WEBVTT
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PROFESSOR: So if you want to
hand in your problem sets, we
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have three handouts.
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The old problem set solutions,
the new problem set, and we're
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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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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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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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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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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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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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Any questions about
these things?
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Homework processes going OK?
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Office hours?
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Solutions?
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Everything seems to be
all right so far?
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Good.
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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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elements that I want
you to know.
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You're responsible only
for what's covered in
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class, by the way.
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We've talked about a lot
of algebraic objects.
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Our main emphasis has been to
get the finite field, and at
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this point, we have
prime fields.
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Fields with a prime number
p of elements.
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And we find that we can
construct such a field by
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taking the integers mod p.
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You can write that
as z mod p z.
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You can equally well consider
this as the equivalence
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classes of integers of the
cosets of pz nz which are each
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identified by a remainder
or residue r between
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0 and p minus 1.
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OK.
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So basically, the elements of
the field are these remainders
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or these cosets.
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There are p of them.
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We do arithmetic by simply
doing mod p addition and
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multiplication.
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So if you understand arithmetic
mod p, you
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understand this field.
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Yes?
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AUDIENCE: Is Fp and
Zp isomorphic?
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PROFESSOR: As an additive group,
it's isomorphic to Zp.
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Zp we use that notation
for the group.
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We use this notation
for the field.
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And I write Zp is isomorphic
to Z mod pZ
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as a quotient group.
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Here I've added in the
multiplication operation mod p
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to make this a field.
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So this is somewhat more than
just a quotient group.
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Sorry.
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The notation is supposed to be
more suggestive than precise.
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This is not a math class.
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I hope it's helpful rather
than otherwise.
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OK.
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So today, we're going to
construct all the rest of the
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finite fields.
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By the way, we showed that these
are the only fields with
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a prime number of elements.
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Today we're going to construct
fields with a prime power
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number of elements in a very
analogous way, and it will
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turn out -- although I'm not
going to prove this -- that
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these are the only
finite fields.
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Well, these are generalization
of this, so all finite fields
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have a prime power number of
elements and are basically
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isomorphic to one
of these fields.
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How do we construct
this field?
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To give you a preview, very
analogously to the way we
00:05:03.660 --> 00:05:05.450
constructed this field.
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You'll see that the character
of the arguments is
00:05:09.130 --> 00:05:10.670
very much the same.
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And that's because there is a
great algebraic similarity
00:05:14.770 --> 00:05:20.030
between the integers and the
polynomials over a field.
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And we'll talk about
that in a second.
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Basically, they're both
countably infinite rings, or
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infinite rings.
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Countably infinite if it's
over a finite field.
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And they both have analogous
factorization properties.
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They're both unique
factorization domains, would
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be one characterization,
algebraically.
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All right.
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How do we construct
this field?
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We're basically going to
construct it by taking the set
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of all polynomials over Fp which
is denoted by Fp square
00:05:56.320 --> 00:05:58.510
brackets x.
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And we're going to take that mod
the set of all polynomials
00:06:03.690 --> 00:06:08.080
that are divisible by G of x, or
G of x times the set of all
00:06:08.080 --> 00:06:08.820
polynomials.
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The set of all multiples
of G of x.
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Or more simply, just
mod G of x.
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Where we're going to take G of
x to be a prime polynomial.
00:06:19.910 --> 00:06:26.060
And I guess I have to add that
the degree of G of x is going
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to be equal to m.
00:06:28.320 --> 00:06:32.080
So we basically need to find a
prime polynomial degree m.
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Then we take the set of all
polynomials modulo this prime
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polynomial.
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They're going to be exactly p
to the m residue classes, or
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equivalence classes, or
remainders modulo g of x.
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And we'll use the arithmetic
operations from mod G of x
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arithmetic, addition mod G of x,
multiplication mod G of x,
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and we'll find that the
resulting object has satisfied
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the axioms of the field.
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OK?
00:07:02.550 --> 00:07:05.026
So that's where we're going.
00:07:05.026 --> 00:07:08.210
You with me?
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OK.
00:07:08.630 --> 00:07:10.710
So let's talk about
factorization.
00:07:10.710 --> 00:07:18.330
And I think it makes it easy to
understand polynomials if
00:07:18.330 --> 00:07:22.530
we understand the analogies
to the integers.
00:07:22.530 --> 00:07:25.960
In particular where I'm headed
is unique factorization.
00:07:28.740 --> 00:07:32.450
Both the integers and the
polynomials have unique
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factorizations.
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The integers into product of
integers, and any polynomial
00:07:38.210 --> 00:07:42.010
is uniquely factorizable into
a product of polynomials.
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Now I wasn't quite precise
when I said that.
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We have to be a little bit
more precise when we talk
00:07:52.560 --> 00:07:53.810
about factorization.
00:07:55.810 --> 00:07:59.600
In particular, we have a
certain kind of trivial
00:07:59.600 --> 00:08:02.870
factorization that
involves units.
00:08:02.870 --> 00:08:06.470
Units, if you remember, are
the invertible elements.
00:08:06.470 --> 00:08:09.870
We're in a ring, so not every
element has an inverse, but
00:08:09.870 --> 00:08:11.120
some of them do.
00:08:13.170 --> 00:08:17.930
And in the integers, what were
the invertible integers under
00:08:17.930 --> 00:08:19.030
multiplication?
00:08:19.030 --> 00:08:22.910
Everything here is about
multiplication, you know?
00:08:22.910 --> 00:08:28.220
A ring is something that
satisfies all of the
00:08:28.220 --> 00:08:32.710
properties of the field,
except that some of the
00:08:32.710 --> 00:08:35.039
elements don't have inverses,
so that division is not
00:08:35.039 --> 00:08:39.850
necessarily well-defined, even
for non-zero elements.
00:08:39.850 --> 00:08:40.250
OK?
00:08:40.250 --> 00:08:41.440
Sorry.
00:08:41.440 --> 00:08:42.919
I think I said that before.
00:08:42.919 --> 00:08:46.430
But we're not emphasizing
that these are rings,
00:08:46.430 --> 00:08:47.515
although they are.
00:08:47.515 --> 00:08:48.775
That's an informal definition.
00:08:51.650 --> 00:08:55.260
So in the integers, of course
12 doesn't have a
00:08:55.260 --> 00:08:58.480
multiplicative inverse,
but it's an integer.
00:08:58.480 --> 00:09:00.805
But which integers do have
multiplicative inverses?
00:09:03.680 --> 00:09:04.930
Plus or minus 1.
00:09:07.980 --> 00:09:11.950
So if an integer is divisible
by n, it's also
00:09:11.950 --> 00:09:14.880
divisible by minus n.
00:09:14.880 --> 00:09:16.350
All right?
00:09:16.350 --> 00:09:18.870
Trivially.
00:09:18.870 --> 00:09:21.370
These are the only ones
in the integers.
00:09:21.370 --> 00:09:24.210
And last time we actually
talked, if I remember
00:09:24.210 --> 00:09:27.320
correctly, about the units
and the polynomials.
00:09:27.320 --> 00:09:29.800
Which polynomials
have inverses?
00:09:33.970 --> 00:09:34.056
Excuse me.
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The degree 0.
00:09:34.930 --> 00:09:36.180
Thank you.
00:09:40.480 --> 00:09:45.760
The degree 0 polynomials have
inverses because they are
00:09:45.760 --> 00:09:48.390
basically the non-zero elements
of the field.
00:09:51.070 --> 00:09:51.320
OK?
00:09:51.320 --> 00:09:55.440
It's slight abuse of notation.
00:09:55.440 --> 00:09:58.500
We identify the field elements
with the degree zero
00:09:58.500 --> 00:09:59.250
polynomials.
00:09:59.250 --> 00:10:03.710
These are polynomials just of
the form F of x equals F_0 a
00:10:03.710 --> 00:10:05.940
constant, where the constant
is non-zero.
00:10:08.570 --> 00:10:08.990
OK.
00:10:08.990 --> 00:10:17.960
So similarly, we will have to
have representatives of
00:10:17.960 --> 00:10:20.370
equivalence classes with
respect to the units.
00:10:23.620 --> 00:10:29.330
So if both plus or minus n
divide something, what we take
00:10:29.330 --> 00:10:34.704
as the representative is we take
the positive integers.
00:10:34.704 --> 00:10:37.690
When we talk about divisibility,
we talk only
00:10:37.690 --> 00:10:42.700
about divisibility by positive
integers or factorization by
00:10:42.700 --> 00:10:43.720
positive integers.
00:10:43.720 --> 00:10:47.650
It's trivial that if something
is divisible by n, it's also
00:10:47.650 --> 00:10:49.680
divisible by minus n.
00:10:49.680 --> 00:10:56.400
Similarly for polynomials, the
representatives are taken as
00:10:56.400 --> 00:10:57.650
mnemonic polynomials.
00:11:03.880 --> 00:11:09.790
And this means that the highest
order term, Fm is
00:11:09.790 --> 00:11:11.040
equal to 1.
00:11:17.620 --> 00:11:23.600
So we can take any polynomial,
and by multiplying it by one
00:11:23.600 --> 00:11:29.050
of these invertible elements,
basically by a non-zero
00:11:29.050 --> 00:11:33.420
constant in the ground field, we
can make the highest order
00:11:33.420 --> 00:11:34.500
term equal to 1.
00:11:34.500 --> 00:11:35.400
Right?
00:11:35.400 --> 00:11:41.990
So if a polynomial is divisible
by F of x, it's also
00:11:41.990 --> 00:11:46.750
divisible by alpha F of x, where
alpha is any non-zero
00:11:46.750 --> 00:11:49.490
field element, right?
00:11:49.490 --> 00:11:54.230
And so we may as well fix the
highest order coefficient
00:11:54.230 --> 00:11:55.610
equal to 1.
00:11:55.610 --> 00:11:59.390
Actually, for some purposes in
the literature, like you've
00:11:59.390 --> 00:12:04.560
probably seen this in filter
design, or we always make the
00:12:04.560 --> 00:12:06.490
lowest order coefficient
equal to one.
00:12:06.490 --> 00:12:08.060
F0 equal to 1.
00:12:08.060 --> 00:12:11.990
You could also adopt that
convention, and say that
00:12:11.990 --> 00:12:16.530
that's going to be a mnemonic
polynomial.
00:12:16.530 --> 00:12:19.120
Here we'll focus on the high
order coefficient, but you
00:12:19.120 --> 00:12:21.720
could do it either way.
00:12:21.720 --> 00:12:25.380
And these both have the nice
property that the product of
00:12:25.380 --> 00:12:28.080
positive integers is
a positive integer.
00:12:28.080 --> 00:12:30.640
The product of monic polynomials
is a monic
00:12:30.640 --> 00:12:33.450
polynomial, right?
00:12:33.450 --> 00:12:36.440
The highest order term of the
product is going to have a
00:12:36.440 --> 00:12:38.880
highest order term equal to 1.
00:12:38.880 --> 00:12:40.520
Or if you chose the
lowest order one,
00:12:40.520 --> 00:12:41.560
that would work, too.
00:12:41.560 --> 00:12:46.920
The lowest order term of the
product would be equal to 1.
00:12:46.920 --> 00:12:47.390
All right.
00:12:47.390 --> 00:12:52.340
So having recognized this,
when we talk about unique
00:12:52.340 --> 00:12:55.930
factorization, just as with the
integers, what we really
00:12:55.930 --> 00:12:58.820
mean is factorization of a
positive integer into a
00:12:58.820 --> 00:13:00.720
product of positive integers.
00:13:00.720 --> 00:13:03.280
It's unique up to units.
00:13:03.280 --> 00:13:09.040
We can always put units either
on the integer that we're
00:13:09.040 --> 00:13:13.730
factoring or on any of the
factors, and we can freely
00:13:13.730 --> 00:13:16.180
multiply any of these things by
units, and it won't affect
00:13:16.180 --> 00:13:18.670
the factorization.
