396
00:22:08,580 --> 00:22:15,150
know for dividing G
of x into F of x.
397
00:22:15,150 --> 00:22:19,920
Basically long division amounts
to taking F of x.
398
00:22:19,920 --> 00:22:27,650
You can always choose some
scalar multiple of G of x such
399
00:22:27,650 --> 00:22:33,360
that F of x minus alpha G of x
has degree less than F of x.
400
00:22:33,360 --> 00:22:37,770
So you pick the top term to
reduce the degree, all right?
401
00:22:37,770 --> 00:22:39,370
Let's take G of x to be monic.
402
00:22:39,370 --> 00:22:42,100
We're only interested in
monic polynomials.
403
00:22:42,100 --> 00:22:45,180
If the top term of F of x is --
well, we're only going to
404
00:22:45,180 --> 00:22:48,520
divide it into monic
polynomials.
405
00:22:48,520 --> 00:22:51,190
But as we go along, we may
get non-monic ones.
406
00:22:51,190 --> 00:22:55,260
So you take the top
term, whatever it
407
00:22:55,260 --> 00:22:56,930
is over here, f(m).
408
00:22:56,930 --> 00:22:59,390
You multiply f(m)
times g of x.
409
00:22:59,390 --> 00:23:02,220
You subtract f(m) g
of x from f of x.
410
00:23:02,220 --> 00:23:04,208
You reduce the degree.
411
00:23:04,208 --> 00:23:05,196
OK?
412
00:23:05,196 --> 00:23:06,446
AUDIENCE: [INAUDIBLE PHRASE].
413
00:23:06,446 --> 00:23:10,730
414
00:23:10,730 --> 00:23:12,680
PROFESSOR: Correct.
415
00:23:12,680 --> 00:23:13,570
Thank you very much.
416
00:23:13,570 --> 00:23:18,740
You need also a term x to
whatever the difference in
417
00:23:18,740 --> 00:23:24,256
degrees is here to move the
degree up to the top.
418
00:23:24,256 --> 00:23:28,803
When we're actually doing long
division, we write f(m) f(m
419
00:23:28,803 --> 00:23:32,220
minus 1) down to f(0).
420
00:23:32,220 --> 00:23:38,310
We divide g(n) down to g(1).
421
00:23:38,310 --> 00:23:42,770
And the first term, g(n) is
going to be equal to 1.
422
00:23:42,770 --> 00:23:45,410
We take f(m) up here.
423
00:23:45,410 --> 00:23:54,189
We implicitly move it over to
get f(m) dot dot dot dot, down
424
00:23:54,189 --> 00:23:55,890
to f(m) alpha.
425
00:23:55,890 --> 00:24:00,060
We subtract, and we're down
to something that only has
426
00:24:00,060 --> 00:24:01,250
degree m minus 1.
427
00:24:01,250 --> 00:24:06,440
That's what polynomial long
division is shorthand for.
428
00:24:06,440 --> 00:24:10,130
429
00:24:10,130 --> 00:24:13,470
So you all know how
to do this.
430
00:24:13,470 --> 00:24:14,010
OK.
431
00:24:14,010 --> 00:24:18,790
Again, similar kind of proof,
that you must be able to get a
432
00:24:18,790 --> 00:24:20,890
remainder in this range,
and furthermore, the
433
00:24:20,890 --> 00:24:22,140
remainder is unique.
434
00:24:22,140 --> 00:24:25,040
435
00:24:25,040 --> 00:24:27,390
You basically can go through
this process.
436
00:24:27,390 --> 00:24:30,220
You reduce the degree by at
least one every time.
437
00:24:30,220 --> 00:24:33,030
438
00:24:33,030 --> 00:24:38,960
Therefore, degree must
eventually be reduced to where
439
00:24:38,960 --> 00:24:41,090
it's less than degree g of x.
440
00:24:41,090 --> 00:24:46,560
At that point, you can't
continue this process.
441
00:24:46,560 --> 00:24:47,640
You're stuck.
442
00:24:47,640 --> 00:24:48,730
And that's your remainder.
443
00:24:48,730 --> 00:24:51,780
There's no way of taking
something of lesser degree of
444
00:24:51,780 --> 00:24:54,770
g of x and then subtracting some
multiple from g of x from
445
00:24:54,770 --> 00:24:58,160
it to still further
reduce the degree.
446
00:24:58,160 --> 00:24:58,570
All right.
447
00:24:58,570 --> 00:25:00,490
So similar proof.
448
00:25:00,490 --> 00:25:03,160
Uniqueness is pretty obvious.
449
00:25:03,160 --> 00:25:10,080
So you get a unique remainder of
lesser degree than g of x.
450
00:25:10,080 --> 00:25:20,810
And so you can reduce any f of
x to some remainder r of x,
451
00:25:20,810 --> 00:25:22,270
which is called f
of x mod g of x.
452
00:25:22,270 --> 00:25:30,840
453
00:25:30,840 --> 00:25:39,490
And if we do arithmetic, the way
we do arithmetic over here
454
00:25:39,490 --> 00:25:43,770
for addition is we just, if we
want to add two remainders mod
455
00:25:43,770 --> 00:25:47,310
n, we take the sum of them, and
then if necessary, reduce
456
00:25:47,310 --> 00:25:49,610
them mod n.
457
00:25:49,610 --> 00:25:50,846
Similarly with multiplication.
458
00:25:50,846 --> 00:25:54,040
If we want to multiply them,
we take the product of two
459
00:25:54,040 --> 00:25:57,320
remainders, and if necessary,
reduce them again to a
460
00:25:57,320 --> 00:26:02,272
legitimate remainder which is
in this range or to mod n.
461
00:26:02,272 --> 00:26:05,270
It's the same over here.
462
00:26:05,270 --> 00:26:10,450
If we want to add two
463
00:26:10,450 --> 00:26:13,490
remainders, that's easy enough.
464
00:26:13,490 --> 00:26:16,160
We can do that and we won't
increase the degree, so we
465
00:26:16,160 --> 00:26:18,850
automatically get something
when we add two remainders
466
00:26:18,850 --> 00:26:24,100
that satisfies this
degree property.
467
00:26:24,100 --> 00:26:26,060
We don't have to reduce
mod g of x.
468
00:26:26,060 --> 00:26:30,060
If we multiply, you do have to
check that when you multiply
469
00:26:30,060 --> 00:26:35,900
two remainders, then all you
need to do is reduce
470
00:26:35,900 --> 00:26:38,050
the mod g of x.
471
00:26:38,050 --> 00:26:41,440
And basically, the assertion
is that --
472
00:26:41,440 --> 00:26:46,390
473
00:26:46,390 --> 00:26:48,570
let's see. r of x, s of x --
474
00:26:48,570 --> 00:26:53,520
475
00:26:53,520 --> 00:26:57,300
it's hard to write this without
being tautological.
476
00:26:57,300 --> 00:26:59,710
r of x, s of x.
477
00:26:59,710 --> 00:27:06,420
Reduced mod n, I'm sorry,
mod g of x.
478
00:27:06,420 --> 00:27:06,670
I'm sorry.
479
00:27:06,670 --> 00:27:08,860
This is not worth writing.
480
00:27:08,860 --> 00:27:11,080
r of x, s of x, mod g of x.
481
00:27:11,080 --> 00:27:15,340
482
00:27:15,340 --> 00:27:18,104
Something like that.
483
00:27:18,104 --> 00:27:18,540
Bah.
484
00:27:18,540 --> 00:27:19,900
It's a total tautology.
485
00:27:19,900 --> 00:27:21,110
Forget it.
486
00:27:21,110 --> 00:27:23,940
Said correctly in the notes.
487
00:27:23,940 --> 00:27:28,710
I'd rather think of it
as residue classes.
488
00:27:28,710 --> 00:27:38,512
If we, say, we talk about this
as a coset, f of x plus r of
489
00:27:38,512 --> 00:27:43,600
x, and we want to multiply
any element.
490
00:27:43,600 --> 00:27:48,160
This coset times any element
of the coset f
491
00:27:48,160 --> 00:27:50,520
of x plus s of x.
492
00:27:50,520 --> 00:27:53,030
493
00:27:53,030 --> 00:27:58,500
Then the result is just
multiplying out symbolically
494
00:27:58,500 --> 00:28:01,300
the polynomials times the
polynomials are the
495
00:28:01,300 --> 00:28:02,810
polynomials.
496
00:28:02,810 --> 00:28:06,180
Any polynomial times a
polynomial is a polynomial.
497
00:28:06,180 --> 00:28:08,840
Plus r of x f of x.
498
00:28:08,840 --> 00:28:11,800
We could get some multiple
of r of x,
499
00:28:11,800 --> 00:28:12,955
some polynomial multiple.
500
00:28:12,955 --> 00:28:16,130
We could get some polynomial
multiple of s of x.
501
00:28:16,130 --> 00:28:19,450
502
00:28:19,450 --> 00:28:22,280
Plus r of x s of x.
503
00:28:22,280 --> 00:28:26,740
504
00:28:26,740 --> 00:28:29,450
But this is a polynomial that's
included in here.
505
00:28:29,450 --> 00:28:30,830
We don't need to
say that again.
506
00:28:30,830 --> 00:28:32,590
Similarly here.
507
00:28:32,590 --> 00:28:39,310
And so this is equal
to f of x.
508
00:28:39,310 --> 00:28:44,360
It's equal to the coset f
of x plus r of x s of x.
509
00:28:44,360 --> 00:28:50,190
510
00:28:50,190 --> 00:28:52,900
But this is another
polynomial that
511
00:28:52,900 --> 00:28:55,090
probably has higher degree.
512
00:28:55,090 --> 00:29:03,160
This is equal to the coset
f of x plus r of x s
513
00:29:03,160 --> 00:29:09,780
of x mod g of x.
514
00:29:09,780 --> 00:29:10,170
OK.
515
00:29:10,170 --> 00:29:17,180
So this tells us that to
multiply cosets, coset with
516
00:29:17,180 --> 00:29:20,270
representative r of x times the
coset with representative
517
00:29:20,270 --> 00:29:26,120
s of x, we're going to get
something in the coset whose
518
00:29:26,120 --> 00:29:31,210
representative is r of
x, s of x mod g of x.
519
00:29:31,210 --> 00:29:36,870
So this is a sketch of a proof
that basically mod g of x
520
00:29:36,870 --> 00:29:39,830
commutes with multiplication.
521
00:29:39,830 --> 00:29:44,760
To multiply r of
x times s of x.
522
00:29:44,760 --> 00:29:45,210
All right.
523
00:29:45,210 --> 00:29:46,460
So --
524
00:29:46,460 --> 00:29:50,390
525
00:29:50,390 --> 00:29:51,640
AUDIENCE: [INAUDIBLE PHRASE].
