I mentioned earlier that most things hold for arbitrary fields. However, I have not bothered to give
any examples of other fields. This is the point of this section. It also turns out that showing
the algebraic numbers are a field can be understood using vector space concepts and
it gives a very convincing application of the abstract theory presented earlier in this
chapter.
Here I will give some basic algebra relating to polynomials. This is interesting for its own sake
but also provides the basis for constructing many different kinds of fields. The first is the
Euclidean algorithm for polynomials.
Definition 7.3.1A polynomialis an expression of the form p
(λ)
= ∑_{k=0}^{n}a_{k}λ^{k}where asusual λ^{0}is defined to equal 1. Two polynomialsare said to be equal if their correspondingcoefficients are the same. Thus, in particular, p
(λ )
= 0 means each of the a_{k} = 0. Anelement of the field λ is said to be a root of thepolynomial if p
(λ)
= 0 in the sense that whenyou plug in λ into the formula and do the indicated operations, you get 0. The degree of anonzero polynomial is the highest exponent appearing on λ. The degree of the zero polynomialp
(λ)
= 0 is not defined. A polynomial of degree n ismonic if the coefficient of λ^{n}is 1. Inany case, this coefficient is called the leading coefficient.
Example 7.3.2Consider the polynomial p
(λ)
= λ^{2}+λ where the coefficients are in ℤ_{2}. Isthis polynomial equal to 0? Not according to the above definition, because its coefficients arenot all equal to 0. However, p
(1)
= p
(0)
= 0 so it sends every element of ℤ_{2}to 0. Note thedistinction between saying it sends everything in the field to 0 with having the polynomial bethe zero polynomial.
The fundamental result is the division theorem for polynomials. It is Lemma 1.10.10 on Page
59. We state it here for convenience.
Lemma 7.3.3Let f
(λ)
and g
(λ)
≠0 be polynomials.Then there exists a polynomial, q
(λ)
suchthat
f (λ) = q(λ)g(λ)+ r(λ)
where the degree of r
(λ)
is less than the degree of g
(λ)
or r
(λ)
= 0. These polynomials q
(λ)
andr
(λ)
are unique.
In what follows, the coefficients of polynomials are in F, a field of scalars which is completely
arbitrary. Think ℝ if you need an example.
Definition 7.3.4A polynomial f is said to dividea polynomial g if g
(λ)
= f
(λ)
r
(λ)
forsome polynomial r
(λ)
. Let
{ϕi(λ)}
be a finite set of polynomials. The greatest commondivisorwill be the monic polynomial q
(λ)
such that q
(λ )
divides each ϕ_{i}
(λ)
and if p
(λ)
divides each ϕ_{i}
(λ)
, then p
(λ)
divides q
(λ)
. The finite set of polynomials
{ϕi}
is said tobe relatively primeif their greatest common divisor is 1. A polynomial f
(λ)
is irreducibleifthere is no polynomial with coefficients in F which divides it except nonzero scalar multiplesof f
(λ)
and constants. In other words, it is not possible to write f
(λ)
= a
(λ)
b
(λ)
whereeach of a
(λ)
,b
(λ)
have degree less than the degree of f
(λ)
.
Proposition 7.3.5The greatest common divisor is unique.
Proof: Suppose both q
(λ)
and q^{′}
(λ)
work. Then q
(λ)
divides q^{′}
(λ)
and the other way around
and so
q′(λ) = q(λ)l(λ),q(λ) = l′(λ) q′(λ)
Therefore, the two must have the same degree. Hence l^{′}
(λ )
,l
(λ)
are both constants. However, this
constant must be 1 because both q
(λ)
and q^{′}
(λ)
are monic. ■
Theorem 7.3.6Let ψ
(λ)
be the greatest commondivisor of
{ϕi(λ)}
, not all of which are zeropolynomials. Then there exist polynomials r_{i}
(λ)
such that
p
ψ(λ) = ∑ ri(λ)ϕi(λ).
i=1
Furthermore, ψ
(λ)
is the monic polynomial of smallest degree which can be written in the aboveform.
Proof: Let S denote the set of monic polynomials which are of the form
∑p
ri(λ)ϕi(λ)
i=1
where r_{i}
(λ )
is a polynomial. Then S≠∅ because some ϕ_{i}
(λ)
≠0. Then let the r_{i} be
chosen such that the degree of the expression ∑_{i=1}^{p}r_{i}
(λ)
ϕ_{i}
(λ)
is as small as possible.
Letting ψ
(λ)
equal this sum, it remains to verify it is the greatest common divisor.
First, does it divide each ϕ_{i}
for a suitable a ∈ F. This is one of the polynomials in S. Therefore, ψ
(λ)
does not have the
smallest degree after all because the degree of ψ_{1}
(λ)
is smaller. This is a contradiction. Therefore,
ψ
(λ)
divides ϕ_{1}
(λ)
. Similarly it divides all the other ϕ_{i}
(λ)
.
If p
(λ)
divides all the ϕ_{i}
(λ)
, then it divides ψ
(λ)
because of the formula for ψ
(λ)
which
equals ∑_{i=1}^{p}r_{i}
(λ )
ϕ_{i}
(λ )
. ■
Lemma 7.3.7Suppose ϕ
(λ )
and ψ
(λ)
are monic polynomials which areirreducible andnot equal. Then they are relatively prime.