00:13:18.670 --> 00:13:21.920
Similarly over here, when
we talk about unique
00:13:21.920 --> 00:13:25.090
factorization, we mean
unique up to units.
00:13:25.090 --> 00:13:27.730
We're basically going to talk
about the factorization of
00:13:27.730 --> 00:13:35.250
monic polynomials into a product
of monic polynomials.
00:13:41.100 --> 00:13:43.560
Not getting a real positive
feeling that everybody's
00:13:43.560 --> 00:13:43.980
following me.
00:13:43.980 --> 00:13:46.020
Would it help if I wrote
down more things,
00:13:46.020 --> 00:13:47.270
or wrote some examples?
00:13:53.670 --> 00:13:57.770
In F2 of x, for instance,
we have --
00:13:57.770 --> 00:13:59.940
well, this isn't a very
good example.
00:13:59.940 --> 00:14:05.070
Let's write R of x.
00:14:05.070 --> 00:14:06.320
OK?
00:14:08.080 --> 00:14:10.910
We're going to talk about
factorizations like this. x
00:14:10.910 --> 00:14:17.900
squared minus 1 equals x
minus 1 times x plus 1.
00:14:17.900 --> 00:14:21.730
That's a factorization of a
monic polynomial into a
00:14:21.730 --> 00:14:27.990
product of monic polynomials
of a lower degree.
00:14:27.990 --> 00:14:29.160
Happens in F2 of x.
00:14:29.160 --> 00:14:32.560
Since there is only one non-zero
field element, then
00:14:32.560 --> 00:14:36.164
all polynomials are monic,
except for the 0 polynomial.
00:14:39.070 --> 00:14:41.540
The only non-zero term we
have play with is one.
00:14:41.540 --> 00:14:45.082
So the highest order non-zero
term is always 1.
00:14:45.082 --> 00:14:48.000
All right.
00:14:48.000 --> 00:14:48.095
OK.
00:14:48.095 --> 00:14:51.440
So that's what we're going to
mean by unique factorization.
00:14:51.440 --> 00:14:55.200
Now, there's one other
qualifier.
00:14:55.200 --> 00:14:56.640
There's some trivial factors.
00:15:04.760 --> 00:15:10.190
We took some care to use the
standard mathematical
00:15:10.190 --> 00:15:13.580
terminology in the notes, so if
it says trivial devisors,
00:15:13.580 --> 00:15:14.330
then that's what I mean.
00:15:14.330 --> 00:15:18.280
What would the trivial divisors
of integer n be?
00:15:23.450 --> 00:15:27.160
1 and n are always going
to divide in.
00:15:27.160 --> 00:15:29.000
And we're really not interested
in those when we
00:15:29.000 --> 00:15:30.250
talk about factorization.
00:15:34.210 --> 00:15:36.820
Similarly, over here, the
trivial divisor of a
00:15:36.820 --> 00:15:41.740
polynomial F of x are
1 and F of x.
00:15:41.740 --> 00:15:43.060
We're not interested in those.
00:15:46.570 --> 00:15:53.340
So when we talk about unique
factorization, we mean up to
00:15:53.340 --> 00:15:59.010
units and nontrivial factors.
00:16:04.100 --> 00:16:07.400
And for the integers, that means
that we're going to talk
00:16:07.400 --> 00:16:13.980
about divisors d, let's
say, such that d is
00:16:13.980 --> 00:16:15.360
between 1 and n.
00:16:21.080 --> 00:16:26.440
And for polynomials, what this
means is we're not interested
00:16:26.440 --> 00:16:29.200
in degree 0 factors.
00:16:29.200 --> 00:16:34.630
The only factor of degree the
same as F of x is going to be
00:16:34.630 --> 00:16:37.070
F of x up to units.
00:16:37.070 --> 00:16:44.010
We're interested in divisors d
of x such that the degree that
00:16:44.010 --> 00:16:50.200
0 is less than the degree of
the divisor less than the
00:16:50.200 --> 00:16:53.685
degree of what it's
dividing into.
00:16:53.685 --> 00:16:54.090
Sorry.
00:16:54.090 --> 00:16:57.290
I'm not defining everything,
but I hope that's clear.
00:16:57.290 --> 00:17:02.480
We're just interested in factors
that have degree less
00:17:02.480 --> 00:17:05.910
than the polynomial that we're
factoring, but we're not
00:17:05.910 --> 00:17:07.310
interested in degree
0 factors.
00:17:10.349 --> 00:17:14.109
So that's what's meant by
unique factorization.
00:17:14.109 --> 00:17:17.585
It also shows you the
analogy in general.
00:17:20.280 --> 00:17:27.269
In the case of integers, the key
thing in a divisor is that
00:17:27.269 --> 00:17:31.670
it have magnitude
between 1 and n.
00:17:31.670 --> 00:17:37.370
The key thing in a polynomial
is it have degree between 0
00:17:37.370 --> 00:17:41.370
and the degree of F of x.
00:17:41.370 --> 00:17:45.790
Basically, we want to factor
something into
00:17:45.790 --> 00:17:47.240
smaller things here.
00:17:47.240 --> 00:17:49.850
And when we say smaller, we
talk about magnitude.
00:17:49.850 --> 00:17:52.680
Here when we say smaller, we're
talking about degrees.
00:17:52.680 --> 00:17:54.650
In general, we go between
these two things.
00:17:54.650 --> 00:17:57.780
The concept of magnitude is
replaced by the concept of
00:17:57.780 --> 00:18:01.350
degree, to say how
big something is.
00:18:05.120 --> 00:18:11.600
In both of these, the key to
all proofs is the Euclidean
00:18:11.600 --> 00:18:12.850
division algorithm.
00:18:24.510 --> 00:18:33.810
Suppose we want to see if
n is a divisor of m.
00:18:33.810 --> 00:18:36.965
I've forgotten what I
put in the notes.
00:18:36.965 --> 00:18:41.380
Then we go through division, and
we find that m is equal to
00:18:41.380 --> 00:18:46.440
q, some quotient times n,
plus a remainder r.
00:18:46.440 --> 00:18:49.190
This is standard grade
school division.
00:18:49.190 --> 00:18:52.520
But it's really the key in the
universe in talking about
00:18:52.520 --> 00:18:55.680
these two domains.
00:18:55.680 --> 00:18:58.440
And this is what we've used,
really, to prove everything
00:18:58.440 --> 00:19:02.270
about the factorization
properties of integers.
00:19:02.270 --> 00:19:07.400
And how would you actually prove
that everything can be
00:19:07.400 --> 00:19:08.050
written this way?
00:19:08.050 --> 00:19:09.510
There's an important caveat.
00:19:09.510 --> 00:19:13.730
That the remainder we can always
choose to have to be in
00:19:13.730 --> 00:19:16.180
the range from 0 less
than r less than or
00:19:16.180 --> 00:19:19.380
equal to n minus 1.
00:19:19.380 --> 00:19:23.920
And the remainder is what
we call m mod n.
00:19:28.340 --> 00:19:31.910
We divide and we get a remainder
that's one of these
00:19:31.910 --> 00:19:37.030
n things, and that's the main
thing we get out of this
00:19:37.030 --> 00:19:44.080
division, is m mod n, which is
equal to the remainder r.
00:19:44.080 --> 00:19:47.070
And there are precisely
n remainders.
00:19:47.070 --> 00:19:49.340
Now, how do you actually
prove this?
00:19:49.340 --> 00:19:51.600
You prove this, if
you want, very
00:19:51.600 --> 00:19:55.310
easily, just by recursion.
00:19:55.310 --> 00:20:03.630
You take m, and you ask, is
it already in this range?
00:20:03.630 --> 00:20:05.740
If it is, you're done.
00:20:05.740 --> 00:20:08.850
m is the remainder
and q is zero.
00:20:08.850 --> 00:20:13.510
If not, then you subtract n from
m, thereby reducing the
00:20:13.510 --> 00:20:14.410
magnitude of m.
00:20:14.410 --> 00:20:17.850
You can use the magnitude as an
indicator of how far you've
00:20:17.850 --> 00:20:19.720
gotten in this process.
00:20:19.720 --> 00:20:25.220
You reduce the magnitude, and
then you ask, is the result in
00:20:25.220 --> 00:20:28.070
this range?
00:20:28.070 --> 00:20:28.180
OK.
00:20:28.180 --> 00:20:32.140
If it's in that range, fine,
you finish this and q is 1.
00:20:32.140 --> 00:20:34.170
Otherwise, you continue.
00:20:34.170 --> 00:20:36.980
And in the recursion, you're
continually reducing the
00:20:36.980 --> 00:20:40.870
magnitude, and it's easy to
show that eventually, the
00:20:40.870 --> 00:20:44.990
magnitude has to fall into this
range and be one of these
00:20:44.990 --> 00:20:50.060
n remainder numbers, and
then you're done.
00:20:50.060 --> 00:20:50.550
OK?
00:20:50.550 --> 00:20:53.290
So it's a descending chain
where the chain
00:20:53.290 --> 00:20:54.880
has a bottom at 0.
00:20:54.880 --> 00:20:58.020
If you start with a positive
integer, you can't go below 0.
00:20:58.020 --> 00:20:59.790
And this is the only way
it can come out.
00:20:59.790 --> 00:21:01.040
Very easy to prove that.
00:21:03.420 --> 00:21:03.850
All right.
00:21:03.850 --> 00:21:09.950
Similarly in polynomials, we get
an analogous expression.
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
00:21:16.340 --> 00:21:21.220
take F of x, and we can always
write this as some quotient
00:21:21.220 --> 00:21:25.900
times G of x plus
some remainder.
00:21:25.900 --> 00:21:28.770
Where the important thing here
about the remainder is that
00:21:28.770 --> 00:21:33.220
the degree of the remainder
is less than the
00:21:33.220 --> 00:21:36.460
degree of G of x.
00:21:36.460 --> 00:21:42.245
And here the remainder is called
F of x mod G of x.
00:21:45.310 --> 00:21:51.750
Just as the remainder here
is called m mod n.
00:21:54.790 --> 00:22:00.190
And there's a unique
remainder.
00:22:00.190 --> 00:22:05.900
And again, how would
you prove this?
00:22:05.900 --> 00:22:08.580
You could just take any long
division algorithm that you
00:22:08.580 --> 00:22:15.150
know for dividing G
of x into F of x.
00:22:15.150 --> 00:22:19.920
Basically long division amounts
to taking F of x.
00:22:19.920 --> 00:22:27.650
You can always choose some
scalar multiple of G of x such
00:22:27.650 --> 00:22:33.360
that F of x minus alpha G of x
has degree less than F of x.
00:22:33.360 --> 00:22:37.770
So you pick the top term to
reduce the degree, all right?
00:22:37.770 --> 00:22:39.370
Let's take G of x to be monic.
00:22:39.370 --> 00:22:42.100
We're only interested in
monic polynomials.
00:22:42.100 --> 00:22:45.180
If the top term of F of x is --
well, we're only going to
00:22:45.180 --> 00:22:48.520
divide it into monic
polynomials.
00:22:48.520 --> 00:22:51.190
But as we go along, we may
get non-monic ones.
00:22:51.190 --> 00:22:55.260
So you take the top
term, whatever it
00:22:55.260 --> 00:22:56.930
is over here, f(m).
00:22:56.930 --> 00:22:59.390
You multiply f(m)
times g of x.
00:22:59.390 --> 00:23:02.220
You subtract f(m) g
of x from f of x.
00:23:02.220 --> 00:23:04.208
You reduce the degree.
00:23:04.208 --> 00:23:05.196
OK?
00:23:05.196 --> 00:23:06.446
AUDIENCE: [INAUDIBLE PHRASE].
00:23:10.730 --> 00:23:12.680
PROFESSOR: Correct.
00:23:12.680 --> 00:23:13.570
Thank you very much.
00:23:13.570 --> 00:23:18.740
You need also a term x to
whatever the difference in
00:23:18.740 --> 00:23:24.256
degrees is here to move the
degree up to the top.