526
00:29:51,640 --> 00:29:58,873
527
00:29:58,873 --> 00:30:00,370
PROFESSOR: Yeah.
528
00:30:00,370 --> 00:30:04,230
You're quite correct.
529
00:30:04,230 --> 00:30:08,220
What I mean is the cosets
of g of x of f of x.
530
00:30:08,220 --> 00:30:10,750
So I now understand all
the blank looks.
531
00:30:10,750 --> 00:30:18,638
532
00:30:18,638 --> 00:30:20,300
Are we better off now?
533
00:30:20,300 --> 00:30:25,120
534
00:30:25,120 --> 00:30:31,000
We have a group consisting of
the set of all multiples of g
535
00:30:31,000 --> 00:30:37,450
of x, which I write as g of x
times all the polynomials.
536
00:30:37,450 --> 00:30:40,900
The cosets of the group
are precisely --
537
00:30:40,900 --> 00:30:44,230
there's one remainder of degree
less than g of x in
538
00:30:44,230 --> 00:30:44,980
every coset.
539
00:30:44,980 --> 00:30:49,270
So this is the representative
of the coset.
540
00:30:49,270 --> 00:30:55,190
This is a sketch of a proof that
multiplication basically
541
00:30:55,190 --> 00:30:58,150
just amounts to multiplying
the remainders.
542
00:30:58,150 --> 00:31:03,010
The representative of the
product coset is the product
543
00:31:03,010 --> 00:31:05,760
of the representatives
modulo g of x.
544
00:31:05,760 --> 00:31:09,170
I think I said it correctly
for once.
545
00:31:09,170 --> 00:31:12,260
Please read the notes if you are
still confused, as I can
546
00:31:12,260 --> 00:31:13,510
see some of you are.
547
00:31:13,510 --> 00:31:24,721
548
00:31:24,721 --> 00:31:26,740
Maybe it would help
to do an example.
549
00:31:26,740 --> 00:31:34,480
550
00:31:34,480 --> 00:31:43,670
Let's take g of x to be equal
to x squared plus x
551
00:31:43,670 --> 00:31:50,480
plus 1 in f2 of x.
552
00:31:50,480 --> 00:31:54,350
It's a binary polynomial.
553
00:31:54,350 --> 00:32:01,370
Then r of x.
554
00:32:01,370 --> 00:32:12,760
The remainders are
equal to 0,1.
555
00:32:12,760 --> 00:32:16,680
This was degree minus infinity,
this is degree 0.
556
00:32:16,680 --> 00:32:18,900
They're x or x plus one.
557
00:32:18,900 --> 00:32:23,260
558
00:32:23,260 --> 00:32:24,500
OK?
559
00:32:24,500 --> 00:32:29,650
These are the possible
remainders when I take any
560
00:32:29,650 --> 00:32:31,620
polynomial modulo g of x.
561
00:32:31,620 --> 00:32:36,080
Divide it by g of x, and I'm
going to get a polynomial in
562
00:32:36,080 --> 00:32:40,340
f2 of x of degree less than or
equal to 1, and those are all
563
00:32:40,340 --> 00:32:44,765
of the polynomials that are
decisively for degree less
564
00:32:44,765 --> 00:32:47,140
than or equal to 1.
565
00:32:47,140 --> 00:32:49,020
All right?
566
00:32:49,020 --> 00:32:52,340
So what's the addition table?
567
00:32:52,340 --> 00:32:54,026
Addition is pretty easy.
568
00:32:54,026 --> 00:33:07,780
569
00:33:07,780 --> 00:33:09,620
0 plus anything is itself.
570
00:33:09,620 --> 00:33:13,820
571
00:33:13,820 --> 00:33:19,070
1 plus 1 in F2 is simply 0.
572
00:33:19,070 --> 00:33:24,440
1 plus x is 1 plus x, sorry,
x plus 1 as I've written.
573
00:33:24,440 --> 00:33:27,750
And 1 plus x plus 1 is x.
574
00:33:27,750 --> 00:33:31,710
x, x plus 1.
575
00:33:31,710 --> 00:33:33,290
x plus x is what?
576
00:33:33,290 --> 00:33:35,890
577
00:33:35,890 --> 00:33:37,136
Hello?
578
00:33:37,136 --> 00:33:37,540
0.
579
00:33:37,540 --> 00:33:38,830
Thank you.
580
00:33:38,830 --> 00:33:42,360
x plus x plus 1?
581
00:33:42,360 --> 00:33:43,392
Thank you.
582
00:33:43,392 --> 00:33:47,080
x plus 1, x.
583
00:33:47,080 --> 00:33:49,770
x plus x plus 1, 1, and 0.
584
00:33:49,770 --> 00:33:52,460
So that's what the addition
table looks like.
585
00:33:52,460 --> 00:33:55,150
586
00:33:55,150 --> 00:34:01,770
Actually you could think of
these as being just written as
587
00:34:01,770 --> 00:34:10,670
binary pairs, 0 0, 0 1, 1 0,
1 1, where this pair is
588
00:34:10,670 --> 00:34:13,760
basically F1 F0.
589
00:34:13,760 --> 00:34:19,110
And then the addition table is
precisely the same as the
590
00:34:19,110 --> 00:34:21,370
addition table for these
binary 2-tuples.
591
00:34:21,370 --> 00:34:26,480
You just add, component-wise,
the lowest order coefficient,
592
00:34:26,480 --> 00:34:28,850
the F1 coefficient.
593
00:34:28,850 --> 00:34:37,780
So addition is just like
addition in F2 squared.
594
00:34:37,780 --> 00:34:41,550
Or z2 squared, if you like.
595
00:34:41,550 --> 00:34:42,000
OK.
596
00:34:42,000 --> 00:34:44,610
That's actually an
additive group.
597
00:34:44,610 --> 00:34:46,080
Check it out.
598
00:34:46,080 --> 00:34:48,219
It's not z4 by the way.
599
00:34:48,219 --> 00:34:54,120
The other abelian group, or the
other group of size four,
600
00:34:54,120 --> 00:34:56,969
sometimes called the Klein
four-group, but it's really
601
00:34:56,969 --> 00:35:00,120
just the addition table for
the set of all binary
602
00:35:00,120 --> 00:35:00,640
[UNINTELLIGIBLE]
603
00:35:00,640 --> 00:35:02,760
two-tuples.
604
00:35:02,760 --> 00:35:04,010
OK?
605
00:35:04,010 --> 00:35:05,440
606
00:35:05,440 --> 00:35:06,690
Multiplication.
607
00:35:06,690 --> 00:35:14,320
608
00:35:14,320 --> 00:35:16,520
One very nice thing about
finite fields is
609
00:35:16,520 --> 00:35:18,470
you can simply --
610
00:35:18,470 --> 00:35:18,800
sorry.
611
00:35:18,800 --> 00:35:21,062
This was supposed to be
addition, this was supposed to
612
00:35:21,062 --> 00:35:22,312
be multiplication.
613
00:35:22,312 --> 00:35:24,390
614
00:35:24,390 --> 00:35:28,750
You know can simply write out
what all the rules are in a
615
00:35:28,750 --> 00:35:30,000
finite space.
616
00:35:30,000 --> 00:35:33,900
617
00:35:33,900 --> 00:35:34,360
OK.
618
00:35:34,360 --> 00:35:35,680
So what's 0 times anything?
619
00:35:35,680 --> 00:35:40,720
620
00:35:40,720 --> 00:35:41,970
What's 1 times anything?
621
00:35:41,970 --> 00:35:44,550
622
00:35:44,550 --> 00:35:45,180
Itself.
623
00:35:45,180 --> 00:35:46,750
1 is the multiplicative
identity.
624
00:35:46,750 --> 00:35:50,720
625
00:35:50,720 --> 00:35:52,080
But you know.
626
00:35:52,080 --> 00:35:54,150
You could formally do this
by doing polynomial
627
00:35:54,150 --> 00:35:55,360
multiplication.
628
00:35:55,360 --> 00:35:55,880
All right.
629
00:35:55,880 --> 00:35:56,850
Here's an interesting one.
630
00:35:56,850 --> 00:35:58,330
What's x times x?
631
00:35:58,330 --> 00:36:02,150
632
00:36:02,150 --> 00:36:04,060
Do x times x.
633
00:36:04,060 --> 00:36:08,280
634
00:36:08,280 --> 00:36:11,700
What is that going
to be equal to?
635
00:36:11,700 --> 00:36:12,500
x plus 1.
636
00:36:12,500 --> 00:36:15,600
How did you do that?
637
00:36:15,600 --> 00:36:17,410
We did the modulo.
638
00:36:17,410 --> 00:36:20,510
First of all, we write that
that's x squared.
639
00:36:20,510 --> 00:36:24,990
But then we have to do x
squared modulo g of x.
640
00:36:24,990 --> 00:36:27,540
x squared plus x plus 1.
641
00:36:27,540 --> 00:36:31,970
So we have to go through a
little long division process.
642
00:36:31,970 --> 00:36:34,720
We have to subtract this
out from that.
643
00:36:34,720 --> 00:36:36,430
And that gives us x plus 1.
644
00:36:36,430 --> 00:36:41,380
645
00:36:41,380 --> 00:36:42,560
So that's the key rule.
646
00:36:42,560 --> 00:36:47,970
Whenever we see something of
degree two or higher, we can
647
00:36:47,970 --> 00:36:51,170
always reduce it by subtracting
out some multiple
648
00:36:51,170 --> 00:36:54,810
of g of x down to something
of lower degree.
649
00:36:54,810 --> 00:36:55,100
Right?
650
00:36:55,100 --> 00:36:57,230
So this is what I've
been talking about.
651
00:36:57,230 --> 00:37:01,100
So x times x is x plus 1.
652
00:37:01,100 --> 00:37:02,975
What's x times x plus 1?
653
00:37:02,975 --> 00:37:08,230
654
00:37:08,230 --> 00:37:09,480
Equals what?
655
00:37:09,480 --> 00:37:12,162
656
00:37:12,162 --> 00:37:12,660
Good.
657
00:37:12,660 --> 00:37:15,780
You can do that in your heads.
658
00:37:15,780 --> 00:37:18,860
And what's x plus 1
times x plus 1?
659
00:37:18,860 --> 00:37:25,160
660
00:37:25,160 --> 00:37:27,810
Remember, we're doing
mod-2 arithmetic in
661
00:37:27,810 --> 00:37:29,760
our base field here.
662
00:37:29,760 --> 00:37:32,170
So this equals what?
663
00:37:32,170 --> 00:37:33,520
x squared plus 1.
664
00:37:33,520 --> 00:37:35,130
We reduce that, we get x.
665
00:37:35,130 --> 00:37:42,970
666
00:37:42,970 --> 00:37:45,560
OK.
667
00:37:45,560 --> 00:37:48,040
Let me check right now.
