# Math 75 Notes

**Proposition 1. **The only units in F[x] are the nonzero constants in F.

The proof is simple: Suppose that A is an element of F[x] that has an inverse.
Write

where we assume that
is nonzero in F. If A has an inverse, then AB = 1 for
some polynomial

B ∈ F[x]. Write

where
is nonzero. In the product AB, the term
shows up with a coefficient
of F,

and one can show (homework!) that
is nonzero since both
and
are. So
AB has degree

k + j. But we were supposed to have that AB = 1. Thus we need k + j = deg 1 = 0,
which

means that k = j = 0, and so both A and B are constant polynomials.

The proposition tells us that we will have to do quite a bit more work to get
the fields we

are after: starting with F and forming F[x] didn't give us any invertible
elements.

Our approach will be the same as the approach taken in elementary number theory
to

constructing the systems Z/(m). That approach begins with defining the notion of
congruence

modulo m. In analogy, we make the following definition: We call two polynomials
A,B ∈ F[x]

congruent modulo M if M divides A - B, i.e., if there is a polynomial Q ∈ F[x]
for which

A - B = MQ:

In this case we write

A ≡ B (mod M):

For example, if F = Z/(2), then x^{3} ≡ x^{2} + x + 1 (mod x + 1) in F[x], because x + 1
divides

the difference x^{3} -(x^{2} +x+1) = x^{3} +x^{2}
+x+1 ∈ F[x]. (One can check this with the
familiar

polynomial long division algorithm.)

If F is a field and M is a polynomial in F[x], then congruence modulo M is an
equivalence

relation. In other words, this relation is...

• reflexive: A ≡ A (mod M) for every A,

• symmetric: if A ≡ B (mod M), then B ≡ A (mod M),

• transitive: if A ≡B (mod M) and B ≡ C (mod M), then A ≡ C (mod M).

Because of this, the ring F[x] is partitioned into equivalence classes. We define
F[x]/(M) as

the set of equivalence classes. For each A ∈ F[x], denote the equivalence class
containing A by

[A]; then

[A] := {B ∈ F[x] : B ≡ A (mod M)};

and F[x]/(M) is exactly the set of these classes [A] as A ranges over F[x].

The systems F[x]/(M) will be our candidates for new fields. For this to make any
sense at

all, we need to define what it means to add and multiply two elements of F[x]. We
make the

only definition that really is sensible: we define

[A] + [B] = [A + B]; and [A] · [B] = [AB]:

There is something to check here. The de nition of addition we've given here
basically says `to

add two equivalence classes, pick an element from each, add those two elements,
and then take

the equivalence class of the sum,' and similarly for multiplication. It is maybe
not so obvious

that no matter how the elements are chosen, one always winds up with the same
result. This

is a consequence of the fact that addition and multiplication respect
congruence: if A_{1} ≡ A_{2}

(mod M) and B_{1} ≡ B_{2} (mod M), then

A_{1} + A_{2} ≡ B_{1} + B_{2} (mod M); and A_{1}A_{2}
≡ B_{1}B_{2} (mod M):

We omit the (pretty easy) proofs of these facts here.

Let us do some examples to get a feel for the arithmetic of the systems F[x]/(M).
For the

rest of this lecture we take F = Z/(5) and M = x^{2} + 1 ∈ F[x], and we try to
understand the

system F[x]/(M).

First let us see if we can do some basic arithmetic. Consider the two elements
[x + 1] and

[x + 3] in F[x]/(M). By definition,

[x + 1] + [x + 3] = [(x + 1) + (x + 3)] = [(2x + 4)];

and

[x + 1][x + 3] = [(x + 1)(x + 3)] = [(x^{2} + 4x + 3)]:

Actually the latter answer can be put in a somewhat simpler form. We have

[x^{2} + 4x + 3] = [x^{2} + 1 + 4x + 2] = [x^{2} + 1] + [4x + 2] = [0] + [4x + 2] = [4x +
2]:

(This calculation of reduction should be viewed as being totally analogous to
(say) the calcu-

lation that 3 · 5 = 15 = 8 in the system Z/(7).) It is somewhat cumbersome to keep
having to

write these brackets, and so we omit them when the system we are working with is
understood.

So, for example, in the future we will just write

(x + 1)(x + 3) = 4x + 2

when we are understood to be working in F[x]/(M) for F = Z/(5) and M = x^{2} + 1.

How many elements are in our system F[x]/(M)? To answer this question it would
be good

if we had a way of telling when two elements are different. For example, which
elements of the

list

0, 1, x, 2x + 2, 3x^{3} + 2, x^{5}

represent distinct elements of F[x]/(M)? We can get some insight into this
question if we recall

that M = x^{2} + 1 is zero in F[x]/(M). So

3x^{3} + 2 = 3x · x^{2} + 2 = 3x(-1) + 2 =
-3x + 2 = 2x + 2

and

x^{5} = x · (x^{2})^{2} = x
· (-1)^{2} = x:

In fact, it's not hard to see that by repeatedly applying the rule that x^{2} =
-1, we can show

that every element of F[x]/(M) has the form a + bx, where a and b belong to F =
Z/(5).

An alternative proof (which gets this result in one fell swoop) is to use long
division for

polynomials. That process provides a constructive proof of the following
fundamental result:

**Theorem 3. **Let F be a field. Let A,B be elements of F[x], and suppose B
≠ 0. Then
there

are Q,R ∈ F[x] with

A = BQ + R with R = 0 or degR < degB:

Taking B = M, this result shows immediately that every polynomial is congruent
to some

polynomial which is either zero, or of degree smaller than that of M. In our
case, where

F = Z/(5) and M = x^{2} + 1, that means that very equivalence class is represented
by a

polynomial either zero or of degree < 2. Again, these are precisely the
polynomials of the form

a + bx.

At this point it is tempting to assert that F[x]/(M) has precisely 5
· 5 = 25
elements: there

are 5 choices of a and 5 choices for b. But perhaps there is some repetition
even among the

elements a+bx? It seems hard to see how this could occur, since the rule x^{2} =
-1 that defines

the system F[x]/(M) doesn't seem to imply any such coincidences. We now prove
that all the

elements a + bx, with a,b ∈ F = Z/(5), do define distinct elements of F[x]/(M),
so that our

system does have 25 elements as claimed. For the proof, notice that if

a + bx = a' + b'x

for a, b, a', b' ∈ F, then

(a - a') + (b - b')x = 0:

Expressed in the language of congruence, this asserts that

a - a' + (b - b')x ≡ 0 (mod x^{2} + 1);

i.e., that

(a - a') + (b - b')x = (x^{2} + 1)Q(x)

for some polynomial Q(x) ∈ F[x]. But if Q(x) is nonzero, then the degrees of the
left and right

hand sides will not match up! So it must be that Q(x) = 0, which implies that a-a'+(b-b')x =

0, which in turn implies that a = a' and b = b'.

The proofs we have just outlined work in general. The corresponding result is as follows:

**Theorem 4.** Let F be a finite field. Let M be a nonzero polynomial in F[x]. Then the
number

of elements in F[x]=(M) is , where q is the number of elements of F and d is
the degree of

M. In fact, every element of F[x]/(M) has the form

where and the elements de fined this way are all distinct.