00:23:24.256 --> 00:23:28.803
When we're actually doing long
division, we write f(m) f(m
00:23:28.803 --> 00:23:32.220
minus 1) down to f(0).
00:23:32.220 --> 00:23:38.310
We divide g(n) down to g(1).
00:23:38.310 --> 00:23:42.770
And the first term, g(n) is
going to be equal to 1.
00:23:42.770 --> 00:23:45.410
We take f(m) up here.
00:23:45.410 --> 00:23:54.189
We implicitly move it over to
get f(m) dot dot dot dot, down
00:23:54.189 --> 00:23:55.890
to f(m) alpha.
00:23:55.890 --> 00:24:00.060
We subtract, and we're down
to something that only has
00:24:00.060 --> 00:24:01.250
degree m minus 1.
00:24:01.250 --> 00:24:06.440
That's what polynomial long
division is shorthand for.
00:24:10.130 --> 00:24:13.470
So you all know how
to do this.
00:24:13.470 --> 00:24:14.010
OK.
00:24:14.010 --> 00:24:18.790
Again, similar kind of proof,
that you must be able to get a
00:24:18.790 --> 00:24:20.890
remainder in this range,
and furthermore, the
00:24:20.890 --> 00:24:22.140
remainder is unique.
00:24:25.040 --> 00:24:27.390
You basically can go through
this process.
00:24:27.390 --> 00:24:30.220
You reduce the degree by at
least one every time.
00:24:33.030 --> 00:24:38.960
Therefore, degree must
eventually be reduced to where
00:24:38.960 --> 00:24:41.090
it's less than degree g of x.
00:24:41.090 --> 00:24:46.560
At that point, you can't
continue this process.
00:24:46.560 --> 00:24:47.640
You're stuck.
00:24:47.640 --> 00:24:48.730
And that's your remainder.
00:24:48.730 --> 00:24:51.780
There's no way of taking
something of lesser degree of
00:24:51.780 --> 00:24:54.770
g of x and then subtracting some
multiple from g of x from
00:24:54.770 --> 00:24:58.160
it to still further
reduce the degree.
00:24:58.160 --> 00:24:58.570
All right.
00:24:58.570 --> 00:25:00.490
So similar proof.
00:25:00.490 --> 00:25:03.160
Uniqueness is pretty obvious.
00:25:03.160 --> 00:25:10.080
So you get a unique remainder of
lesser degree than g of x.
00:25:10.080 --> 00:25:20.810
And so you can reduce any f of
x to some remainder r of x,
00:25:20.810 --> 00:25:22.270
which is called f
of x mod g of x.
00:25:30.840 --> 00:25:39.490
And if we do arithmetic, the way
we do arithmetic over here
00:25:39.490 --> 00:25:43.770
for addition is we just, if we
want to add two remainders mod
00:25:43.770 --> 00:25:47.310
n, we take the sum of them, and
then if necessary, reduce
00:25:47.310 --> 00:25:49.610
them mod n.
00:25:49.610 --> 00:25:50.846
Similarly with multiplication.
00:25:50.846 --> 00:25:54.040
If we want to multiply them,
we take the product of two
00:25:54.040 --> 00:25:57.320
remainders, and if necessary,
reduce them again to a
00:25:57.320 --> 00:26:02.272
legitimate remainder which is
in this range or to mod n.
00:26:02.272 --> 00:26:05.270
It's the same over here.
00:26:05.270 --> 00:26:10.450
If we want to add two
00:26:10.450 --> 00:26:13.490
remainders, that's easy enough.
00:26:13.490 --> 00:26:16.160
We can do that and we won't
increase the degree, so we
00:26:16.160 --> 00:26:18.850
automatically get something
when we add two remainders
00:26:18.850 --> 00:26:24.100
that satisfies this
degree property.
00:26:24.100 --> 00:26:26.060
We don't have to reduce
mod g of x.
00:26:26.060 --> 00:26:30.060
If we multiply, you do have to
check that when you multiply
00:26:30.060 --> 00:26:35.900
two remainders, then all you
need to do is reduce
00:26:35.900 --> 00:26:38.050
the mod g of x.
00:26:38.050 --> 00:26:41.440
And basically, the assertion
is that --
00:26:46.390 --> 00:26:48.570
let's see. r of x, s of x --
00:26:53.520 --> 00:26:57.300
it's hard to write this without
being tautological.
00:26:57.300 --> 00:26:59.710
r of x, s of x.
00:26:59.710 --> 00:27:06.420
Reduced mod n, I'm sorry,
mod g of x.
00:27:06.420 --> 00:27:06.670
I'm sorry.
00:27:06.670 --> 00:27:08.860
This is not worth writing.
00:27:08.860 --> 00:27:11.080
r of x, s of x, mod g of x.
00:27:15.340 --> 00:27:18.104
Something like that.
00:27:18.104 --> 00:27:18.540
Bah.
00:27:18.540 --> 00:27:19.900
It's a total tautology.
00:27:19.900 --> 00:27:21.110
Forget it.
00:27:21.110 --> 00:27:23.940
Said correctly in the notes.
00:27:23.940 --> 00:27:28.710
I'd rather think of it
as residue classes.
00:27:28.710 --> 00:27:38.512
If we, say, we talk about this
as a coset, f of x plus r of
00:27:38.512 --> 00:27:43.600
x, and we want to multiply
any element.
00:27:43.600 --> 00:27:48.160
This coset times any element
of the coset f
00:27:48.160 --> 00:27:50.520
of x plus s of x.
00:27:53.030 --> 00:27:58.500
Then the result is just
multiplying out symbolically
00:27:58.500 --> 00:28:01.300
the polynomials times the
polynomials are the
00:28:01.300 --> 00:28:02.810
polynomials.
00:28:02.810 --> 00:28:06.180
Any polynomial times a
polynomial is a polynomial.
00:28:06.180 --> 00:28:08.840
Plus r of x f of x.
00:28:08.840 --> 00:28:11.800
We could get some multiple
of r of x,
00:28:11.800 --> 00:28:12.955
some polynomial multiple.
00:28:12.955 --> 00:28:16.130
We could get some polynomial
multiple of s of x.
00:28:19.450 --> 00:28:22.280
Plus r of x s of x.
00:28:26.740 --> 00:28:29.450
But this is a polynomial that's
included in here.
00:28:29.450 --> 00:28:30.830
We don't need to
say that again.
00:28:30.830 --> 00:28:32.590
Similarly here.
00:28:32.590 --> 00:28:39.310
And so this is equal
to f of x.
00:28:39.310 --> 00:28:44.360
It's equal to the coset f
of x plus r of x s of x.
00:28:50.190 --> 00:28:52.900
But this is another
polynomial that
00:28:52.900 --> 00:28:55.090
probably has higher degree.
00:28:55.090 --> 00:29:03.160
This is equal to the coset
f of x plus r of x s
00:29:03.160 --> 00:29:09.780
of x mod g of x.
00:29:09.780 --> 00:29:10.170
OK.
00:29:10.170 --> 00:29:17.180
So this tells us that to
multiply cosets, coset with
00:29:17.180 --> 00:29:20.270
representative r of x times the
coset with representative
00:29:20.270 --> 00:29:26.120
s of x, we're going to get
something in the coset whose
00:29:26.120 --> 00:29:31.210
representative is r of
x, s of x mod g of x.
00:29:31.210 --> 00:29:36.870
So this is a sketch of a proof
that basically mod g of x
00:29:36.870 --> 00:29:39.830
commutes with multiplication.
00:29:39.830 --> 00:29:44.760
To multiply r of
x times s of x.
00:29:44.760 --> 00:29:45.210
All right.
00:29:45.210 --> 00:29:46.460
So --
00:29:50.390 --> 00:29:51.640
AUDIENCE: [INAUDIBLE PHRASE].
00:29:58.873 --> 00:30:00.370
PROFESSOR: Yeah.
00:30:00.370 --> 00:30:04.230
You're quite correct.
00:30:04.230 --> 00:30:08.220
What I mean is the cosets
of g of x of f of x.
00:30:08.220 --> 00:30:10.750
So I now understand all
the blank looks.
00:30:18.638 --> 00:30:20.300
Are we better off now?
00:30:25.120 --> 00:30:31.000
We have a group consisting of
the set of all multiples of g
00:30:31.000 --> 00:30:37.450
of x, which I write as g of x
times all the polynomials.
00:30:37.450 --> 00:30:40.900
The cosets of the group
are precisely --
00:30:40.900 --> 00:30:44.230
there's one remainder of degree
less than g of x in
00:30:44.230 --> 00:30:44.980
every coset.
00:30:44.980 --> 00:30:49.270
So this is the representative
of the coset.
00:30:49.270 --> 00:30:55.190
This is a sketch of a proof that
multiplication basically
00:30:55.190 --> 00:30:58.150
just amounts to multiplying
the remainders.
00:30:58.150 --> 00:31:03.010
The representative of the
product coset is the product
00:31:03.010 --> 00:31:05.760
of the representatives
modulo g of x.
00:31:05.760 --> 00:31:09.170
I think I said it correctly
for once.
00:31:09.170 --> 00:31:12.260
Please read the notes if you are
still confused, as I can
00:31:12.260 --> 00:31:13.510
see some of you are.
00:31:24.721 --> 00:31:26.740
Maybe it would help
to do an example.
00:31:34.480 --> 00:31:43.670
Let's take g of x to be equal
to x squared plus x
00:31:43.670 --> 00:31:50.480
plus 1 in f2 of x.
00:31:50.480 --> 00:31:54.350
It's a binary polynomial.
00:31:54.350 --> 00:32:01.370
Then r of x.
00:32:01.370 --> 00:32:12.760
The remainders are
equal to 0,1.
00:32:12.760 --> 00:32:16.680
This was degree minus infinity,
this is degree 0.
00:32:16.680 --> 00:32:18.900
They're x or x plus one.
00:32:23.260 --> 00:32:24.500
OK?
00:32:24.500 --> 00:32:29.650
These are the possible
remainders when I take any
00:32:29.650 --> 00:32:31.620
polynomial modulo g of x.
00:32:31.620 --> 00:32:36.080
Divide it by g of x, and I'm
going to get a polynomial in
00:32:36.080 --> 00:32:40.340
f2 of x of degree less than or
equal to 1, and those are all
00:32:40.340 --> 00:32:44.765
of the polynomials that are
decisively for degree less
00:32:44.765 --> 00:32:47.140
than or equal to 1.
00:32:47.140 --> 00:32:49.020
All right?
00:32:49.020 --> 00:32:52.340
So what's the addition table?
00:32:52.340 --> 00:32:54.026
Addition is pretty easy.
00:33:07.780 --> 00:33:09.620
0 plus anything is itself.
00:33:13.820 --> 00:33:19.070
1 plus 1 in F2 is simply 0.
00:33:19.070 --> 00:33:24.440
1 plus x is 1 plus x, sorry,
x plus 1 as I've written.
00:33:24.440 --> 00:33:27.750
And 1 plus x plus 1 is x.
00:33:27.750 --> 00:33:31.710
x, x plus 1.
00:33:31.710 --> 00:33:33.290
x plus x is what?
00:33:35.890 --> 00:33:37.136
Hello?
00:33:37.136 --> 00:33:37.540
0.
00:33:37.540 --> 00:33:38.830
Thank you.
00:33:38.830 --> 00:33:42.360
x plus x plus 1?
00:33:42.360 --> 00:33:43.392
Thank you.
00:33:43.392 --> 00:33:47.080
x plus 1, x.
00:33:47.080 --> 00:33:49.770
x plus x plus 1, 1, and 0.
00:33:49.770 --> 00:33:52.460
So that's what the addition
table looks like.
00:33:55.150 --> 00:34:01.770
Actually you could think of
these as being just written as
00:34:01.770 --> 00:34:10.670
binary pairs, 0 0, 0 1, 1 0,
1 1, where this pair is
00:34:10.670 --> 00:34:13.760
basically F1 F0.