668
00:37:48,040 --> 00:37:51,920
Is this a field?
669
00:37:51,920 --> 00:37:54,760
These four elements with these
rules for addition and
670
00:37:54,760 --> 00:37:56,260
multiplication.
671
00:37:56,260 --> 00:37:57,510
Does that form a field?
672
00:37:57,510 --> 00:38:03,370
673
00:38:03,370 --> 00:38:07,230
What do I have to check, apart
from formalities like the
674
00:38:07,230 --> 00:38:10,520
distributive law?
675
00:38:10,520 --> 00:38:12,370
Which follows from the
distributive law for
676
00:38:12,370 --> 00:38:14,690
polynomials.
677
00:38:14,690 --> 00:38:18,432
That's always going to hold
through mod x arithmetic.
678
00:38:18,432 --> 00:38:19,860
What do I have to check?
679
00:38:19,860 --> 00:38:21,730
What are my field axioms?
680
00:38:21,730 --> 00:38:23,800
Anybody?
681
00:38:23,800 --> 00:38:26,920
Closure under multiplication.
682
00:38:26,920 --> 00:38:29,610
683
00:38:29,610 --> 00:38:33,720
That's getting towards a very
crisp statement of the --
684
00:38:33,720 --> 00:38:36,060
I'm looking for two
group axioms.
685
00:38:36,060 --> 00:38:39,850
One has to do with something we
have to check for addition.
686
00:38:39,850 --> 00:38:41,760
Something has to do with
something we have to check for
687
00:38:41,760 --> 00:38:43,010
multiplication.
688
00:38:43,010 --> 00:38:46,190
689
00:38:46,190 --> 00:38:47,230
Inverses?
690
00:38:47,230 --> 00:38:48,480
AUDIENCE: [INAUDIBLE PHRASE].
691
00:38:48,480 --> 00:38:52,450
692
00:38:52,450 --> 00:38:53,130
PROFESSOR: OK.
693
00:38:53,130 --> 00:38:56,840
Between the two of you I heard
the two answers that I want.
694
00:38:56,840 --> 00:38:58,620
We have to check that
this forms an
695
00:38:58,620 --> 00:39:01,930
abelian group under addition.
696
00:39:01,930 --> 00:39:04,310
So we have to check that the
addition table is the addition
697
00:39:04,310 --> 00:39:08,170
table of an abelian group.
698
00:39:08,170 --> 00:39:10,850
And under multiplication, we
have to check that the
699
00:39:10,850 --> 00:39:14,410
non-zero elements form
a billion group.
700
00:39:14,410 --> 00:39:22,000
So just this part of the table
has to form an abelian group,
701
00:39:22,000 --> 00:39:25,830
and both these have to have
an identity, of course.
702
00:39:25,830 --> 00:39:34,480
But the identity in mod g of x
arithmetic is always going to
703
00:39:34,480 --> 00:39:37,430
be 0 for addition and it's
always going to be 1 for
704
00:39:37,430 --> 00:39:38,680
multiplication.
705
00:39:38,680 --> 00:39:42,930
706
00:39:42,930 --> 00:39:43,350
All right.
707
00:39:43,350 --> 00:39:44,990
So I check this.
708
00:39:44,990 --> 00:39:46,180
Is this a group table?
709
00:39:46,180 --> 00:39:49,080
Basically, I just have to check
whether every row and
710
00:39:49,080 --> 00:39:55,080
column is a permutation
of the elements.
711
00:39:55,080 --> 00:39:57,180
And it is.
712
00:39:57,180 --> 00:40:02,810
And 0 acts as 0 should act.
713
00:40:02,810 --> 00:40:06,600
Has the additive identity.
714
00:40:06,600 --> 00:40:08,450
All right?
715
00:40:08,450 --> 00:40:12,290
Here what I have to check is
that the nonzero elements,
716
00:40:12,290 --> 00:40:21,070
these three, form an abelian
group under multiplication.
717
00:40:21,070 --> 00:40:25,830
Well, there really is only
one group of size three.
718
00:40:25,830 --> 00:40:29,790
It is isomorphic to Z3.
719
00:40:29,790 --> 00:40:31,710
If I replace --
720
00:40:31,710 --> 00:40:38,360
let's remember what Z3 looks
like under addition.
721
00:40:38,360 --> 00:40:45,270
This looks like 0 1 2,
0 1 2, 0 1 2, 0 1 2.
722
00:40:45,270 --> 00:40:46,270
It's mod-3.
723
00:40:46,270 --> 00:40:52,420
1 plus 1 is 2, 1 plus 2 is 3,
which is 0, 1 plus 2 is 3,
724
00:40:52,420 --> 00:40:56,360
which is 0, 2 plus 2 equals 1.
725
00:40:56,360 --> 00:40:58,710
So gee whiz.
726
00:40:58,710 --> 00:41:03,930
This is isomorphic to that if I
relabel 1 by 0, x by 1, and
727
00:41:03,930 --> 00:41:06,475
x plus 1 by 2.
728
00:41:06,475 --> 00:41:09,700
That's the only thing
it could be.
729
00:41:09,700 --> 00:41:14,690
The only group table in which
every row and column is a
730
00:41:14,690 --> 00:41:16,025
permutation of every other.
731
00:41:16,025 --> 00:41:18,620
732
00:41:18,620 --> 00:41:19,870
OK?
733
00:41:19,870 --> 00:41:21,730
734
00:41:21,730 --> 00:41:26,430
So we verified that we
now have a finite
735
00:41:26,430 --> 00:41:32,540
field with four elements.
736
00:41:32,540 --> 00:41:36,470
737
00:41:36,470 --> 00:41:37,985
Prime power number
of elements.
738
00:41:37,985 --> 00:41:41,020
739
00:41:41,020 --> 00:41:43,480
Right?
740
00:41:43,480 --> 00:41:48,310
The elements of my field are
these four remainders, or you
741
00:41:48,310 --> 00:41:50,340
can think of them as
representatives for their
742
00:41:50,340 --> 00:41:53,020
cosets, modulo g of x.
743
00:41:53,020 --> 00:42:02,410
The addition rule is addition
modulo g of x, and the
744
00:42:02,410 --> 00:42:07,720
multiplication rule is
multiplication modulo g of x.
745
00:42:07,720 --> 00:42:10,920
And it satisfies the field
axioms, therefore, it's a
746
00:42:10,920 --> 00:42:12,333
finite field.
747
00:42:12,333 --> 00:42:13,750
All right?
748
00:42:13,750 --> 00:42:16,470
I can add, subtract.
749
00:42:16,470 --> 00:42:19,360
Addition and subtraction
basically looks like addition
750
00:42:19,360 --> 00:42:23,170
and subtraction of binary
two-tuples, just
751
00:42:23,170 --> 00:42:24,680
component-wise.
752
00:42:24,680 --> 00:42:27,320
Multiplication is a little
bit more mysterious
753
00:42:27,320 --> 00:42:29,045
right now, but it works.
754
00:42:29,045 --> 00:42:33,370
755
00:42:33,370 --> 00:42:35,203
Let me tell you where we're
going to go on multiplication.
756
00:42:35,203 --> 00:42:40,690
757
00:42:40,690 --> 00:42:47,010
In this case, I can write x
plus 1 in a different way.
758
00:42:47,010 --> 00:42:52,530
I note that x plus 1 is
equal to x squared.
759
00:42:52,530 --> 00:42:54,240
All right?
760
00:42:54,240 --> 00:43:03,840
So let me write a little
log table over here for
761
00:43:03,840 --> 00:43:05,135
multiplication purposes.
762
00:43:05,135 --> 00:43:08,260
763
00:43:08,260 --> 00:43:12,320
I'm going to write x --
764
00:43:12,320 --> 00:43:13,570
I'm going to call that alpha.
765
00:43:13,570 --> 00:43:17,620
766
00:43:17,620 --> 00:43:20,590
And x plus 1 is equal
to x squared,
767
00:43:20,590 --> 00:43:21,925
or it's alpha squared.
768
00:43:21,925 --> 00:43:25,700
769
00:43:25,700 --> 00:43:26,950
What's alpha cubed?
770
00:43:26,950 --> 00:43:33,210
771
00:43:33,210 --> 00:43:38,636
Alpha cubed is x times
this again.
772
00:43:38,636 --> 00:43:43,050
Let me look in the table. x
times x plus 1 is equal to 1.
773
00:43:43,050 --> 00:43:45,940
So 1 equals alpha cubed.
774
00:43:45,940 --> 00:43:49,590
Or I could write that
as alpha to 0.
775
00:43:49,590 --> 00:43:53,730
If I multiply by x again,
I just cycle.
776
00:43:53,730 --> 00:43:55,990
So I'm going to get a
cyclic group here.
777
00:43:55,990 --> 00:43:58,910
778
00:43:58,910 --> 00:43:59,720
And now I'm going to write the
779
00:43:59,720 --> 00:44:01,405
multiplication table as follows.
780
00:44:01,405 --> 00:44:04,630
781
00:44:04,630 --> 00:44:07,310
I'm going to write the elements
of the group as 1,
782
00:44:07,310 --> 00:44:12,915
alpha, alpha squared, 0, 1,
alpha, alpha squared.
783
00:44:12,915 --> 00:44:17,910
784
00:44:17,910 --> 00:44:20,060
Again, 0 times anything is 0.
785
00:44:20,060 --> 00:44:23,130
We never have to worry
about that.
786
00:44:23,130 --> 00:44:25,040
1, alpha, alpha squared.
787
00:44:25,040 --> 00:44:29,330
788
00:44:29,330 --> 00:44:31,540
Alpha times alpha is
alpha squared.
789
00:44:31,540 --> 00:44:34,610
Alpha times alpha squared
is alpha cubed.
790
00:44:34,610 --> 00:44:37,520
But that's equal to 1.
791
00:44:37,520 --> 00:44:38,860
Same here.
792
00:44:38,860 --> 00:44:40,680
Alpha squared times
alpha squared --
793
00:44:40,680 --> 00:44:43,750
what's that?
794
00:44:43,750 --> 00:44:47,360
Alpha to the fourth, but what
does alpha fourth equal to if
795
00:44:47,360 --> 00:44:49,280
alpha cubed is equal to 1?
796
00:44:49,280 --> 00:44:51,930
797
00:44:51,930 --> 00:44:54,830
This has to be equal to alpha.
798
00:44:54,830 --> 00:44:59,560
Point is, because of this
relationship here, I can
799
00:44:59,560 --> 00:45:03,420
always reduce the exponents
modulo 3.
800
00:45:03,420 --> 00:45:08,270
I've basically got a little
multiplicative cyclic group of
801
00:45:08,270 --> 00:45:12,830
order three that's, of course,
isomorphic to the
802
00:45:12,830 --> 00:45:14,230
additive group, Z3.