Proof:Suppose η
(λ)
is a nonconstant polynomial. If η
(λ)
divides ϕ
(λ)
, then since ϕ
(λ)
is
irreducible, η
(λ)
equals aϕ
(λ)
for some a ∈ F. If η
(λ)
divides ψ
(λ)
then it must be of the form
bψ
(λ )
for some b ∈ F and so it follows
a-
ψ (λ) = b ϕ(λ)
but both ψ
(λ)
and ϕ
(λ)
are monic polynomials which implies a = b and so ψ
(λ)
= ϕ
(λ )
. This is
assumed not to happen. It follows the only polynomials which divide both ψ
(λ)
and ϕ
(λ )
are
constants and so the two polynomials are relatively prime. Thus a polynomial which divides them
both must be a constant, and if it is monic, then it must be 1. Thus 1 is the greatest common
divisor. ■
Lemma 7.3.8Let ψ
(λ)
be anirreducible monic polynomial not equal to 1 which divides
∏p
ϕi(λ)ki ,ki a positive integer,
i=1
where each ϕ_{i}
(λ)
is an irreducible monic polynomial not equal to 1. Then ψ
(λ)
equals someϕ_{i}
(λ)
.
Proof :Say ψ
(λ)
l
(λ)
= ∏_{i=1}^{p}ϕ_{i}
(λ )
^{ki}. Suppose ψ
(λ)
≠ϕ_{i}
(λ)
for all i. Then by Lemma
7.3.7, there exist polynomials m_{i}
p p
ψ(λ)n (λ) ≡ ψ(λ)l(λ)∏ n (λ)ki = ∏ (n (λ)ϕ (λ))ki
i=1 i i=1 i i
∏p
= (1 − ψ(λ)mi (λ ))ki = 1+ g (λ )ψ(λ)
i=1
for a polynomial g
(λ)
. Thus
1 = ψ (λ )(n (λ)− g(λ))
which is impossible because ψ
(λ)
is not equal to 1. ■
Now here is a simple lemma about canceling monic polynomials.
Lemma 7.3.9Suppose p
(λ)
is a monic polynomial and q
(λ)
is a polynomial suchthat
p(λ)q(λ) = 0.
Then q
(λ)
= 0. Also if
p (λ )q1 (λ ) = p(λ)q2(λ)
then q_{1}
(λ)
= q_{2}
(λ)
.
Proof: Let
∑k ∑n
p(λ) = pjλj,q(λ) = qiλi,pk = 1.
j=1 i=1
Then the product equals
∑k ∑n i+j
pjqiλ .
j=1i=1
Then look at those terms involving λ^{k+n}. This is p_{k}q_{n}λ^{k+n} and is given to be 0. Since p_{k} = 1, it
follows q_{n} = 0. Thus
k n−1
∑ ∑ pjqiλi+j = 0.
j=1i=1
Then consider the term involving λ^{n−1+k} and conclude that since p_{k} = 1, it follows
q_{n−1} = 0. Continuing this way, each q_{i} = 0. This proves the first part. The second follows
from
p(λ)(q1(λ)− q2(λ)) = 0.■
The following is the analog of the fundamental theorem of arithmetic for polynomials.
Theorem 7.3.10Let f
(λ )
be a nonconstantpolynomial with coefficients in F. Then thereis some a ∈ F such that f
(λ)
= a∏_{i=1}^{n}ϕ_{i}
(λ)
where ϕ_{i}
(λ)
is an irreducible nonconstantmonic polynomial and repeats are allowed. Furthermore, this factorization is unique in thesense that any two of these factorizations have the same nonconstant factors in the product,possibly in different order and the same constant a.
Proof:That such a factorization exists is obvious. If f
(λ)
is irreducible, you are done. Factor
out the leading coefficient. If not, then f
(λ)
= aϕ_{1}
(λ)
ϕ_{2}
(λ)
where these are monic polynomials.
Continue doing this with the ϕ_{i} and eventually arrive at a factorization of the desired
form.
It remains to argue the factorization is unique except for order of the factors. Suppose
∏n m∏
a ϕi(λ ) = b ψi(λ)
i=1 i=1
where the ϕ_{i}
(λ)
and the ψ_{i}
(λ)
are all irreducible monic nonconstant polynomials and a,b ∈ F. If
n > m, then by Lemma 7.3.8, each ψ_{i}
(λ)
equals one of the ϕ_{j}
(λ)
. By the above cancellation
lemma, Lemma 7.3.9, you can cancel all these ψ_{i}
(λ)
with appropriate ϕ_{j}
(λ)
and obtain a
contradiction because the resulting polynomials on either side would have different degrees.
Similarly, it cannot happen that n < m. It follows n = m and the two products consist of the same
polynomials. Then it follows a = b. ■
The following corollary will be well used. This corollary seems rather believable but does
require a proof.
Corollary 7.3.11Let q
(λ)
= ∏_{i=1}^{p}ϕ_{i}
(λ)
^{ki}where the k_{i}are positive integers and the ϕ_{i}
(λ)
areirreducible monic polynomials. Suppose also that p
(λ)
is a monic polynomial which divides q
(λ)
.Then
∏p ri
p(λ) = ϕi(λ)
i=1
where r_{i}is a nonnegative integer no larger than k_{i}.