00:34:13.760 --> 00:34:19.110
And then the addition table is
precisely the same as the
00:34:19.110 --> 00:34:21.370
addition table for these
binary 2-tuples.
00:34:21.370 --> 00:34:26.480
You just add, component-wise,
the lowest order coefficient,
00:34:26.480 --> 00:34:28.850
the F1 coefficient.
00:34:28.850 --> 00:34:37.780
So addition is just like
addition in F2 squared.
00:34:37.780 --> 00:34:41.550
Or z2 squared, if you like.
00:34:41.550 --> 00:34:42.000
OK.
00:34:42.000 --> 00:34:44.610
That's actually an
additive group.
00:34:44.610 --> 00:34:46.080
Check it out.
00:34:46.080 --> 00:34:48.219
It's not z4 by the way.
00:34:48.219 --> 00:34:54.120
The other abelian group, or the
other group of size four,
00:34:54.120 --> 00:34:56.969
sometimes called the Klein
four-group, but it's really
00:34:56.969 --> 00:35:00.120
just the addition table for
the set of all binary
00:35:00.120 --> 00:35:00.640
[UNINTELLIGIBLE]
00:35:00.640 --> 00:35:02.760
two-tuples.
00:35:02.760 --> 00:35:04.010
OK?
00:35:05.440 --> 00:35:06.690
Multiplication.
00:35:14.320 --> 00:35:16.520
One very nice thing about
finite fields is
00:35:16.520 --> 00:35:18.470
you can simply --
00:35:18.470 --> 00:35:18.800
sorry.
00:35:18.800 --> 00:35:21.062
This was supposed to be
addition, this was supposed to
00:35:21.062 --> 00:35:22.312
be multiplication.
00:35:24.390 --> 00:35:28.750
You know can simply write out
what all the rules are in a
00:35:28.750 --> 00:35:30.000
finite space.
00:35:33.900 --> 00:35:34.360
OK.
00:35:34.360 --> 00:35:35.680
So what's 0 times anything?
00:35:40.720 --> 00:35:41.970
What's 1 times anything?
00:35:44.550 --> 00:35:45.180
Itself.
00:35:45.180 --> 00:35:46.750
1 is the multiplicative
identity.
00:35:50.720 --> 00:35:52.080
But you know.
00:35:52.080 --> 00:35:54.150
You could formally do this
by doing polynomial
00:35:54.150 --> 00:35:55.360
multiplication.
00:35:55.360 --> 00:35:55.880
All right.
00:35:55.880 --> 00:35:56.850
Here's an interesting one.
00:35:56.850 --> 00:35:58.330
What's x times x?
00:36:02.150 --> 00:36:04.060
Do x times x.
00:36:08.280 --> 00:36:11.700
What is that going
to be equal to?
00:36:11.700 --> 00:36:12.500
x plus 1.
00:36:12.500 --> 00:36:15.600
How did you do that?
00:36:15.600 --> 00:36:17.410
We did the modulo.
00:36:17.410 --> 00:36:20.510
First of all, we write that
that's x squared.
00:36:20.510 --> 00:36:24.990
But then we have to do x
squared modulo g of x.
00:36:24.990 --> 00:36:27.540
x squared plus x plus 1.
00:36:27.540 --> 00:36:31.970
So we have to go through a
little long division process.
00:36:31.970 --> 00:36:34.720
We have to subtract this
out from that.
00:36:34.720 --> 00:36:36.430
And that gives us x plus 1.
00:36:41.380 --> 00:36:42.560
So that's the key rule.
00:36:42.560 --> 00:36:47.970
Whenever we see something of
degree two or higher, we can
00:36:47.970 --> 00:36:51.170
always reduce it by subtracting
out some multiple
00:36:51.170 --> 00:36:54.810
of g of x down to something
of lower degree.
00:36:54.810 --> 00:36:55.100
Right?
00:36:55.100 --> 00:36:57.230
So this is what I've
been talking about.
00:36:57.230 --> 00:37:01.100
So x times x is x plus 1.
00:37:01.100 --> 00:37:02.975
What's x times x plus 1?
00:37:08.230 --> 00:37:09.480
Equals what?
00:37:12.162 --> 00:37:12.660
Good.
00:37:12.660 --> 00:37:15.780
You can do that in your heads.
00:37:15.780 --> 00:37:18.860
And what's x plus 1
times x plus 1?
00:37:25.160 --> 00:37:27.810
Remember, we're doing
mod-2 arithmetic in
00:37:27.810 --> 00:37:29.760
our base field here.
00:37:29.760 --> 00:37:32.170
So this equals what?
00:37:32.170 --> 00:37:33.520
x squared plus 1.
00:37:33.520 --> 00:37:35.130
We reduce that, we get x.
00:37:42.970 --> 00:37:45.560
OK.
00:37:45.560 --> 00:37:48.040
Let me check right now.
00:37:48.040 --> 00:37:51.920
Is this a field?
00:37:51.920 --> 00:37:54.760
These four elements with these
rules for addition and
00:37:54.760 --> 00:37:56.260
multiplication.
00:37:56.260 --> 00:37:57.510
Does that form a field?
00:38:03.370 --> 00:38:07.230
What do I have to check, apart
from formalities like the
00:38:07.230 --> 00:38:10.520
distributive law?
00:38:10.520 --> 00:38:12.370
Which follows from the
distributive law for
00:38:12.370 --> 00:38:14.690
polynomials.
00:38:14.690 --> 00:38:18.432
That's always going to hold
through mod x arithmetic.
00:38:18.432 --> 00:38:19.860
What do I have to check?
00:38:19.860 --> 00:38:21.730
What are my field axioms?
00:38:21.730 --> 00:38:23.800
Anybody?
00:38:23.800 --> 00:38:26.920
Closure under multiplication.
00:38:29.610 --> 00:38:33.720
That's getting towards a very
crisp statement of the --
00:38:33.720 --> 00:38:36.060
I'm looking for two
group axioms.
00:38:36.060 --> 00:38:39.850
One has to do with something we
have to check for addition.
00:38:39.850 --> 00:38:41.760
Something has to do with
something we have to check for
00:38:41.760 --> 00:38:43.010
multiplication.
00:38:46.190 --> 00:38:47.230
Inverses?
00:38:47.230 --> 00:38:48.480
AUDIENCE: [INAUDIBLE PHRASE].
00:38:52.450 --> 00:38:53.130
PROFESSOR: OK.
00:38:53.130 --> 00:38:56.840
Between the two of you I heard
the two answers that I want.
00:38:56.840 --> 00:38:58.620
We have to check that
this forms an
00:38:58.620 --> 00:39:01.930
abelian group under addition.
00:39:01.930 --> 00:39:04.310
So we have to check that the
addition table is the addition
00:39:04.310 --> 00:39:08.170
table of an abelian group.
00:39:08.170 --> 00:39:10.850
And under multiplication, we
have to check that the
00:39:10.850 --> 00:39:14.410
non-zero elements form
a billion group.
00:39:14.410 --> 00:39:22.000
So just this part of the table
has to form an abelian group,
00:39:22.000 --> 00:39:25.830
and both these have to have
an identity, of course.
00:39:25.830 --> 00:39:34.480
But the identity in mod g of x
arithmetic is always going to
00:39:34.480 --> 00:39:37.430
be 0 for addition and it's
always going to be 1 for
00:39:37.430 --> 00:39:38.680
multiplication.
00:39:42.930 --> 00:39:43.350
All right.
00:39:43.350 --> 00:39:44.990
So I check this.
00:39:44.990 --> 00:39:46.180
Is this a group table?
00:39:46.180 --> 00:39:49.080
Basically, I just have to check
whether every row and
00:39:49.080 --> 00:39:55.080
column is a permutation
of the elements.
00:39:55.080 --> 00:39:57.180
And it is.
00:39:57.180 --> 00:40:02.810
And 0 acts as 0 should act.
00:40:02.810 --> 00:40:06.600
Has the additive identity.
00:40:06.600 --> 00:40:08.450
All right?
00:40:08.450 --> 00:40:12.290
Here what I have to check is
that the nonzero elements,
00:40:12.290 --> 00:40:21.070
these three, form an abelian
group under multiplication.
00:40:21.070 --> 00:40:25.830
Well, there really is only
one group of size three.
00:40:25.830 --> 00:40:29.790
It is isomorphic to Z3.
00:40:29.790 --> 00:40:31.710
If I replace --
00:40:31.710 --> 00:40:38.360
let's remember what Z3 looks
like under addition.
00:40:38.360 --> 00:40:45.270
This looks like 0 1 2,
0 1 2, 0 1 2, 0 1 2.
00:40:45.270 --> 00:40:46.270
It's mod-3.
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,
00:40:52.420 --> 00:40:56.360
which is 0, 2 plus 2 equals 1.
00:40:56.360 --> 00:40:58.710
So gee whiz.
00:40:58.710 --> 00:41:03.930
This is isomorphic to that if I
relabel 1 by 0, x by 1, and
00:41:03.930 --> 00:41:06.475
x plus 1 by 2.
00:41:06.475 --> 00:41:09.700
That's the only thing
it could be.
00:41:09.700 --> 00:41:14.690
The only group table in which
every row and column is a
00:41:14.690 --> 00:41:16.025
permutation of every other.
00:41:18.620 --> 00:41:19.870
OK?
00:41:21.730 --> 00:41:26.430
So we verified that we
now have a finite
00:41:26.430 --> 00:41:32.540
field with four elements.
00:41:36.470 --> 00:41:37.985
Prime power number
of elements.
00:41:41.020 --> 00:41:43.480
Right?
00:41:43.480 --> 00:41:48.310
The elements of my field are
these four remainders, or you
00:41:48.310 --> 00:41:50.340
can think of them as
representatives for their
00:41:50.340 --> 00:41:53.020
cosets, modulo g of x.
00:41:53.020 --> 00:42:02.410
The addition rule is addition
modulo g of x, and the
00:42:02.410 --> 00:42:07.720
multiplication rule is
multiplication modulo g of x.
00:42:07.720 --> 00:42:10.920
And it satisfies the field
axioms, therefore, it's a
00:42:10.920 --> 00:42:12.333
finite field.
00:42:12.333 --> 00:42:13.750
All right?
00:42:13.750 --> 00:42:16.470
I can add, subtract.
00:42:16.470 --> 00:42:19.360
Addition and subtraction
basically looks like addition
00:42:19.360 --> 00:42:23.170
and subtraction of binary
two-tuples, just
00:42:23.170 --> 00:42:24.680
component-wise.
00:42:24.680 --> 00:42:27.320
Multiplication is a little
bit more mysterious
00:42:27.320 --> 00:42:29.045
right now, but it works.
00:42:33.370 --> 00:42:35.203
Let me tell you where we're
going to go on multiplication.
00:42:40.690 --> 00:42:47.010
In this case, I can write x
plus 1 in a different way.
00:42:47.010 --> 00:42:52.530
I note that x plus 1 is
equal to x squared.
00:42:52.530 --> 00:42:54.240
All right?
00:42:54.240 --> 00:43:03.840
So let me write a little
log table over here for
00:43:03.840 --> 00:43:05.135
multiplication purposes.
00:43:08.260 --> 00:43:12.320
I'm going to write x --
00:43:12.320 --> 00:43:13.570
I'm going to call that alpha.
00:43:17.620 --> 00:43:20.590
And x plus 1 is equal
to x squared,
00:43:20.590 --> 00:43:21.925
or it's alpha squared.
00:43:25.700 --> 00:43:26.950
What's alpha cubed?
00:43:33.210 --> 00:43:38.636
Alpha cubed is x times
this again.
00:43:38.636 --> 00:43:43.050
Let me look in the table. x
times x plus 1 is equal to 1.
00:43:43.050 --> 00:43:45.940
So 1 equals alpha cubed.
00:43:45.940 --> 00:43:49.590
Or I could write that
as alpha to 0.
00:43:49.590 --> 00:43:53.730
If I multiply by x again,
I just cycle.