803
00:45:14,230 --> 00:45:17,770
804
00:45:17,770 --> 00:45:22,240
So there are two ways I
can do multiplication.
805
00:45:22,240 --> 00:45:25,560
One is, I can do it by
this mod g of x way.
806
00:45:25,560 --> 00:45:32,435
I can represent things by these
which basically stand
807
00:45:32,435 --> 00:45:37,010
for polynomials of degree
one or less.
808
00:45:37,010 --> 00:45:40,540
And I can multiply two of these
by simply going through
809
00:45:40,540 --> 00:45:43,040
standard polynomial
multiplication over the
810
00:45:43,040 --> 00:45:44,850
appropriate field.
811
00:45:44,850 --> 00:45:47,870
And then I'll likely get some
powers of x squared or higher,
812
00:45:47,870 --> 00:45:51,850
and I reduce those to modulo
x squared plus x plus 1.
813
00:45:51,850 --> 00:45:53,820
That's legitimate and that
gives me this statement.
814
00:45:53,820 --> 00:45:56,460
815
00:45:56,460 --> 00:45:58,820
Well, the other thing I can do
is for multiplication, I can
816
00:45:58,820 --> 00:46:00,990
have a different
representation.
817
00:46:00,990 --> 00:46:03,980
This is basically just
a log table.
818
00:46:03,980 --> 00:46:06,280
OK?
819
00:46:06,280 --> 00:46:11,740
For each, I would have in my
little computer a separate
820
00:46:11,740 --> 00:46:16,790
table where I'd write that this
corresponds to 0, 0 1
821
00:46:16,790 --> 00:46:21,230
corresponds to 1, 1 0
corresponds to alpha, and 1 1
822
00:46:21,230 --> 00:46:24,430
corresponds to alpha squared.
823
00:46:24,430 --> 00:46:27,170
Or of course, I would just
represent these by their
824
00:46:27,170 --> 00:46:31,810
exponents that have some
special value for 0.
825
00:46:31,810 --> 00:46:34,940
And then all I have to do is
add exponents modulo 3 for
826
00:46:34,940 --> 00:46:37,360
multiplication.
827
00:46:37,360 --> 00:46:45,700
So log of x is 1, log of x plus
1 is 2, log of 1 is 0,
828
00:46:45,700 --> 00:46:54,220
and then I just use this for my
multiplication operation,
829
00:46:54,220 --> 00:46:58,990
or equivalently this cyclic
multiplicative group.
830
00:46:58,990 --> 00:47:02,590
Then I go through some inverse
log operation to get back to
831
00:47:02,590 --> 00:47:05,926
the other representation,
if I wanted to.
832
00:47:05,926 --> 00:47:08,760
And in fact, in finite field
arithmetic, this is what's
833
00:47:08,760 --> 00:47:09,550
typically done.
834
00:47:09,550 --> 00:47:13,140
There's just a little table
lookup such that you can go
835
00:47:13,140 --> 00:47:16,450
back and forth between this
representation, which we use
836
00:47:16,450 --> 00:47:19,380
for addition, and this
representation, which we use
837
00:47:19,380 --> 00:47:20,630
for multiplication.
838
00:47:20,630 --> 00:47:24,188
839
00:47:24,188 --> 00:47:25,170
Yeah?
840
00:47:25,170 --> 00:47:26,420
AUDIENCE: [INAUDIBLE PHRASE].
841
00:47:26,420 --> 00:47:30,140
842
00:47:30,140 --> 00:47:32,410
PROFESSOR: You have to represent
it as being special
843
00:47:32,410 --> 00:47:34,130
in some way.
844
00:47:34,130 --> 00:47:43,010
So if, in fact, literally, I
made log x equal to 1, log x
845
00:47:43,010 --> 00:47:50,060
plus 1 equal to 2, log
1 equal to 0 --
846
00:47:50,060 --> 00:47:52,860
one thing I suggest in the
notes, you can make log 0
847
00:47:52,860 --> 00:47:55,190
equal to minus infinity.
848
00:47:55,190 --> 00:47:57,620
That will always work.
849
00:47:57,620 --> 00:48:03,220
If you ever multiply
by 0, you'll be
850
00:48:03,220 --> 00:48:04,450
adding minus infinity.
851
00:48:04,450 --> 00:48:06,750
The result will be minus
infinity, so the
852
00:48:06,750 --> 00:48:08,020
inverse log is 0.
853
00:48:08,020 --> 00:48:10,770
So that's one way
you can do it.
854
00:48:10,770 --> 00:48:13,540
Or you could do what you do in
ordinary real and complex
855
00:48:13,540 --> 00:48:14,160
arithmetic.
856
00:48:14,160 --> 00:48:18,340
You could just, say, have
some special routine for
857
00:48:18,340 --> 00:48:21,040
multiplication by
0 is always 0.
858
00:48:21,040 --> 00:48:24,300
Division by 0 is illegal.
859
00:48:24,300 --> 00:48:25,550
Put out some error message.
860
00:48:25,550 --> 00:48:29,400
861
00:48:29,400 --> 00:48:29,770
All right?
862
00:48:29,770 --> 00:48:32,820
So that's how you actually build
a little finite field
863
00:48:32,820 --> 00:48:34,500
computer for this
finite field.
864
00:48:34,500 --> 00:48:38,270
865
00:48:38,270 --> 00:48:39,520
OK.
866
00:48:39,520 --> 00:48:42,050
867
00:48:42,050 --> 00:48:45,005
Now, I chose this advisedly.
868
00:48:45,005 --> 00:48:51,700
869
00:48:51,700 --> 00:48:57,240
Suppose I have chosen g of x
equal to x squared plus 1.
870
00:48:57,240 --> 00:49:00,980
871
00:49:00,980 --> 00:49:01,640
OK?
872
00:49:01,640 --> 00:49:06,230
I can do mod g of x arithmetic
for x squared plus 1.
873
00:49:06,230 --> 00:49:09,070
Again, my four field elements
are going to be the four
874
00:49:09,070 --> 00:49:13,370
binary polynomials of
degree 1 or less.
875
00:49:13,370 --> 00:49:15,750
The addition table is going
to be exactly the same.
876
00:49:15,750 --> 00:49:16,750
So let's just write the
877
00:49:16,750 --> 00:49:18,440
multiplication table over here.
878
00:49:18,440 --> 00:49:23,150
879
00:49:23,150 --> 00:49:30,680
0, 1, x, x plus 1, 0,
1, x, x plus 1.
880
00:49:30,680 --> 00:49:35,930
0 times anything is 0 in
ordinary polynomial
881
00:49:35,930 --> 00:49:39,360
multiplication, and therefore
also in mod g of x
882
00:49:39,360 --> 00:49:40,450
for any g of x.
883
00:49:40,450 --> 00:49:43,020
1 times anything is itself.
884
00:49:43,020 --> 00:49:45,600
No problem there.
885
00:49:45,600 --> 00:49:49,190
x, x plus 1.
886
00:49:49,190 --> 00:49:54,090
So we again have to give a
little bit care to x squared.
887
00:49:54,090 --> 00:49:57,260
What's that going
to be equal to?
888
00:49:57,260 --> 00:49:58,910
1.
889
00:49:58,910 --> 00:50:07,870
And x times x plus 1
is equal to what?
890
00:50:07,870 --> 00:50:14,200
891
00:50:14,200 --> 00:50:16,360
x squared plus x is
equal to what?
892
00:50:16,360 --> 00:50:23,570
893
00:50:23,570 --> 00:50:25,940
Looks like there's a problem.
894
00:50:25,940 --> 00:50:30,050
x plus 1 times x squared x plus
1 is equal to x squared
895
00:50:30,050 --> 00:50:33,080
plus 1, what's that equal?
896
00:50:33,080 --> 00:50:34,320
0.
897
00:50:34,320 --> 00:50:35,580
Yuck.
898
00:50:35,580 --> 00:50:36,830
Didn't work.
899
00:50:36,830 --> 00:50:41,540
900
00:50:41,540 --> 00:50:43,070
What's the essential
problem here?
901
00:50:43,070 --> 00:50:46,998
902
00:50:46,998 --> 00:50:47,490
Yeah.
903
00:50:47,490 --> 00:50:51,650
This clearly is not a group.
904
00:50:51,650 --> 00:50:54,070
It's not even closed under
multiplication.
905
00:50:54,070 --> 00:50:58,790
Because x plus 1 times x
plus 1 is equal to 0.
906
00:50:58,790 --> 00:51:01,640
The essential problem is that
there are two polynomials of
907
00:51:01,640 --> 00:51:08,590
degree less than two whose
product is x squared plus 1.
908
00:51:08,590 --> 00:51:11,880
In other words, this
is factorizable.
909
00:51:11,880 --> 00:51:17,860
910
00:51:17,860 --> 00:51:18,340
OK?
911
00:51:18,340 --> 00:51:20,050
In F2 of x.
912
00:51:20,050 --> 00:51:22,730
913
00:51:22,730 --> 00:51:23,330
All right?
914
00:51:23,330 --> 00:51:31,200
So whereas x squared
plus x plus 1.
915
00:51:31,200 --> 00:51:34,250
Does that have any factors
of degree 1 or less?
916
00:51:34,250 --> 00:51:36,750
Any nontrivial factors
of degree 1 or less?
917
00:51:36,750 --> 00:51:39,610
918
00:51:39,610 --> 00:51:43,050
Well, basically we proved that
it didn't, when we wrote this
919
00:51:43,050 --> 00:51:45,870
multiplication table.
920
00:51:45,870 --> 00:51:52,620
We tried all products of degree
1 or less polynomials
921
00:51:52,620 --> 00:51:56,160
where they're both non-zero,
and we never got 0.
922
00:51:56,160 --> 00:52:00,030
923
00:52:00,030 --> 00:52:08,520
So this will work if and only if
g of x is irreducible, has
924
00:52:08,520 --> 00:52:11,980
no nontrivial factors, is
a prime polynomial.
925
00:52:11,980 --> 00:52:14,600
These are all equivalent
in F2 of x.
926
00:52:14,600 --> 00:52:17,300
927
00:52:17,300 --> 00:52:22,730
There's this distinction in
other fields that irreducible
928
00:52:22,730 --> 00:52:25,060
just means has no nontrivial
factors.
929
00:52:25,060 --> 00:52:27,370
Prime means there's a monic
polynomial with
930
00:52:27,370 --> 00:52:29,460
no non-trivial factors.
931
00:52:29,460 --> 00:52:36,800
So that's what I
said back here.
932
00:52:36,800 --> 00:52:40,850
That the way we're ultimately
going to have to construct
933
00:52:40,850 --> 00:52:45,510
finite fields is take the
polynomials in Fp, Fp of x --
934
00:52:45,510 --> 00:52:47,530
we've been looking
at F2 of x --
935
00:52:47,530 --> 00:52:50,150
modulo a prime polynomial
g of x.