00:43:53.730 --> 00:43:55.990
So I'm going to get a
cyclic group here.
00:43:58.910 --> 00:43:59.720
And now I'm going to write the
00:43:59.720 --> 00:44:01.405
multiplication table as follows.
00:44:04.630 --> 00:44:07.310
I'm going to write the elements
of the group as 1,
00:44:07.310 --> 00:44:12.915
alpha, alpha squared, 0, 1,
alpha, alpha squared.
00:44:17.910 --> 00:44:20.060
Again, 0 times anything is 0.
00:44:20.060 --> 00:44:23.130
We never have to worry
about that.
00:44:23.130 --> 00:44:25.040
1, alpha, alpha squared.
00:44:29.330 --> 00:44:31.540
Alpha times alpha is
alpha squared.
00:44:31.540 --> 00:44:34.610
Alpha times alpha squared
is alpha cubed.
00:44:34.610 --> 00:44:37.520
But that's equal to 1.
00:44:37.520 --> 00:44:38.860
Same here.
00:44:38.860 --> 00:44:40.680
Alpha squared times
alpha squared --
00:44:40.680 --> 00:44:43.750
what's that?
00:44:43.750 --> 00:44:47.360
Alpha to the fourth, but what
does alpha fourth equal to if
00:44:47.360 --> 00:44:49.280
alpha cubed is equal to 1?
00:44:51.930 --> 00:44:54.830
This has to be equal to alpha.
00:44:54.830 --> 00:44:59.560
Point is, because of this
relationship here, I can
00:44:59.560 --> 00:45:03.420
always reduce the exponents
modulo 3.
00:45:03.420 --> 00:45:08.270
I've basically got a little
multiplicative cyclic group of
00:45:08.270 --> 00:45:12.830
order three that's, of course,
isomorphic to the
00:45:12.830 --> 00:45:14.230
additive group, Z3.
00:45:17.770 --> 00:45:22.240
So there are two ways I
can do multiplication.
00:45:22.240 --> 00:45:25.560
One is, I can do it by
this mod g of x way.
00:45:25.560 --> 00:45:32.435
I can represent things by these
which basically stand
00:45:32.435 --> 00:45:37.010
for polynomials of degree
one or less.
00:45:37.010 --> 00:45:40.540
And I can multiply two of these
by simply going through
00:45:40.540 --> 00:45:43.040
standard polynomial
multiplication over the
00:45:43.040 --> 00:45:44.850
appropriate field.
00:45:44.850 --> 00:45:47.870
And then I'll likely get some
powers of x squared or higher,
00:45:47.870 --> 00:45:51.850
and I reduce those to modulo
x squared plus x plus 1.
00:45:51.850 --> 00:45:53.820
That's legitimate and that
gives me this statement.
00:45:56.460 --> 00:45:58.820
Well, the other thing I can do
is for multiplication, I can
00:45:58.820 --> 00:46:00.990
have a different
representation.
00:46:00.990 --> 00:46:03.980
This is basically just
a log table.
00:46:03.980 --> 00:46:06.280
OK?
00:46:06.280 --> 00:46:11.740
For each, I would have in my
little computer a separate
00:46:11.740 --> 00:46:16.790
table where I'd write that this
corresponds to 0, 0 1
00:46:16.790 --> 00:46:21.230
corresponds to 1, 1 0
corresponds to alpha, and 1 1
00:46:21.230 --> 00:46:24.430
corresponds to alpha squared.
00:46:24.430 --> 00:46:27.170
Or of course, I would just
represent these by their
00:46:27.170 --> 00:46:31.810
exponents that have some
special value for 0.
00:46:31.810 --> 00:46:34.940
And then all I have to do is
add exponents modulo 3 for
00:46:34.940 --> 00:46:37.360
multiplication.
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,
00:46:45.700 --> 00:46:54.220
and then I just use this for my
multiplication operation,
00:46:54.220 --> 00:46:58.990
or equivalently this cyclic
multiplicative group.
00:46:58.990 --> 00:47:02.590
Then I go through some inverse
log operation to get back to
00:47:02.590 --> 00:47:05.926
the other representation,
if I wanted to.
00:47:05.926 --> 00:47:08.760
And in fact, in finite field
arithmetic, this is what's
00:47:08.760 --> 00:47:09.550
typically done.
00:47:09.550 --> 00:47:13.140
There's just a little table
lookup such that you can go
00:47:13.140 --> 00:47:16.450
back and forth between this
representation, which we use
00:47:16.450 --> 00:47:19.380
for addition, and this
representation, which we use
00:47:19.380 --> 00:47:20.630
for multiplication.
00:47:24.188 --> 00:47:25.170
Yeah?
00:47:25.170 --> 00:47:26.420
AUDIENCE: [INAUDIBLE PHRASE].
00:47:30.140 --> 00:47:32.410
PROFESSOR: You have to represent
it as being special
00:47:32.410 --> 00:47:34.130
in some way.
00:47:34.130 --> 00:47:43.010
So if, in fact, literally, I
made log x equal to 1, log x
00:47:43.010 --> 00:47:50.060
plus 1 equal to 2, log
1 equal to 0 --
00:47:50.060 --> 00:47:52.860
one thing I suggest in the
notes, you can make log 0
00:47:52.860 --> 00:47:55.190
equal to minus infinity.
00:47:55.190 --> 00:47:57.620
That will always work.
00:47:57.620 --> 00:48:03.220
If you ever multiply
by 0, you'll be
00:48:03.220 --> 00:48:04.450
adding minus infinity.
00:48:04.450 --> 00:48:06.750
The result will be minus
infinity, so the
00:48:06.750 --> 00:48:08.020
inverse log is 0.
00:48:08.020 --> 00:48:10.770
So that's one way
you can do it.
00:48:10.770 --> 00:48:13.540
Or you could do what you do in
ordinary real and complex
00:48:13.540 --> 00:48:14.160
arithmetic.
00:48:14.160 --> 00:48:18.340
You could just, say, have
some special routine for
00:48:18.340 --> 00:48:21.040
multiplication by
0 is always 0.
00:48:21.040 --> 00:48:24.300
Division by 0 is illegal.
00:48:24.300 --> 00:48:25.550
Put out some error message.
00:48:29.400 --> 00:48:29.770
All right?
00:48:29.770 --> 00:48:32.820
So that's how you actually build
a little finite field
00:48:32.820 --> 00:48:34.500
computer for this
finite field.
00:48:38.270 --> 00:48:39.520
OK.
00:48:42.050 --> 00:48:45.005
Now, I chose this advisedly.
00:48:51.700 --> 00:48:57.240
Suppose I have chosen g of x
equal to x squared plus 1.
00:49:00.980 --> 00:49:01.640
OK?
00:49:01.640 --> 00:49:06.230
I can do mod g of x arithmetic
for x squared plus 1.
00:49:06.230 --> 00:49:09.070
Again, my four field elements
are going to be the four
00:49:09.070 --> 00:49:13.370
binary polynomials of
degree 1 or less.
00:49:13.370 --> 00:49:15.750
The addition table is going
to be exactly the same.
00:49:15.750 --> 00:49:16.750
So let's just write the
00:49:16.750 --> 00:49:18.440
multiplication table over here.
00:49:23.150 --> 00:49:30.680
0, 1, x, x plus 1, 0,
1, x, x plus 1.
00:49:30.680 --> 00:49:35.930
0 times anything is 0 in
ordinary polynomial
00:49:35.930 --> 00:49:39.360
multiplication, and therefore
also in mod g of x
00:49:39.360 --> 00:49:40.450
for any g of x.
00:49:40.450 --> 00:49:43.020
1 times anything is itself.
00:49:43.020 --> 00:49:45.600
No problem there.
00:49:45.600 --> 00:49:49.190
x, x plus 1.
00:49:49.190 --> 00:49:54.090
So we again have to give a
little bit care to x squared.
00:49:54.090 --> 00:49:57.260
What's that going
to be equal to?
00:49:57.260 --> 00:49:58.910
1.
00:49:58.910 --> 00:50:07.870
And x times x plus 1
is equal to what?
00:50:14.200 --> 00:50:16.360
x squared plus x is
equal to what?
00:50:23.570 --> 00:50:25.940
Looks like there's a problem.
00:50:25.940 --> 00:50:30.050
x plus 1 times x squared x plus
1 is equal to x squared
00:50:30.050 --> 00:50:33.080
plus 1, what's that equal?
00:50:33.080 --> 00:50:34.320
0.
00:50:34.320 --> 00:50:35.580
Yuck.
00:50:35.580 --> 00:50:36.830
Didn't work.
00:50:41.540 --> 00:50:43.070
What's the essential
problem here?
00:50:46.998 --> 00:50:47.490
Yeah.
00:50:47.490 --> 00:50:51.650
This clearly is not a group.
00:50:51.650 --> 00:50:54.070
It's not even closed under
multiplication.
00:50:54.070 --> 00:50:58.790
Because x plus 1 times x
plus 1 is equal to 0.
00:50:58.790 --> 00:51:01.640
The essential problem is that
there are two polynomials of
00:51:01.640 --> 00:51:08.590
degree less than two whose
product is x squared plus 1.
00:51:08.590 --> 00:51:11.880
In other words, this
is factorizable.
00:51:17.860 --> 00:51:18.340
OK?
00:51:18.340 --> 00:51:20.050
In F2 of x.
00:51:22.730 --> 00:51:23.330
All right?
00:51:23.330 --> 00:51:31.200
So whereas x squared
plus x plus 1.
00:51:31.200 --> 00:51:34.250
Does that have any factors
of degree 1 or less?
00:51:34.250 --> 00:51:36.750
Any nontrivial factors
of degree 1 or less?
00:51:39.610 --> 00:51:43.050
Well, basically we proved that
it didn't, when we wrote this
00:51:43.050 --> 00:51:45.870
multiplication table.
00:51:45.870 --> 00:51:52.620
We tried all products of degree
1 or less polynomials
00:51:52.620 --> 00:51:56.160
where they're both non-zero,
and we never got 0.
00:52:00.030 --> 00:52:08.520
So this will work if and only if
g of x is irreducible, has
00:52:08.520 --> 00:52:11.980
no nontrivial factors, is
a prime polynomial.
00:52:11.980 --> 00:52:14.600
These are all equivalent
in F2 of x.
00:52:17.300 --> 00:52:22.730
There's this distinction in
other fields that irreducible
00:52:22.730 --> 00:52:25.060
just means has no nontrivial
factors.
00:52:25.060 --> 00:52:27.370
Prime means there's a monic
polynomial with
00:52:27.370 --> 00:52:29.460
no non-trivial factors.
00:52:29.460 --> 00:52:36.800
So that's what I
said back here.
00:52:36.800 --> 00:52:40.850
That the way we're ultimately
going to have to construct
00:52:40.850 --> 00:52:45.510
finite fields is take the
polynomials in Fp, Fp of x --
00:52:45.510 --> 00:52:47.530
we've been looking
at F2 of x --
00:52:47.530 --> 00:52:50.150
modulo a prime polynomial
g of x.
00:52:57.450 --> 00:53:03.560
So what we're going to need to
find is prime polynomials.
00:53:03.560 --> 00:53:04.300
Let's see.
00:53:04.300 --> 00:53:11.690
Can I already prove that this is
going to work for any prime
00:53:11.690 --> 00:53:12.940
polynomial?
00:53:19.400 --> 00:53:26.160
So I'm going to force the g of
x to be equal to a prime
00:53:26.160 --> 00:53:33.546
polynomial in Fp of
x of degree m.
00:53:33.546 --> 00:53:44.360
All right
00:53:44.360 --> 00:53:48.135
For example, x squared plus x
plus 1 is a prime polynomial.
00:53:54.360 --> 00:54:15.250
And I'm going to ask if the
remainders mod g of x form a
00:54:15.250 --> 00:54:21.980
field under mod g
of x arithmetic.
00:54:33.370 --> 00:54:34.620
OK.
00:54:36.290 --> 00:54:39.660
Let's flow this out
a little bit.
00:54:39.660 --> 00:54:46.740
Again, what are the remainders
going to be of any polynomial
00:54:46.740 --> 00:54:47.990
of degree m?
00:54:54.450 --> 00:55:00.750
So this is basically going to
be the polynomials of degree
00:55:00.750 --> 00:55:06.915
less than m in Fp of x.
00:55:11.000 --> 00:55:12.620
How many of them are
there, by the way?
00:55:19.620 --> 00:55:20.390
p to the m.
00:55:20.390 --> 00:55:30.390
So the size of this
is p to the m.
00:55:30.390 --> 00:55:34.200
And one of the representations
for these polynomials is just
00:55:34.200 --> 00:55:40.350
to write out F0, F1,
up to m minus 1.
00:55:40.350 --> 00:55:49.350
So this just basically looks
like the polynomial m-tuples.
00:55:49.350 --> 00:55:59.420
Set of F0, F1 up to F minus 1,
for each of these an element
00:55:59.420 --> 00:56:00.670
of the field.
00:56:05.230 --> 00:56:05.830
OK?
00:56:05.830 --> 00:56:09.030
I can make a one-to-one
correspondence between the
00:56:09.030 --> 00:56:12.930
polynomials of degree less
than m and the set of all
00:56:12.930 --> 00:56:16.130
coefficient m-tuples over Fp.
00:56:16.130 --> 00:56:19.260
These are just m-tuples
over Fp.
00:56:19.260 --> 00:56:20.660
So there are p to
the m of them.
00:56:23.430 --> 00:56:29.580
And so it's a finite set,
size p to the m.
00:56:29.580 --> 00:56:33.018
Does it form a field under
mod g of x arithmetic?
00:56:33.018 --> 00:56:34.268
AUDIENCE: [INAUDIBLE PHRASE].
00:56:42.490 --> 00:56:42.920
PROFESSOR: Correct.
00:56:42.920 --> 00:56:44.320
And that's a homework problem.
00:56:48.450 --> 00:56:51.970
So I'm not going to do it in
class, but I will sketch here
00:56:51.970 --> 00:56:53.400
how it's going to be done.
00:56:53.400 --> 00:56:56.360
But I'm glad that you
instantly see that.
00:56:56.360 --> 00:56:59.030
Because again, we just model
everything we do on what we
00:56:59.030 --> 00:57:01.160
did for integers.
00:57:01.160 --> 00:57:04.650
If you remember how we proved
it for integers.
00:57:04.650 --> 00:57:08.820
First of all, we really need
to check two things.
00:57:08.820 --> 00:57:10.070
Addition.
00:57:12.680 --> 00:57:15.620
Just as we did for this specific
example up here, we
00:57:15.620 --> 00:57:20.550
first have to check that the
addition table is that of an
00:57:20.550 --> 00:57:25.072
abelian group, that these
supposed field elements form
00:57:25.072 --> 00:57:29.440
an abelian group
under addition.
00:57:29.440 --> 00:57:32.810
And we've already observed
several times that addition is
00:57:32.810 --> 00:57:38.460
just basically component-wise
addition of these m-tuples.
00:57:38.460 --> 00:57:40.210
So it's just like
vector addition.
00:57:43.270 --> 00:57:47.390
And vector addition, of course,
00:57:47.390 --> 00:57:48.560
has the group property.
00:57:48.560 --> 00:57:54.230
So addition is basically just
like vector addition.
00:57:54.230 --> 00:58:00.370
Or I could write, perhaps more
precisely, Zp over m.
00:58:00.370 --> 00:58:03.540
Distinction without a
difference, really.
00:58:03.540 --> 00:58:11.145
So it's just component-wise
addition of the coefficients.
00:58:13.760 --> 00:58:16.510
That's how we do polynomial
addition.
00:58:16.510 --> 00:58:20.780
And it will give us a remainder
that has a degree of
00:58:20.780 --> 00:58:23.850
less than m, so we don't have
to have ever reduce it.
00:58:23.850 --> 00:58:28.970
Modulo g of x, we just simply
add component-wise.
00:58:28.970 --> 00:58:29.420
OK.
00:58:29.420 --> 00:58:31.180
So that's easy to verify.
00:58:31.180 --> 00:58:34.180
The addition table is always
going to be OK.
00:58:34.180 --> 00:58:36.150
So what do we have
to prove now?
00:58:36.150 --> 00:58:37.400
We have to --
00:58:40.029 --> 00:58:41.250
what am I going to call these?
00:58:41.250 --> 00:58:42.680
Rg of x.
00:58:42.680 --> 00:58:45.420
The remainders mod g of x.
00:58:45.420 --> 00:58:55.070
For multiplication, we have to
prove that Rg of x star --
00:58:55.070 --> 00:59:08.350
the non-zero polynomials form an
abelian group, which, as I
00:59:08.350 --> 00:59:09.640
say, it's a homework problem.
00:59:09.640 --> 00:59:11.090
Let me sketch the proof.
00:59:11.090 --> 00:59:13.560
It's precisely analogous
to the proof that
00:59:13.560 --> 00:59:18.060
we made for Zp star.
00:59:18.060 --> 00:59:21.980
We basically have to
check closure.
00:59:21.980 --> 00:59:27.070
If we multiply two non-zero
polynomials, do we get another
00:59:27.070 --> 00:59:28.320
non-zero polynomial?
00:59:32.820 --> 00:59:37.130
Asking another way, is it
possible to multiply two
00:59:37.130 --> 00:59:44.410
polynomials of degree less than
m and get a result which
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
00:59:50.910 --> 00:59:53.870
multiple of g of x?
00:59:53.870 --> 00:59:59.050
And it's, I think, easy to
convince yourself if this has
00:59:59.050 --> 01:00:01.460
no factors --
01:00:01.460 --> 01:00:04.910
its only factors are
itself and 1, no
01:00:04.910 --> 01:00:07.860
non-trivial factors --
01:00:07.860 --> 01:00:14.510
then you can't multiply two
lesser degree polynomials and
01:00:14.510 --> 01:00:19.090
get either g of x, of course,
or any multiple of g of x.
01:00:19.090 --> 01:00:21.800
And it's an exercise
for the student to
01:00:21.800 --> 01:00:24.060
write that proof out.
01:00:24.060 --> 01:00:24.440
OK?
01:00:24.440 --> 01:00:27.400
But it's just exactly analogous
to the proof that
01:00:27.400 --> 01:00:32.920
you can't multiply two integers
less than a prime p
01:00:32.920 --> 01:00:35.790
and get a multiple of p.
01:00:35.790 --> 01:00:40.472
So use that as a model, if you
want to, in your proof.
01:00:40.472 --> 01:00:44.070
That of course depends
on this being prime.
01:00:44.070 --> 01:00:49.210
If this is factorizable,
non-prime, then of course
01:00:49.210 --> 01:00:52.020
there are going to be two
nontrivial factors, two
01:00:52.020 --> 01:00:58.380
remainders of degree less than
m, whose product is equal to g
01:00:58.380 --> 01:01:01.060
of x itself, if it's
factorizable.
01:01:01.060 --> 01:01:04.040
So you can't possibly
get closure.
01:01:04.040 --> 01:01:07.850
So this is why non-prime
polynomials don't work, just
01:01:07.850 --> 01:01:14.310
like non-prime integers don't
work when we constructed Fp.
01:01:14.310 --> 01:01:17.580
Exactly analogous reasons.
01:01:17.580 --> 01:01:19.940
All right.
01:01:19.940 --> 01:01:25.760
Second question is, when we go
through this multiplication,
01:01:25.760 --> 01:01:34.640
suppose we take r
of x times --
01:01:34.640 --> 01:01:36.430
let's construct a multiplication
table.
01:01:36.430 --> 01:01:37.720
Let's take a particular row.
01:01:37.720 --> 01:01:42.340
Let's take r of x times all
the non-zero things.
01:01:42.340 --> 01:01:45.630
Can we possibly get
any repeats?
01:01:45.630 --> 01:01:51.950
Can r of x times s of x equal
r of x times t of x?
01:01:51.950 --> 01:01:54.280
And again, it's easy to convince
yourself that that's
01:01:54.280 --> 01:01:55.020
not possible.
01:01:55.020 --> 01:01:59.010
If that were possible, then r of
x times s of x minus t of x
01:01:59.010 --> 01:02:01.210
would be equal to 0.
01:02:01.210 --> 01:02:04.630
And so we're back to where
we were before --
01:02:04.630 --> 01:02:05.740
mod g of x.
01:02:05.740 --> 01:02:08.170
And this is impossible.
01:02:08.170 --> 01:02:10.260
These are both degrees
less than m.
01:02:10.260 --> 01:02:13.210
We can't multiply two things
of lower degree together to
01:02:13.210 --> 01:02:14.965
get a multiple of g of x.
01:02:14.965 --> 01:02:17.060
So it can't equal 0.
01:02:17.060 --> 01:02:19.880
That means there can't
be any repeats.
01:02:19.880 --> 01:02:22.390
That means that each row
is a permutation
01:02:22.390 --> 01:02:24.050
of every other row.
01:02:24.050 --> 01:02:25.180
Similarly for columns.
01:02:25.180 --> 01:02:27.500
If you want to do that,
all you have to
01:02:27.500 --> 01:02:29.780
actually prove is one.
01:02:29.780 --> 01:02:39.070
So every row or column is a
permutations of every other
01:02:39.070 --> 01:02:44.100
one, if we just look at the
non-zero polynomials, star.
01:02:44.100 --> 01:02:54.240
And therefore, this forms an
abelian group whose identity
01:02:54.240 --> 01:02:55.490
is always one.
01:02:58.520 --> 01:03:00.880
Just as we proved up here.
01:03:00.880 --> 01:03:06.522
So this depends on
irreducibility.
01:03:06.522 --> 01:03:17.605
Depends on no nontrivial
factors of g of x.
01:03:24.990 --> 01:03:25.380
OK.
01:03:25.380 --> 01:03:26.640
And that's all we
have to check.
01:03:29.730 --> 01:03:35.350
So I claim that by this process,
I can, given a prime
01:03:35.350 --> 01:03:39.200
polynomial in Fp of x of degree
m, I can construct a
01:03:39.200 --> 01:03:46.235
finite field with p to the m
elements, that the addition
01:03:46.235 --> 01:03:49.360
and multiplication rules can
be taken as mod g of x
01:03:49.360 --> 01:03:53.700
arithmetic, and they will
satisfy the field axioms.
01:03:53.700 --> 01:03:58.550
So you can now construct a
finite field for any prime
01:03:58.550 --> 01:04:03.160
power p to the m, right?
01:04:03.160 --> 01:04:07.250
There's actually still
a hole in this.
01:04:07.250 --> 01:04:08.500
AUDIENCE: [INAUDIBLE PHRASE].
01:04:13.040 --> 01:04:14.680
PROFESSOR: Define prime
polynomial?
01:04:14.680 --> 01:04:16.315
The term, or --
01:04:16.315 --> 01:04:18.800
AUDIENCE: Define the
[UNINTELLIGIBLE] of degree m.
01:04:18.800 --> 01:04:21.280
PROFESSOR: Correct.
01:04:21.280 --> 01:04:24.225
Is there going to be a prime
polynomial of every degree?
01:04:28.990 --> 01:04:30.240
I don't know.
01:04:33.410 --> 01:04:37.230
Suppose you want to define an
irreducible polynomial over,
01:04:37.230 --> 01:04:44.390
say, F2 of x or F3
of x of degree 4.
01:04:44.390 --> 01:04:45.640
Could you do that?
01:04:49.460 --> 01:04:50.710
AUDIENCE: [INAUDIBLE PHRASE].
01:05:09.520 --> 01:05:10.180
PROFESSOR: Beautiful.
01:05:10.180 --> 01:05:11.920
I mean, that's an excellent
suggestion.