936
00:52:50,150 --> 00:52:57,450
937
00:52:57,450 --> 00:53:03,560
So what we're going to need to
find is prime polynomials.
938
00:53:03,560 --> 00:53:04,300
Let's see.
939
00:53:04,300 --> 00:53:11,690
Can I already prove that this is
going to work for any prime
940
00:53:11,690 --> 00:53:12,940
polynomial?
941
00:53:12,940 --> 00:53:19,400
942
00:53:19,400 --> 00:53:26,160
So I'm going to force the g of
x to be equal to a prime
943
00:53:26,160 --> 00:53:33,546
polynomial in Fp of
x of degree m.
944
00:53:33,546 --> 00:53:44,360
All right
945
00:53:44,360 --> 00:53:48,135
For example, x squared plus x
plus 1 is a prime polynomial.
946
00:53:48,135 --> 00:53:54,360
947
00:53:54,360 --> 00:54:15,250
And I'm going to ask if the
remainders mod g of x form a
948
00:54:15,250 --> 00:54:21,980
field under mod g
of x arithmetic.
949
00:54:21,980 --> 00:54:33,370
950
00:54:33,370 --> 00:54:34,620
OK.
951
00:54:34,620 --> 00:54:36,290
952
00:54:36,290 --> 00:54:39,660
Let's flow this out
a little bit.
953
00:54:39,660 --> 00:54:46,740
Again, what are the remainders
going to be of any polynomial
954
00:54:46,740 --> 00:54:47,990
of degree m?
955
00:54:47,990 --> 00:54:54,450
956
00:54:54,450 --> 00:55:00,750
So this is basically going to
be the polynomials of degree
957
00:55:00,750 --> 00:55:06,915
less than m in Fp of x.
958
00:55:06,915 --> 00:55:11,000
959
00:55:11,000 --> 00:55:12,620
How many of them are
there, by the way?
960
00:55:12,620 --> 00:55:19,620
961
00:55:19,620 --> 00:55:20,390
p to the m.
962
00:55:20,390 --> 00:55:30,390
So the size of this
is p to the m.
963
00:55:30,390 --> 00:55:34,200
And one of the representations
for these polynomials is just
964
00:55:34,200 --> 00:55:40,350
to write out F0, F1,
up to m minus 1.
965
00:55:40,350 --> 00:55:49,350
So this just basically looks
like the polynomial m-tuples.
966
00:55:49,350 --> 00:55:59,420
Set of F0, F1 up to F minus 1,
for each of these an element
967
00:55:59,420 --> 00:56:00,670
of the field.
968
00:56:00,670 --> 00:56:05,230
969
00:56:05,230 --> 00:56:05,830
OK?
970
00:56:05,830 --> 00:56:09,030
I can make a one-to-one
correspondence between the
971
00:56:09,030 --> 00:56:12,930
polynomials of degree less
than m and the set of all
972
00:56:12,930 --> 00:56:16,130
coefficient m-tuples over Fp.
973
00:56:16,130 --> 00:56:19,260
These are just m-tuples
over Fp.
974
00:56:19,260 --> 00:56:20,660
So there are p to
the m of them.
975
00:56:20,660 --> 00:56:23,430
976
00:56:23,430 --> 00:56:29,580
And so it's a finite set,
size p to the m.
977
00:56:29,580 --> 00:56:33,018
Does it form a field under
mod g of x arithmetic?
978
00:56:33,018 --> 00:56:34,268
AUDIENCE: [INAUDIBLE PHRASE].
979
00:56:34,268 --> 00:56:42,490
980
00:56:42,490 --> 00:56:42,920
PROFESSOR: Correct.
981
00:56:42,920 --> 00:56:44,320
And that's a homework problem.
982
00:56:44,320 --> 00:56:48,450
983
00:56:48,450 --> 00:56:51,970
So I'm not going to do it in
class, but I will sketch here
984
00:56:51,970 --> 00:56:53,400
how it's going to be done.
985
00:56:53,400 --> 00:56:56,360
But I'm glad that you
instantly see that.
986
00:56:56,360 --> 00:56:59,030
Because again, we just model
everything we do on what we
987
00:56:59,030 --> 00:57:01,160
did for integers.
988
00:57:01,160 --> 00:57:04,650
If you remember how we proved
it for integers.
989
00:57:04,650 --> 00:57:08,820
First of all, we really need
to check two things.
990
00:57:08,820 --> 00:57:10,070
Addition.
991
00:57:10,070 --> 00:57:12,680
992
00:57:12,680 --> 00:57:15,620
Just as we did for this specific
example up here, we
993
00:57:15,620 --> 00:57:20,550
first have to check that the
addition table is that of an
994
00:57:20,550 --> 00:57:25,072
abelian group, that these
supposed field elements form
995
00:57:25,072 --> 00:57:29,440
an abelian group
under addition.
996
00:57:29,440 --> 00:57:32,810
And we've already observed
several times that addition is
997
00:57:32,810 --> 00:57:38,460
just basically component-wise
addition of these m-tuples.
998
00:57:38,460 --> 00:57:40,210
So it's just like
vector addition.
999
00:57:40,210 --> 00:57:43,270
1000
00:57:43,270 --> 00:57:47,390
And vector addition, of course,
1001
00:57:47,390 --> 00:57:48,560
has the group property.
1002
00:57:48,560 --> 00:57:54,230
So addition is basically just
like vector addition.
1003
00:57:54,230 --> 00:58:00,370
Or I could write, perhaps more
precisely, Zp over m.
1004
00:58:00,370 --> 00:58:03,540
Distinction without a
difference, really.
1005
00:58:03,540 --> 00:58:11,145
So it's just component-wise
addition of the coefficients.
1006
00:58:11,145 --> 00:58:13,760
1007
00:58:13,760 --> 00:58:16,510
That's how we do polynomial
addition.
1008
00:58:16,510 --> 00:58:20,780
And it will give us a remainder
that has a degree of
1009
00:58:20,780 --> 00:58:23,850
less than m, so we don't have
to have ever reduce it.
1010
00:58:23,850 --> 00:58:28,970
Modulo g of x, we just simply
add component-wise.
1011
00:58:28,970 --> 00:58:29,420
OK.
1012
00:58:29,420 --> 00:58:31,180
So that's easy to verify.
1013
00:58:31,180 --> 00:58:34,180
The addition table is always
going to be OK.
1014
00:58:34,180 --> 00:58:36,150
So what do we have
to prove now?
1015
00:58:36,150 --> 00:58:37,400
We have to --
1016
00:58:37,400 --> 00:58:40,029
1017
00:58:40,029 --> 00:58:41,250
what am I going to call these?
1018
00:58:41,250 --> 00:58:42,680
Rg of x.
1019
00:58:42,680 --> 00:58:45,420
The remainders mod g of x.
1020
00:58:45,420 --> 00:58:55,070
For multiplication, we have to
prove that Rg of x star --
1021
00:58:55,070 --> 00:59:08,350
the non-zero polynomials form an
abelian group, which, as I
1022
00:59:08,350 --> 00:59:09,640
say, it's a homework problem.
1023
00:59:09,640 --> 00:59:11,090
Let me sketch the proof.
1024
00:59:11,090 --> 00:59:13,560
It's precisely analogous
to the proof that
1025
00:59:13,560 --> 00:59:18,060
we made for Zp star.
1026
00:59:18,060 --> 00:59:21,980
We basically have to
check closure.
1027
00:59:21,980 --> 00:59:27,070
If we multiply two non-zero
polynomials, do we get another
1028
00:59:27,070 --> 00:59:28,320
non-zero polynomial?
1029
00:59:28,320 --> 00:59:32,820
1030
00:59:32,820 --> 00:59:37,130
Asking another way, is it
possible to multiply two
1031
00:59:37,130 --> 00:59:44,410
polynomials of degree less than
m and get a result which
1032
00:59:44,410 --> 00:59:50,910
is a multiple of g of x, which
is either equal to g of x or a
1033
00:59:50,910 --> 00:59:53,870
multiple of g of x?
1034
00:59:53,870 --> 00:59:59,050
And it's, I think, easy to
convince yourself if this has
1035
00:59:59,050 --> 01:00:01,460
no factors --
1036
01:00:01,460 --> 01:00:04,910
its only factors are
itself and 1, no
1037
01:00:04,910 --> 01:00:07,860
non-trivial factors --
1038
01:00:07,860 --> 01:00:14,510
then you can't multiply two
lesser degree polynomials and
1039
01:00:14,510 --> 01:00:19,090
get either g of x, of course,
or any multiple of g of x.
1040
01:00:19,090 --> 01:00:21,800
And it's an exercise
for the student to
1041
01:00:21,800 --> 01:00:24,060
write that proof out.
1042
01:00:24,060 --> 01:00:24,440
OK?
1043
01:00:24,440 --> 01:00:27,400
But it's just exactly analogous
to the proof that
1044
01:00:27,400 --> 01:00:32,920
you can't multiply two integers
less than a prime p
1045
01:00:32,920 --> 01:00:35,790
and get a multiple of p.
1046
01:00:35,790 --> 01:00:40,472
So use that as a model, if you
want to, in your proof.
1047
01:00:40,472 --> 01:00:44,070
That of course depends
on this being prime.
1048
01:00:44,070 --> 01:00:49,210
If this is factorizable,
non-prime, then of course
1049
01:00:49,210 --> 01:00:52,020
there are going to be two
nontrivial factors, two
1050
01:00:52,020 --> 01:00:58,380
remainders of degree less than
m, whose product is equal to g
1051
01:00:58,380 --> 01:01:01,060
of x itself, if it's
factorizable.
1052
01:01:01,060 --> 01:01:04,040
So you can't possibly
get closure.
1053
01:01:04,040 --> 01:01:07,850
So this is why non-prime
polynomials don't work, just
1054
01:01:07,850 --> 01:01:14,310
like non-prime integers don't
work when we constructed Fp.
1055
01:01:14,310 --> 01:01:17,580
Exactly analogous reasons.
1056
01:01:17,580 --> 01:01:19,940
All right.
1057
01:01:19,940 --> 01:01:25,760
Second question is, when we go
through this multiplication,
1058
01:01:25,760 --> 01:01:34,640
suppose we take r
of x times --
1059
01:01:34,640 --> 01:01:36,430
let's construct a multiplication
table.
1060
01:01:36,430 --> 01:01:37,720
Let's take a particular row.
1061
01:01:37,720 --> 01:01:42,340
Let's take r of x times all
the non-zero things.
1062
01:01:42,340 --> 01:01:45,630
Can we possibly get
any repeats?
1063
01:01:45,630 --> 01:01:51,950
Can r of x times s of x equal
r of x times t of x?
1064
01:01:51,950 --> 01:01:54,280
And again, it's easy to convince
yourself that that's
1065
01:01:54,280 --> 01:01:55,020
not possible.
1066
01:01:55,020 --> 01:01:59,010
If that were possible, then r of
x times s of x minus t of x
1067
01:01:59,010 --> 01:02:01,210
would be equal to 0.
1068
01:02:01,210 --> 01:02:04,630
And so we're back to where
we were before --
1069
01:02:04,630 --> 01:02:05,740
mod g of x.
1070
01:02:05,740 --> 01:02:08,170
And this is impossible.
1071
01:02:08,170 --> 01:02:10,260
These are both degrees
less than m.
1072
01:02:10,260 --> 01:02:13,210
We can't multiply two things
of lower degree together to
1073
01:02:13,210 --> 01:02:14,965
get a multiple of g of x.
1074
01:02:14,965 --> 01:02:17,060
So it can't equal 0.
1075
01:02:17,060 --> 01:02:19,880
That means there can't
be any repeats.
1076
01:02:19,880 --> 01:02:22,390
That means that each row
is a permutation
1077
01:02:22,390 --> 01:02:24,050
of every other row.
1078
01:02:24,050 --> 01:02:25,180
Similarly for columns.
1079
01:02:25,180 --> 01:02:27,500
If you want to do that,
all you have to
1080
01:02:27,500 --> 01:02:29,780
actually prove is one.
1081
01:02:29,780 --> 01:02:39,070
So every row or column is a
permutations of every other
1082
01:02:39,070 --> 01:02:44,100
one, if we just look at the
non-zero polynomials, star.
1083
01:02:44,100 --> 01:02:54,240
And therefore, this forms an
abelian group whose identity
1084
01:02:54,240 --> 01:02:55,490
is always one.
1085
01:02:55,490 --> 01:02:58,520
1086
01:02:58,520 --> 01:03:00,880
Just as we proved up here.
1087
01:03:00,880 --> 01:03:06,522
So this depends on
irreducibility.
1088
01:03:06,522 --> 01:03:17,605
Depends on no nontrivial
factors of g of x.
1089
01:03:17,605 --> 01:03:24,990
1090
01:03:24,990 --> 01:03:25,380
OK.
1091
01:03:25,380 --> 01:03:26,640
And that's all we
have to check.
1092
01:03:26,640 --> 01:03:29,730
1093
01:03:29,730 --> 01:03:35,350
So I claim that by this process,
I can, given a prime
1094
01:03:35,350 --> 01:03:39,200
polynomial in Fp of x of degree
m, I can construct a
1095
01:03:39,200 --> 01:03:46,235
finite field with p to the m
elements, that the addition
1096
01:03:46,235 --> 01:03:49,360
and multiplication rules can
be taken as mod g of x
1097
01:03:49,360 --> 01:03:53,700
arithmetic, and they will
satisfy the field axioms.
1098
01:03:53,700 --> 01:03:58,550
So you can now construct a
finite field for any prime
1099
01:03:58,550 --> 01:04:03,160
power p to the m, right?
1100
01:04:03,160 --> 01:04:07,250
There's actually still
a hole in this.
1101
01:04:07,250 --> 01:04:08,500
AUDIENCE: [INAUDIBLE PHRASE].
1102
01:04:08,500 --> 01:04:13,040
1103
01:04:13,040 --> 01:04:14,680
PROFESSOR: Define prime
polynomial?
1104
01:04:14,680 --> 01:04:16,315
The term, or --
1105
01:04:16,315 --> 01:04:18,800
AUDIENCE: Define the
[UNINTELLIGIBLE] of degree m.
1106
01:04:18,800 --> 01:04:21,280
PROFESSOR: Correct.
1107
01:04:21,280 --> 01:04:24,225
Is there going to be a prime
polynomial of every degree?
1108
01:04:24,225 --> 01:04:28,990
1109
01:04:28,990 --> 01:04:30,240
I don't know.
1110
01:04:30,240 --> 01:04:33,410
1111
01:04:33,410 --> 01:04:37,230
Suppose you want to define an
irreducible polynomial over,
1112
01:04:37,230 --> 01:04:44,390
say, F2 of x or F3
of x of degree 4.
1113
01:04:44,390 --> 01:04:45,640
Could you do that?
1114
01:04:45,640 --> 01:04:49,460
1115
01:04:49,460 --> 01:04:50,710
AUDIENCE: [INAUDIBLE PHRASE].
1116
01:04:50,710 --> 01:05:09,520
1117
01:05:09,520 --> 01:05:10,180
PROFESSOR: Beautiful.
1118
01:05:10,180 --> 01:05:11,920
I mean, that's an excellent
suggestion.
1119
01:05:11,920 --> 01:05:25,110
So the question is, does there
exist a prime polynomial in Fp
1120
01:05:25,110 --> 01:05:28,390
of x of every degree?
1121
01:05:28,390 --> 01:05:33,680
1122
01:05:33,680 --> 01:05:34,945
Or of a given degree?
1123
01:05:34,945 --> 01:05:42,330
1124
01:05:42,330 --> 01:05:46,140
And there are various ways of
attacking this question.
1125
01:05:46,140 --> 01:05:49,240
First of all, I'll tell
you the answer is yes.
1126
01:05:49,240 --> 01:05:52,830
1127
01:05:52,830 --> 01:05:57,870
For every p and every m, there
does exist a prime polynomial.
1128
01:05:57,870 --> 01:05:59,790
Which is fortunate.
1129
01:05:59,790 --> 01:06:04,530
So from that, we conclude there
is a finite field with p
1130
01:06:04,530 --> 01:06:07,500
to the m elements for every
prime p and every m greater
1131
01:06:07,500 --> 01:06:08,750
than or equal to 1.
1132
01:06:08,750 --> 01:06:11,680
1133
01:06:11,680 --> 01:06:17,410
Now, how might we prove that?
1134
01:06:17,410 --> 01:06:21,840
One is, look it up on Google.
1135
01:06:21,840 --> 01:06:26,920
1136
01:06:26,920 --> 01:06:31,960
You can certainly formulate a
question that will lead you to
1137
01:06:31,960 --> 01:06:35,910
a webpage that will have a
listing of all the prime
1138
01:06:35,910 --> 01:06:39,800
polynomials of all degrees over
any field that you're
1139
01:06:39,800 --> 01:06:42,060
interested in.
1140
01:06:42,060 --> 01:06:45,830
So perhaps that will
suffice for you.
1141
01:06:45,830 --> 01:06:47,441
Two.
1142
01:06:47,441 --> 01:06:50,280
What I'm going to talk about
is the sieve method.
1143
01:06:50,280 --> 01:06:56,460
1144
01:06:56,460 --> 01:06:58,120
Three.
1145
01:06:58,120 --> 01:07:12,520
You could do a bound on the
number of polynomials of each
1146
01:07:12,520 --> 01:07:19,540
degree in d, and show that
it's greater than
1147
01:07:19,540 --> 01:07:22,240
or equal to 1, always.
1148
01:07:22,240 --> 01:07:26,340
And this is done in the notes.
1149
01:07:26,340 --> 01:07:29,800
Section 7.9 I think.
1150
01:07:29,800 --> 01:07:34,550
Or four, you can do what Mr.
Agarwal has suggested.
1151
01:07:34,550 --> 01:07:43,250
You can actually do the closed
form combinatoric
1152
01:07:43,250 --> 01:07:52,310
formula, which --
1153
01:07:52,310 --> 01:07:54,310
I haven't done any number
theory here.
1154
01:07:54,310 --> 01:07:56,230
There's a little bit of
elementary number
1155
01:07:56,230 --> 01:07:58,150
theory in the notes.
1156
01:07:58,150 --> 01:08:02,050
Euler numbers, this
sort of thing.
1157
01:08:02,050 --> 01:08:12,330
We get formulas for the number
of integers degree d that --
1158
01:08:12,330 --> 01:08:15,920
well, number of integers that
have multiplicative orders d
1159
01:08:15,920 --> 01:08:17,700
mod n, and so forth.
1160
01:08:17,700 --> 01:08:23,520
It's a lovely combinatoric
field.
1161
01:08:23,520 --> 01:08:26,960
There is a lovely closed form
for this that you get from the
1162
01:08:26,960 --> 01:08:30,140
Mobius inversion formula.
1163
01:08:30,140 --> 01:08:33,439
And it can be found in
combinatoric books.
1164
01:08:33,439 --> 01:08:40,440
I know it's in Berlekamp's
Algebraic Coding Theory book.
1165
01:08:40,440 --> 01:08:42,330
And it's extremely pretty.
1166
01:08:42,330 --> 01:08:45,750
And then of course, given the
formula, you have to prove to
1167
01:08:45,750 --> 01:08:48,620
all of the --
1168
01:08:48,620 --> 01:08:51,120
again, n of d is greater
than or equal to 1.
1169
01:08:51,120 --> 01:08:54,880
But there is a closed form
expression for it that you get
1170
01:08:54,880 --> 01:08:55,880
out of combinatorics.
1171
01:08:55,880 --> 01:08:58,420
We're not going to
do that here.
1172
01:08:58,420 --> 01:09:00,180
I'll be satisfied --
1173
01:09:00,180 --> 01:09:03,330
well, this is the real
engineering solution here.
1174
01:09:03,330 --> 01:09:05,484
This is the mathematical
engineering solution.
1175
01:09:05,484 --> 01:09:07,990
1176
01:09:07,990 --> 01:09:13,800
And what do I mean by
the sieve method?
1177
01:09:13,800 --> 01:09:17,850
Again, take the analogy
with the integers.
1178
01:09:17,850 --> 01:09:21,080
One of the first mathematical
accomplishments was
1179
01:09:21,080 --> 01:09:25,630
Eratosthenes' sieve for
finding prime numbers.
1180
01:09:25,630 --> 01:09:27,660
How does it work?
1181
01:09:27,660 --> 01:09:30,430
You start to write
down all the --
1182
01:09:30,430 --> 01:09:30,920
well.
1183
01:09:30,920 --> 01:09:34,939
Imagine first writing down
all the integers.
1184
01:09:34,939 --> 01:09:35,720
All right?
1185
01:09:35,720 --> 01:09:36,630
Cross off 1.
1186
01:09:36,630 --> 01:09:39,240
Start with 2.
1187
01:09:39,240 --> 01:09:40,080
All right.
1188
01:09:40,080 --> 01:09:42,700
So the first prime is 2.
1189
01:09:42,700 --> 01:09:44,890
Then you cross off all
multiples of 2.
1190
01:09:44,890 --> 01:09:47,710
4,6,8, so forth.
1191
01:09:47,710 --> 01:09:49,979
OK?
1192
01:09:49,979 --> 01:09:52,460
So what's the next number that
you haven't crossed off?
1193
01:09:52,460 --> 01:09:54,370
It's 3, so that's
the next prime.