01:05:11.920 --> 01:05:25.110
So the question is, does there
exist a prime polynomial in Fp
01:05:25.110 --> 01:05:28.390
of x of every degree?
01:05:33.680 --> 01:05:34.945
Or of a given degree?
01:05:42.330 --> 01:05:46.140
And there are various ways of
attacking this question.
01:05:46.140 --> 01:05:49.240
First of all, I'll tell
you the answer is yes.
01:05:52.830 --> 01:05:57.870
For every p and every m, there
does exist a prime polynomial.
01:05:57.870 --> 01:05:59.790
Which is fortunate.
01:05:59.790 --> 01:06:04.530
So from that, we conclude there
is a finite field with p
01:06:04.530 --> 01:06:07.500
to the m elements for every
prime p and every m greater
01:06:07.500 --> 01:06:08.750
than or equal to 1.
01:06:11.680 --> 01:06:17.410
Now, how might we prove that?
01:06:17.410 --> 01:06:21.840
One is, look it up on Google.
01:06:26.920 --> 01:06:31.960
You can certainly formulate a
question that will lead you to
01:06:31.960 --> 01:06:35.910
a webpage that will have a
listing of all the prime
01:06:35.910 --> 01:06:39.800
polynomials of all degrees over
any field that you're
01:06:39.800 --> 01:06:42.060
interested in.
01:06:42.060 --> 01:06:45.830
So perhaps that will
suffice for you.
01:06:45.830 --> 01:06:47.441
Two.
01:06:47.441 --> 01:06:50.280
What I'm going to talk about
is the sieve method.
01:06:56.460 --> 01:06:58.120
Three.
01:06:58.120 --> 01:07:12.520
You could do a bound on the
number of polynomials of each
01:07:12.520 --> 01:07:19.540
degree in d, and show that
it's greater than
01:07:19.540 --> 01:07:22.240
or equal to 1, always.
01:07:22.240 --> 01:07:26.340
And this is done in the notes.
01:07:26.340 --> 01:07:29.800
Section 7.9 I think.
01:07:29.800 --> 01:07:34.550
Or four, you can do what Mr.
Agarwal has suggested.
01:07:34.550 --> 01:07:43.250
You can actually do the closed
form combinatoric
01:07:43.250 --> 01:07:52.310
formula, which --
01:07:52.310 --> 01:07:54.310
I haven't done any number
theory here.
01:07:54.310 --> 01:07:56.230
There's a little bit of
elementary number
01:07:56.230 --> 01:07:58.150
theory in the notes.
01:07:58.150 --> 01:08:02.050
Euler numbers, this
sort of thing.
01:08:02.050 --> 01:08:12.330
We get formulas for the number
of integers degree d that --
01:08:12.330 --> 01:08:15.920
well, number of integers that
have multiplicative orders d
01:08:15.920 --> 01:08:17.700
mod n, and so forth.
01:08:17.700 --> 01:08:23.520
It's a lovely combinatoric
field.
01:08:23.520 --> 01:08:26.960
There is a lovely closed form
for this that you get from the
01:08:26.960 --> 01:08:30.140
Mobius inversion formula.
01:08:30.140 --> 01:08:33.439
And it can be found in
combinatoric books.
01:08:33.439 --> 01:08:40.440
I know it's in Berlekamp's
Algebraic Coding Theory book.
01:08:40.440 --> 01:08:42.330
And it's extremely pretty.
01:08:42.330 --> 01:08:45.750
And then of course, given the
formula, you have to prove to
01:08:45.750 --> 01:08:48.620
all of the --
01:08:48.620 --> 01:08:51.120
again, n of d is greater
than or equal to 1.
01:08:51.120 --> 01:08:54.880
But there is a closed form
expression for it that you get
01:08:54.880 --> 01:08:55.880
out of combinatorics.
01:08:55.880 --> 01:08:58.420
We're not going to
do that here.
01:08:58.420 --> 01:09:00.180
I'll be satisfied --
01:09:00.180 --> 01:09:03.330
well, this is the real
engineering solution here.
01:09:03.330 --> 01:09:05.484
This is the mathematical
engineering solution.
01:09:07.990 --> 01:09:13.800
And what do I mean by
the sieve method?
01:09:13.800 --> 01:09:17.850
Again, take the analogy
with the integers.
01:09:17.850 --> 01:09:21.080
One of the first mathematical
accomplishments was
01:09:21.080 --> 01:09:25.630
Eratosthenes' sieve for
finding prime numbers.
01:09:25.630 --> 01:09:27.660
How does it work?
01:09:27.660 --> 01:09:30.430
You start to write
down all the --
01:09:30.430 --> 01:09:30.920
well.
01:09:30.920 --> 01:09:34.939
Imagine first writing down
all the integers.
01:09:34.939 --> 01:09:35.720
All right?
01:09:35.720 --> 01:09:36.630
Cross off 1.
01:09:36.630 --> 01:09:39.240
Start with 2.
01:09:39.240 --> 01:09:40.080
All right.
01:09:40.080 --> 01:09:42.700
So the first prime is 2.
01:09:42.700 --> 01:09:44.890
Then you cross off all
multiples of 2.
01:09:44.890 --> 01:09:47.710
4,6,8, so forth.
01:09:47.710 --> 01:09:49.979
OK?
01:09:49.979 --> 01:09:52.460
So what's the next number that
you haven't crossed off?
01:09:52.460 --> 01:09:54.370
It's 3, so that's
the next prime.
01:09:54.370 --> 01:09:57.370
You cross off all the
multiples of 3.
01:09:57.370 --> 01:10:00.480
3,6,9.
01:10:00.480 --> 01:10:01.790
So forth.
01:10:01.790 --> 01:10:03.900
15.
01:10:03.900 --> 01:10:07.970
And thereby you continue.
01:10:07.970 --> 01:10:15.590
So the steps are, find the
next integer on the list.
01:10:15.590 --> 01:10:17.460
That's going to be a prime,
because it won't have been
01:10:17.460 --> 01:10:19.370
crossed off by any
previous steps.
01:10:19.370 --> 01:10:23.970
It's not a multiple of any
integer of lower degree.
01:10:23.970 --> 01:10:28.240
Then cross off all of its
multiples, up to however long
01:10:28.240 --> 01:10:30.870
your scribe has written
this on the tablet.
01:10:30.870 --> 01:10:35.770
And this way, you can find
the primes up to
01:10:35.770 --> 01:10:37.960
any number you want.
01:10:37.960 --> 01:10:41.580
Gets kind of tedious after a
while, but you can certainly
01:10:41.580 --> 01:10:45.210
find all the primes
up to 100 in a few
01:10:45.210 --> 01:10:48.092
minutes doing this, right?
01:10:48.092 --> 01:10:52.250
Well, it's the same for
integers, and it's the same
01:10:52.250 --> 01:10:54.260
for polynomials.
01:10:54.260 --> 01:11:02.256
So let's, for instance,
do a polynomial sieve.
01:11:05.335 --> 01:11:10.750
Of course, what we're most
interested in is the
01:11:10.750 --> 01:11:13.670
polynomials with binary
coefficients, F2 of x.
01:11:16.470 --> 01:11:18.500
And how do we do it?
01:11:18.500 --> 01:11:21.430
Let's write down --
01:11:21.430 --> 01:11:24.490
let's forget about 0 and 1.
01:11:24.490 --> 01:11:28.280
Those are not considered
to be prime.
01:11:28.280 --> 01:11:32.770
Let's start with the degree
1 polynomials.
01:11:32.770 --> 01:11:35.770
What are the degree
one polynomials?
01:11:35.770 --> 01:11:38.505
They're x, x plus 1.
01:11:42.300 --> 01:11:43.550
Are these factorizable?
01:11:48.720 --> 01:11:52.160
Obviously their only factors
are one and themselves.
01:11:52.160 --> 01:11:54.460
They have no nontrivial
factors.
01:11:54.460 --> 01:11:55.710
So these are primes.
01:12:00.210 --> 01:12:01.460
OK.
01:12:03.310 --> 01:12:11.995
So now let's write down all the
polynomials of degree two.
01:12:17.550 --> 01:12:22.700
I'm sort of doing this in
an interleaved manner.
01:12:22.700 --> 01:12:25.950
There are going to be two of
degree 1, four of degree 2.
01:12:25.950 --> 01:12:28.330
All I'm going to do is -- well,
the only polynomials are
01:12:28.330 --> 01:12:30.410
monic in F2 of x, so I don't
have to make that
01:12:30.410 --> 01:12:32.350
qualification.
01:12:32.350 --> 01:12:32.700
All right.
01:12:32.700 --> 01:12:35.000
So here are the four
of degree 2, right?
01:12:38.150 --> 01:12:40.800
Now I go through with my sieve
and we take out all
01:12:40.800 --> 01:12:42.560
multiples of x.
01:12:42.560 --> 01:12:48.040
Multiples of x are polynomials
with 0 constant term, right?
01:12:48.040 --> 01:12:49.960
A lowest order term.
01:12:49.960 --> 01:12:52.380
So obviously this is a multiple
of x, this is a
01:12:52.380 --> 01:12:53.630
multiple of x.
01:12:56.580 --> 01:12:58.060
Multiples of x plus 1.
01:12:58.060 --> 01:13:00.590
How do you recognize those
over the binary field?
01:13:05.450 --> 01:13:12.380
This is basically the polynomial
whose root is 1.
01:13:12.380 --> 01:13:14.050
That means the mod-2 sum of the
01:13:14.050 --> 01:13:16.420
coefficients is equal to 0.
01:13:16.420 --> 01:13:20.200
So any polynomial that has an
even number of non-zero
01:13:20.200 --> 01:13:23.500
coefficients is divisible
by x plus 1.
01:13:23.500 --> 01:13:24.920
Did you all get that?
01:13:24.920 --> 01:13:29.270
If not, try that at home.
01:13:29.270 --> 01:13:31.510
That's the easy way to recognize
whether something is
01:13:31.510 --> 01:13:32.750
a multiple of x plus 1.
01:13:32.750 --> 01:13:36.640
It has an even number of
non-zero coefficients.
01:13:36.640 --> 01:13:38.530
So this is a multiple
of x plus 1.
01:13:38.530 --> 01:13:40.380
We wrote it out explicitly.
01:13:40.380 --> 01:13:46.600
It's x plus 1 squared in F2
of x, and this is not.
01:13:46.600 --> 01:13:58.410
So there is only one prime
polynomial over F2 of x
01:13:58.410 --> 01:13:59.690
that's degree 2.
01:13:59.690 --> 01:14:03.480
So this is our only possible
choice if we want to construct
01:14:03.480 --> 01:14:05.985
a finite field with
four elements,
01:14:05.985 --> 01:14:07.235
due to the two elements.
01:14:09.380 --> 01:14:09.870
OK.
01:14:09.870 --> 01:14:13.200
So that might begin
to get us scared.
01:14:13.200 --> 01:14:15.080
Is it possible that
there are no prime
01:14:15.080 --> 01:14:18.260
polynomials of degree three?
01:14:18.260 --> 01:14:20.390
Well, we had two
here, one here.
01:14:20.390 --> 01:14:21.875
Is it going down?
01:14:21.875 --> 01:14:25.710
Let's write down the polynomials
of degree 3.
01:14:25.710 --> 01:14:33.400
x third, x third plus 1, x to
the three plus x, to the three
01:14:33.400 --> 01:14:38.225
plus x plus 1, x to the three
plus x squared, x to the three
01:14:38.225 --> 01:14:44.760
plus x squared plus 1, x to the
three plus x squared plus
01:14:44.760 --> 01:14:50.620
x, x to the three plus x
squared plus x plus 1.
01:14:50.620 --> 01:14:52.130
Eight of them.
01:14:52.130 --> 01:14:54.370
So that's working
in our favor.
01:14:54.370 --> 01:14:57.810
There's double the number
of polynomials as we
01:14:57.810 --> 01:15:01.300
go up one in degree.
01:15:01.300 --> 01:15:03.430
And again, let's go
through the sieve.