1194
01:09:54,370 --> 01:09:57,370
You cross off all the
multiples of 3.
1195
01:09:57,370 --> 01:10:00,480
3,6,9.
1196
01:10:00,480 --> 01:10:01,790
So forth.
1197
01:10:01,790 --> 01:10:03,900
15.
1198
01:10:03,900 --> 01:10:07,970
And thereby you continue.
1199
01:10:07,970 --> 01:10:15,590
So the steps are, find the
next integer on the list.
1200
01:10:15,590 --> 01:10:17,460
That's going to be a prime,
because it won't have been
1201
01:10:17,460 --> 01:10:19,370
crossed off by any
previous steps.
1202
01:10:19,370 --> 01:10:23,970
It's not a multiple of any
integer of lower degree.
1203
01:10:23,970 --> 01:10:28,240
Then cross off all of its
multiples, up to however long
1204
01:10:28,240 --> 01:10:30,870
your scribe has written
this on the tablet.
1205
01:10:30,870 --> 01:10:35,770
And this way, you can find
the primes up to
1206
01:10:35,770 --> 01:10:37,960
any number you want.
1207
01:10:37,960 --> 01:10:41,580
Gets kind of tedious after a
while, but you can certainly
1208
01:10:41,580 --> 01:10:45,210
find all the primes
up to 100 in a few
1209
01:10:45,210 --> 01:10:48,092
minutes doing this, right?
1210
01:10:48,092 --> 01:10:52,250
Well, it's the same for
integers, and it's the same
1211
01:10:52,250 --> 01:10:54,260
for polynomials.
1212
01:10:54,260 --> 01:11:02,256
So let's, for instance,
do a polynomial sieve.
1213
01:11:02,256 --> 01:11:05,335
1214
01:11:05,335 --> 01:11:10,750
Of course, what we're most
interested in is the
1215
01:11:10,750 --> 01:11:13,670
polynomials with binary
coefficients, F2 of x.
1216
01:11:13,670 --> 01:11:16,470
1217
01:11:16,470 --> 01:11:18,500
And how do we do it?
1218
01:11:18,500 --> 01:11:21,430
Let's write down --
1219
01:11:21,430 --> 01:11:24,490
let's forget about 0 and 1.
1220
01:11:24,490 --> 01:11:28,280
Those are not considered
to be prime.
1221
01:11:28,280 --> 01:11:32,770
Let's start with the degree
1 polynomials.
1222
01:11:32,770 --> 01:11:35,770
What are the degree
one polynomials?
1223
01:11:35,770 --> 01:11:38,505
They're x, x plus 1.
1224
01:11:38,505 --> 01:11:42,300
1225
01:11:42,300 --> 01:11:43,550
Are these factorizable?
1226
01:11:43,550 --> 01:11:48,720
1227
01:11:48,720 --> 01:11:52,160
Obviously their only factors
are one and themselves.
1228
01:11:52,160 --> 01:11:54,460
They have no nontrivial
factors.
1229
01:11:54,460 --> 01:11:55,710
So these are primes.
1230
01:11:55,710 --> 01:12:00,210
1231
01:12:00,210 --> 01:12:01,460
OK.
1232
01:12:01,460 --> 01:12:03,310
1233
01:12:03,310 --> 01:12:11,995
So now let's write down all the
polynomials of degree two.
1234
01:12:11,995 --> 01:12:17,550
1235
01:12:17,550 --> 01:12:22,700
I'm sort of doing this in
an interleaved manner.
1236
01:12:22,700 --> 01:12:25,950
There are going to be two of
degree 1, four of degree 2.
1237
01:12:25,950 --> 01:12:28,330
All I'm going to do is -- well,
the only polynomials are
1238
01:12:28,330 --> 01:12:30,410
monic in F2 of x, so I don't
have to make that
1239
01:12:30,410 --> 01:12:32,350
qualification.
1240
01:12:32,350 --> 01:12:32,700
All right.
1241
01:12:32,700 --> 01:12:35,000
So here are the four
of degree 2, right?
1242
01:12:35,000 --> 01:12:38,150
1243
01:12:38,150 --> 01:12:40,800
Now I go through with my sieve
and we take out all
1244
01:12:40,800 --> 01:12:42,560
multiples of x.
1245
01:12:42,560 --> 01:12:48,040
Multiples of x are polynomials
with 0 constant term, right?
1246
01:12:48,040 --> 01:12:49,960
A lowest order term.
1247
01:12:49,960 --> 01:12:52,380
So obviously this is a multiple
of x, this is a
1248
01:12:52,380 --> 01:12:53,630
multiple of x.
1249
01:12:53,630 --> 01:12:56,580
1250
01:12:56,580 --> 01:12:58,060
Multiples of x plus 1.
1251
01:12:58,060 --> 01:13:00,590
How do you recognize those
over the binary field?
1252
01:13:00,590 --> 01:13:05,450
1253
01:13:05,450 --> 01:13:12,380
This is basically the polynomial
whose root is 1.
1254
01:13:12,380 --> 01:13:14,050
That means the mod-2 sum of the
1255
01:13:14,050 --> 01:13:16,420
coefficients is equal to 0.
1256
01:13:16,420 --> 01:13:20,200
So any polynomial that has an
even number of non-zero
1257
01:13:20,200 --> 01:13:23,500
coefficients is divisible
by x plus 1.
1258
01:13:23,500 --> 01:13:24,920
Did you all get that?
1259
01:13:24,920 --> 01:13:29,270
If not, try that at home.
1260
01:13:29,270 --> 01:13:31,510
That's the easy way to recognize
whether something is
1261
01:13:31,510 --> 01:13:32,750
a multiple of x plus 1.
1262
01:13:32,750 --> 01:13:36,640
It has an even number of
non-zero coefficients.
1263
01:13:36,640 --> 01:13:38,530
So this is a multiple
of x plus 1.
1264
01:13:38,530 --> 01:13:40,380
We wrote it out explicitly.
1265
01:13:40,380 --> 01:13:46,600
It's x plus 1 squared in F2
of x, and this is not.
1266
01:13:46,600 --> 01:13:58,410
So there is only one prime
polynomial over F2 of x
1267
01:13:58,410 --> 01:13:59,690
that's degree 2.
1268
01:13:59,690 --> 01:14:03,480
So this is our only possible
choice if we want to construct
1269
01:14:03,480 --> 01:14:05,985
a finite field with
four elements,
1270
01:14:05,985 --> 01:14:07,235
due to the two elements.
1271
01:14:07,235 --> 01:14:09,380
1272
01:14:09,380 --> 01:14:09,870
OK.
1273
01:14:09,870 --> 01:14:13,200
So that might begin
to get us scared.
1274
01:14:13,200 --> 01:14:15,080
Is it possible that
there are no prime
1275
01:14:15,080 --> 01:14:18,260
polynomials of degree three?
1276
01:14:18,260 --> 01:14:20,390
Well, we had two
here, one here.
1277
01:14:20,390 --> 01:14:21,875
Is it going down?
1278
01:14:21,875 --> 01:14:25,710
Let's write down the polynomials
of degree 3.
1279
01:14:25,710 --> 01:14:33,400
x third, x third plus 1, x to
the three plus x, to the three
1280
01:14:33,400 --> 01:14:38,225
plus x plus 1, x to the three
plus x squared, x to the three
1281
01:14:38,225 --> 01:14:44,760
plus x squared plus 1, x to the
three plus x squared plus
1282
01:14:44,760 --> 01:14:50,620
x, x to the three plus x
squared plus x plus 1.
1283
01:14:50,620 --> 01:14:52,130
Eight of them.
1284
01:14:52,130 --> 01:14:54,370
So that's working
in our favor.
1285
01:14:54,370 --> 01:14:57,810
There's double the number
of polynomials as we
1286
01:14:57,810 --> 01:15:01,300
go up one in degree.
1287
01:15:01,300 --> 01:15:03,430
And again, let's go
through the sieve.
1288
01:15:03,430 --> 01:15:11,990
Which are multiples of x,
non-zero constant term?
1289
01:15:11,990 --> 01:15:16,960
Which are multiples of x plus
1, even number of non-zero
1290
01:15:16,960 --> 01:15:18,210
coefficients?
1291
01:15:18,210 --> 01:15:21,450
1292
01:15:21,450 --> 01:15:23,950
If you doubt that,
write it out.
1293
01:15:23,950 --> 01:15:27,180
Which are multiples of x
squared plus x plus 1?
1294
01:15:27,180 --> 01:15:29,690
Well, they're going to have to
be x squared plus x plus 1
1295
01:15:29,690 --> 01:15:33,600
times x plus 1 or times x, so
we've already got them.
1296
01:15:33,600 --> 01:15:36,370
If we take this times itself,
we're going to get a
1297
01:15:36,370 --> 01:15:38,260
polynomial of degree 4.
1298
01:15:38,260 --> 01:15:41,000
Not going to be on this list.
1299
01:15:41,000 --> 01:15:43,020
So we have to multiply
this by these.
1300
01:15:43,020 --> 01:15:44,160
We've already got those.
1301
01:15:44,160 --> 01:15:46,530
So whew.
1302
01:15:46,530 --> 01:15:47,520
We found two.
1303
01:15:47,520 --> 01:15:50,190
We could use either of these
to construct a finite field
1304
01:15:50,190 --> 01:15:51,440
with eight elements.
1305
01:15:51,440 --> 01:15:56,040
1306
01:15:56,040 --> 01:15:56,810
And so forth.
1307
01:15:56,810 --> 01:15:59,780
And it turns out as you go up
to higher degrees, that the
1308
01:15:59,780 --> 01:16:02,760
number now increases very
nicely, and there's no problem
1309
01:16:02,760 --> 01:16:05,420
finding a prime polynomial
of each degree.
1310
01:16:05,420 --> 01:16:09,270
And in fact, you could be cute
about it and try to find one
1311
01:16:09,270 --> 01:16:13,860
with only three non-zero terms,
or has some other nice
1312
01:16:13,860 --> 01:16:17,760
property that makes it easy
to calculate with.
1313
01:16:17,760 --> 01:16:20,480
And so forth.
1314
01:16:20,480 --> 01:16:24,240
So do you understand
the sieve method?
1315
01:16:24,240 --> 01:16:28,410
If you do, then I believe
you could find a --
1316
01:16:28,410 --> 01:16:34,070
suppose you want to construct
a field with 64 elements.
1317
01:16:34,070 --> 01:16:36,510
What do you need?
1318
01:16:36,510 --> 01:16:38,900
64 is 2 to the sixth, so
you're going to need a
1319
01:16:38,900 --> 01:16:42,260
polynomial with p equals
2 and m equals 6.
1320
01:16:42,260 --> 01:16:47,540
You're going to need a binary
polynomial of degree 6 that is
1321
01:16:47,540 --> 01:16:49,300
prime, irreducible.
1322
01:16:49,300 --> 01:16:52,230
1323
01:16:52,230 --> 01:16:54,770
And again, in a few minutes
by going through the sieve
1324
01:16:54,770 --> 01:16:57,460
process, you can quickly find
one, or you could look it up
1325
01:16:57,460 --> 01:17:01,260
in Google, or in Peterson's
book, or any algebraic coding
1326
01:17:01,260 --> 01:17:05,490
theory book, or probably
a lot of other places.
1327
01:17:05,490 --> 01:17:07,810
So this is a practical solution
for the problem for
1328
01:17:07,810 --> 01:17:09,120
any given p and m.
1329
01:17:09,120 --> 01:17:13,430
Of course, it hardly proves
that there's one of every
1330
01:17:13,430 --> 01:17:15,730
degree, because they're
kind of an
1331
01:17:15,730 --> 01:17:18,190
infinite number of degrees.
1332
01:17:18,190 --> 01:17:19,780
Yeah?
1333
01:17:19,780 --> 01:17:24,566
AUDIENCE: So there's two order
3 polynomials that are prime.
1334
01:17:24,566 --> 01:17:27,372
Will it generate two separate
fields, or are they going to
1335
01:17:27,372 --> 01:17:29,450
be isomorphic to each other?
1336
01:17:29,450 --> 01:17:32,020
PROFESSOR: Great question.
1337
01:17:32,020 --> 01:17:35,820
All fields with p to the m
elements are isomorphic to
1338
01:17:35,820 --> 01:17:37,730
each other.
1339
01:17:37,730 --> 01:17:39,240
That's proved in the notes.
1340
01:17:39,240 --> 01:17:40,490
I'm not going to
do it in class.
1341
01:17:40,490 --> 01:17:42,810
1342
01:17:42,810 --> 01:17:45,410
And the other thing that's
proved in the notes is the
1343
01:17:45,410 --> 01:17:48,200
analog to Zp.
1344
01:17:48,200 --> 01:17:52,790
There are no other finite fields
with other than p to
1345
01:17:52,790 --> 01:17:55,920
the m number of elements.
1346
01:17:55,920 --> 01:17:56,230
OK?
1347
01:17:56,230 --> 01:17:58,520
So this is it.
1348
01:17:58,520 --> 01:18:01,870
Now, if you explicitly write
out these two, and
1349
01:18:01,870 --> 01:18:04,570
write out their --
1350
01:18:04,570 --> 01:18:07,960
well, the addition tables are
always look the same, because
1351
01:18:07,960 --> 01:18:11,630
it's always just binary
three-tuples, in this case.
1352
01:18:11,630 --> 01:18:15,170
But multiplication tables are
going to look different.
1353
01:18:15,170 --> 01:18:16,483
But there is some isomorphism.
1354
01:18:16,483 --> 01:18:19,720
1355
01:18:19,720 --> 01:18:25,820
x and 1 may be equivalent to x
squared plus 1 on the other
1356
01:18:25,820 --> 01:18:27,490
one, or something.
1357
01:18:27,490 --> 01:18:31,040
But if you go through that
isomorphism, you'll find that
1358
01:18:31,040 --> 01:18:33,270
the field tables are the same.
1359
01:18:33,270 --> 01:18:36,220
1360
01:18:36,220 --> 01:18:38,790
Actually, I guess I'm going to
prove that, because I'm going
1361
01:18:38,790 --> 01:18:41,860
to prove that the
multiplication
1362
01:18:41,860 --> 01:18:45,390
table is always cyclic.
1363
01:18:45,390 --> 01:18:50,530
That the group to which it's
isomorphic is z mod p
1364
01:18:50,530 --> 01:18:53,590
to the m minus 1.
1365
01:18:53,590 --> 01:18:56,260
Just as it was to Z3 here.
1366
01:18:56,260 --> 01:18:59,080
And I guess that's sufficient
with the addition table
1367
01:18:59,080 --> 01:19:00,310
isomorphism.
1368
01:19:00,310 --> 01:19:02,140
I guess you have to prove
they're equivalent.
1369
01:19:02,140 --> 01:19:04,140
You saw it in a different
way in the notes.
1370
01:19:04,140 --> 01:19:10,360
But you're going to see that all
the multiplication group
1371
01:19:10,360 --> 01:19:14,900
is always a cyclic group, with
p to the m minus 1 elements,
1372
01:19:14,900 --> 01:19:17,530
and that goes a long way towards
suggesting that these
1373
01:19:17,530 --> 01:19:20,150
are always going to be
isomorphic to each other.
1374
01:19:20,150 --> 01:19:20,846
Yeah?
1375
01:19:20,846 --> 01:19:23,230
AUDIENCE: Identify
roots, right?
1376
01:19:23,230 --> 01:19:24,480
PROFESSOR: Identify roots?
1377
01:19:24,480 --> 01:19:27,460
1378
01:19:27,460 --> 01:19:29,340
If I understand what you're
saying, that's basically the
1379
01:19:29,340 --> 01:19:31,010
way it's done in the notes.
1380
01:19:31,010 --> 01:19:37,230
You first show there's always
going to be some primitive
1381
01:19:37,230 --> 01:19:41,330
element that generates
the cyclic group.
1382
01:19:41,330 --> 01:19:45,070
Some single generator such that
alpha alpha squared, so
1383
01:19:45,070 --> 01:19:50,170
forth, is the entire
non-zero set.
1384
01:19:50,170 --> 01:19:54,600
You're going to show that
alpha has some minimal
1385
01:19:54,600 --> 01:19:59,950
polynomial, and the set of all
linear combinations of powers
1386
01:19:59,950 --> 01:20:04,350
of alpha is basically equal
to the whole field.
1387
01:20:04,350 --> 01:20:10,970
And so this allows you to
establish the isomorphism.
1388
01:20:10,970 --> 01:20:15,250
and I think that's what
you're suggesting.
1389
01:20:15,250 --> 01:20:17,950
So you're well equipped to read
the notes, but I'm not
1390
01:20:17,950 --> 01:20:19,200
going to do that in class.
1391
01:20:19,200 --> 01:20:23,520
1392
01:20:23,520 --> 01:20:24,370
Yeah.
1393
01:20:24,370 --> 01:20:28,340
I'm very interested
in your questions.
1394
01:20:28,340 --> 01:20:32,320
Please ask more, as many
questions as you like.
1395
01:20:32,320 --> 01:20:35,230
What I'm getting from it is,
you know, you all come from
1396
01:20:35,230 --> 01:20:36,170
different backgrounds.
1397
01:20:36,170 --> 01:20:40,350
Some of you have seen this in
perhaps a math context, or
1398
01:20:40,350 --> 01:20:43,360
some other context, or you've
seen parts of it, or some of
1399
01:20:43,360 --> 01:20:45,650
the words are familiar.
1400
01:20:45,650 --> 01:20:49,050
And of course, there are many
different ways to present this
1401
01:20:49,050 --> 01:20:52,320
and to make the proofs
and so forth.
1402
01:20:52,320 --> 01:20:56,770
So I'm trying to pick a line
that works for the particular
1403
01:20:56,770 --> 01:20:58,030
results that I want to get to.
1404
01:20:58,030 --> 01:21:00,560
I don't think you would do
anything much different if you
1405
01:21:00,560 --> 01:21:03,950
wanted to develop
finite fields.
1406
01:21:03,950 --> 01:21:09,275
But the class has a very
different set of backgrounds,
1407
01:21:09,275 --> 01:21:10,910
and I'm trying to reach
all of you.
1408
01:21:10,910 --> 01:21:14,940
1409
01:21:14,940 --> 01:21:16,540
Don't be alarmed if
you've never seen
1410
01:21:16,540 --> 01:21:17,930
anything like this before.
1411
01:21:17,930 --> 01:21:21,560
You're not way behind everybody
else, either.
1412
01:21:21,560 --> 01:21:26,310
I think it's pretty easy
to understand.
1413
01:21:26,310 --> 01:21:29,900
Maybe it would take you a
couple hours longer than
1414
01:21:29,900 --> 01:21:31,470
somebody who has more
background, but
1415
01:21:31,470 --> 01:21:32,720
not more than that.
1416
01:21:32,720 --> 01:21:35,990
1417
01:21:35,990 --> 01:21:37,830
OK.
1418
01:21:37,830 --> 01:21:42,360
So that's how we construct
finite fields.
1419
01:21:42,360 --> 01:21:45,940
You have an example of it.
1420
01:21:45,940 --> 01:21:46,440
Goodness.
1421
01:21:46,440 --> 01:21:48,740
Is it really 11 o'clock?
1422
01:21:48,740 --> 01:21:49,520
OK.
1423
01:21:49,520 --> 01:21:51,390
So I'm not even --
1424
01:21:51,390 --> 01:21:56,010
so I've simply asserted,
but not proved.
1425
01:21:56,010 --> 01:21:59,020
You saw in this case that the
multiplicative group was a
1426
01:21:59,020 --> 01:22:00,770
cyclic group.
1427
01:22:00,770 --> 01:22:04,530
And we could always, for
multiplication, represent
1428
01:22:04,530 --> 01:22:06,800
grouped elements by
this log table.
1429
01:22:06,800 --> 01:22:08,590
So I'd hoped to prove
that in class.
1430
01:22:08,590 --> 01:22:11,860
I'm not going to be able
to prove that in class.
1431
01:22:11,860 --> 01:22:16,400
And I'll simply ask you
to read that, too.
1432
01:22:16,400 --> 01:22:20,370
This is important for working
with finite fields, because it
1433
01:22:20,370 --> 01:22:23,500
is the way, probably the
preferred way, to implement
1434
01:22:23,500 --> 01:22:24,750
multiplication.
1435
01:22:24,750 --> 01:22:26,910
1436
01:22:26,910 --> 01:22:31,180
So you ought to be thinking of
how you would program up a
1437
01:22:31,180 --> 01:22:34,810
finite field multiplier.
1438
01:22:34,810 --> 01:22:36,700
One way is polynomial
multiplication.
1439
01:22:36,700 --> 01:22:41,200
The other way is just use the
fact that the multiplicative
1440
01:22:41,200 --> 01:22:43,160
group is cyclic.
1441
01:22:43,160 --> 01:22:44,580
And then it's easy.
1442
01:22:44,580 --> 01:22:50,810
Just add exponents and reduce
modulo q minus 1.
1443
01:22:50,810 --> 01:22:51,120
OK.
1444
01:22:51,120 --> 01:22:54,700
I'm sorry not to have had a
chance to go over that.
1445
01:22:54,700 --> 01:22:57,510
Next time, Ralf Koetter will
start to get into chapter
1446
01:22:57,510 --> 01:22:58,760
eight, Reed-Solomon codes.
1447
01:22:58,760 --> 01:23:10,676