01:15:03.430 --> 01:15:11.990
Which are multiples of x,
non-zero constant term?
01:15:11.990 --> 01:15:16.960
Which are multiples of x plus
1, even number of non-zero
01:15:16.960 --> 01:15:18.210
coefficients?
01:15:21.450 --> 01:15:23.950
If you doubt that,
write it out.
01:15:23.950 --> 01:15:27.180
Which are multiples of x
squared plus x plus 1?
01:15:27.180 --> 01:15:29.690
Well, they're going to have to
be x squared plus x plus 1
01:15:29.690 --> 01:15:33.600
times x plus 1 or times x, so
we've already got them.
01:15:33.600 --> 01:15:36.370
If we take this times itself,
we're going to get a
01:15:36.370 --> 01:15:38.260
polynomial of degree 4.
01:15:38.260 --> 01:15:41.000
Not going to be on this list.
01:15:41.000 --> 01:15:43.020
So we have to multiply
this by these.
01:15:43.020 --> 01:15:44.160
We've already got those.
01:15:44.160 --> 01:15:46.530
So whew.
01:15:46.530 --> 01:15:47.520
We found two.
01:15:47.520 --> 01:15:50.190
We could use either of these
to construct a finite field
01:15:50.190 --> 01:15:51.440
with eight elements.
01:15:56.040 --> 01:15:56.810
And so forth.
01:15:56.810 --> 01:15:59.780
And it turns out as you go up
to higher degrees, that the
01:15:59.780 --> 01:16:02.760
number now increases very
nicely, and there's no problem
01:16:02.760 --> 01:16:05.420
finding a prime polynomial
of each degree.
01:16:05.420 --> 01:16:09.270
And in fact, you could be cute
about it and try to find one
01:16:09.270 --> 01:16:13.860
with only three non-zero terms,
or has some other nice
01:16:13.860 --> 01:16:17.760
property that makes it easy
to calculate with.
01:16:17.760 --> 01:16:20.480
And so forth.
01:16:20.480 --> 01:16:24.240
So do you understand
the sieve method?
01:16:24.240 --> 01:16:28.410
If you do, then I believe
you could find a --
01:16:28.410 --> 01:16:34.070
suppose you want to construct
a field with 64 elements.
01:16:34.070 --> 01:16:36.510
What do you need?
01:16:36.510 --> 01:16:38.900
64 is 2 to the sixth, so
you're going to need a
01:16:38.900 --> 01:16:42.260
polynomial with p equals
2 and m equals 6.
01:16:42.260 --> 01:16:47.540
You're going to need a binary
polynomial of degree 6 that is
01:16:47.540 --> 01:16:49.300
prime, irreducible.
01:16:52.230 --> 01:16:54.770
And again, in a few minutes
by going through the sieve
01:16:54.770 --> 01:16:57.460
process, you can quickly find
one, or you could look it up
01:16:57.460 --> 01:17:01.260
in Google, or in Peterson's
book, or any algebraic coding
01:17:01.260 --> 01:17:05.490
theory book, or probably
a lot of other places.
01:17:05.490 --> 01:17:07.810
So this is a practical solution
for the problem for
01:17:07.810 --> 01:17:09.120
any given p and m.
01:17:09.120 --> 01:17:13.430
Of course, it hardly proves
that there's one of every
01:17:13.430 --> 01:17:15.730
degree, because they're
kind of an
01:17:15.730 --> 01:17:18.190
infinite number of degrees.
01:17:18.190 --> 01:17:19.780
Yeah?
01:17:19.780 --> 01:17:24.566
AUDIENCE: So there's two order
3 polynomials that are prime.
01:17:24.566 --> 01:17:27.372
Will it generate two separate
fields, or are they going to
01:17:27.372 --> 01:17:29.450
be isomorphic to each other?
01:17:29.450 --> 01:17:32.020
PROFESSOR: Great question.
01:17:32.020 --> 01:17:35.820
All fields with p to the m
elements are isomorphic to
01:17:35.820 --> 01:17:37.730
each other.
01:17:37.730 --> 01:17:39.240
That's proved in the notes.
01:17:39.240 --> 01:17:40.490
I'm not going to
do it in class.
01:17:42.810 --> 01:17:45.410
And the other thing that's
proved in the notes is the
01:17:45.410 --> 01:17:48.200
analog to Zp.
01:17:48.200 --> 01:17:52.790
There are no other finite fields
with other than p to
01:17:52.790 --> 01:17:55.920
the m number of elements.
01:17:55.920 --> 01:17:56.230
OK?
01:17:56.230 --> 01:17:58.520
So this is it.
01:17:58.520 --> 01:18:01.870
Now, if you explicitly write
out these two, and
01:18:01.870 --> 01:18:04.570
write out their --
01:18:04.570 --> 01:18:07.960
well, the addition tables are
always look the same, because
01:18:07.960 --> 01:18:11.630
it's always just binary
three-tuples, in this case.
01:18:11.630 --> 01:18:15.170
But multiplication tables are
going to look different.
01:18:15.170 --> 01:18:16.483
But there is some isomorphism.
01:18:19.720 --> 01:18:25.820
x and 1 may be equivalent to x
squared plus 1 on the other
01:18:25.820 --> 01:18:27.490
one, or something.
01:18:27.490 --> 01:18:31.040
But if you go through that
isomorphism, you'll find that
01:18:31.040 --> 01:18:33.270
the field tables are the same.
01:18:36.220 --> 01:18:38.790
Actually, I guess I'm going to
prove that, because I'm going
01:18:38.790 --> 01:18:41.860
to prove that the
multiplication
01:18:41.860 --> 01:18:45.390
table is always cyclic.
01:18:45.390 --> 01:18:50.530
That the group to which it's
isomorphic is z mod p
01:18:50.530 --> 01:18:53.590
to the m minus 1.
01:18:53.590 --> 01:18:56.260
Just as it was to Z3 here.
01:18:56.260 --> 01:18:59.080
And I guess that's sufficient
with the addition table
01:18:59.080 --> 01:19:00.310
isomorphism.
01:19:00.310 --> 01:19:02.140
I guess you have to prove
they're equivalent.
01:19:02.140 --> 01:19:04.140
You saw it in a different
way in the notes.
01:19:04.140 --> 01:19:10.360
But you're going to see that all
the multiplication group
01:19:10.360 --> 01:19:14.900
is always a cyclic group, with
p to the m minus 1 elements,
01:19:14.900 --> 01:19:17.530
and that goes a long way towards
suggesting that these
01:19:17.530 --> 01:19:20.150
are always going to be
isomorphic to each other.
01:19:20.150 --> 01:19:20.846
Yeah?
01:19:20.846 --> 01:19:23.230
AUDIENCE: Identify
roots, right?
01:19:23.230 --> 01:19:24.480
PROFESSOR: Identify roots?
01:19:27.460 --> 01:19:29.340
If I understand what you're
saying, that's basically the
01:19:29.340 --> 01:19:31.010
way it's done in the notes.
01:19:31.010 --> 01:19:37.230
You first show there's always
going to be some primitive
01:19:37.230 --> 01:19:41.330
element that generates
the cyclic group.
01:19:41.330 --> 01:19:45.070
Some single generator such that
alpha alpha squared, so
01:19:45.070 --> 01:19:50.170
forth, is the entire
non-zero set.
01:19:50.170 --> 01:19:54.600
You're going to show that
alpha has some minimal
01:19:54.600 --> 01:19:59.950
polynomial, and the set of all
linear combinations of powers
01:19:59.950 --> 01:20:04.350
of alpha is basically equal
to the whole field.
01:20:04.350 --> 01:20:10.970
And so this allows you to
establish the isomorphism.
01:20:10.970 --> 01:20:15.250
and I think that's what
you're suggesting.
01:20:15.250 --> 01:20:17.950
So you're well equipped to read
the notes, but I'm not
01:20:17.950 --> 01:20:19.200
going to do that in class.
01:20:23.520 --> 01:20:24.370
Yeah.
01:20:24.370 --> 01:20:28.340
I'm very interested
in your questions.
01:20:28.340 --> 01:20:32.320
Please ask more, as many
questions as you like.
01:20:32.320 --> 01:20:35.230
What I'm getting from it is,
you know, you all come from
01:20:35.230 --> 01:20:36.170
different backgrounds.
01:20:36.170 --> 01:20:40.350
Some of you have seen this in
perhaps a math context, or
01:20:40.350 --> 01:20:43.360
some other context, or you've
seen parts of it, or some of
01:20:43.360 --> 01:20:45.650
the words are familiar.
01:20:45.650 --> 01:20:49.050
And of course, there are many
different ways to present this
01:20:49.050 --> 01:20:52.320
and to make the proofs
and so forth.
01:20:52.320 --> 01:20:56.770
So I'm trying to pick a line
that works for the particular
01:20:56.770 --> 01:20:58.030
results that I want to get to.
01:20:58.030 --> 01:21:00.560
I don't think you would do
anything much different if you
01:21:00.560 --> 01:21:03.950
wanted to develop
finite fields.
01:21:03.950 --> 01:21:09.275
But the class has a very
different set of backgrounds,
01:21:09.275 --> 01:21:10.910
and I'm trying to reach
all of you.
01:21:14.940 --> 01:21:16.540
Don't be alarmed if
you've never seen
01:21:16.540 --> 01:21:17.930
anything like this before.
01:21:17.930 --> 01:21:21.560
You're not way behind everybody
else, either.
01:21:21.560 --> 01:21:26.310
I think it's pretty easy
to understand.
01:21:26.310 --> 01:21:29.900
Maybe it would take you a
couple hours longer than
01:21:29.900 --> 01:21:31.470
somebody who has more
background, but
01:21:31.470 --> 01:21:32.720
not more than that.
01:21:35.990 --> 01:21:37.830
OK.
01:21:37.830 --> 01:21:42.360
So that's how we construct
finite fields.
01:21:42.360 --> 01:21:45.940
You have an example of it.
01:21:45.940 --> 01:21:46.440
Goodness.
01:21:46.440 --> 01:21:48.740
Is it really 11 o'clock?
01:21:48.740 --> 01:21:49.520
OK.
01:21:49.520 --> 01:21:51.390
So I'm not even --
01:21:51.390 --> 01:21:56.010
so I've simply asserted,
but not proved.
01:21:56.010 --> 01:21:59.020
You saw in this case that the
multiplicative group was a
01:21:59.020 --> 01:22:00.770
cyclic group.
01:22:00.770 --> 01:22:04.530
And we could always, for
multiplication, represent
01:22:04.530 --> 01:22:06.800
grouped elements by
this log table.
01:22:06.800 --> 01:22:08.590
So I'd hoped to prove
that in class.
01:22:08.590 --> 01:22:11.860
I'm not going to be able
to prove that in class.
01:22:11.860 --> 01:22:16.400
And I'll simply ask you
to read that, too.
01:22:16.400 --> 01:22:20.370
This is important for working
with finite fields, because it
01:22:20.370 --> 01:22:23.500
is the way, probably the
preferred way, to implement
01:22:23.500 --> 01:22:24.750
multiplication.
01:22:26.910 --> 01:22:31.180
So you ought to be thinking of
how you would program up a
01:22:31.180 --> 01:22:34.810
finite field multiplier.
01:22:34.810 --> 01:22:36.700
One way is polynomial
multiplication.
01:22:36.700 --> 01:22:41.200
The other way is just use the
fact that the multiplicative
01:22:41.200 --> 01:22:43.160
group is cyclic.
01:22:43.160 --> 01:22:44.580
And then it's easy.
01:22:44.580 --> 01:22:50.810
Just add exponents and reduce
modulo q minus 1.
01:22:50.810 --> 01:22:51.120
OK.
01:22:51.120 --> 01:22:54.700
I'm sorry not to have had a
chance to go over that.
01:22:54.700 --> 01:22:57.510
Next time, Ralf Koetter will
start to get into chapter
01:22:57.510 --> 01:22:58.760
eight, Reed-Solomon codes.