s with rational
can be difficult and Gauss's Lemma is a helpful tool for this problem.
It is used implicitly in
Theorem A polynomial with integer coefficients that is
irreducible in Z[x] is irreducible in
Here's an example to illustrate the theorem. Consider
f(x)=x3 - x2 - x - 1.
We are going to show that this polynomial is irreducible
in Q[x]. Suppose that f(x)
factorizes nontrivially in Z[x]. Since the highest
common factor of its coefficients is 1 such a factorization has to
be as g(x)h(x) where g and h both have degree <3.
x3 - x2 - x - 1 = (ax + b)(cx2 + dx + e).
Equating coefficients of
x3 and x0
we get ac=1 and be=-1 so we see that
a and b both have to be 1 or -1. It follows that
f(x) has 1 or -1 as a zero but f(1) and f(-1)
are both nonzero.
This contradiction shows that f(x) is irreducible in
Z[x] and hence in Q[x]
Another application of the theorem is Eisenstein's irreducibility criterion.
We will prove a more general result than the theorem. First we need some
definitions. We work with R[x] the polynomial ring
over a unique factorization domain (UFD). Let f(x) in
R[x], then we can find a highest common factor
c, say, of the coefficients of f(x). Thus
f=cg where g is in R[x] and is such that
the highest common factor of its coefficient is 1. Such a polynomial
is called primitive. The constant c is called the
content of f. Note that c is only unique
up to associates (so should really be called a content).
Gauss's Lemma Let R be a UFD and let
f,g in R[x] be primitive. Then so
Write fg=ch with h primitive and c the content
of h. Suppose that p is a prime in R that divides
c. Since f is primitive p cannot divide all its
coefficients so choose fi to be the first one
not divisible by p. Likewise choose gj to
be the first coefficient of g not divisible by p.
Now consider the coefficient of xi+j in
fg=ch. We have
chi+j = (f0gi+j + ... + figj +...+ fi+jg0).
Now because of the way fi and gj
were chosen all the terms on the right hand side except
figj are divisible by p. But p
divides the right hand side by assumption. So we deduce that
p|figj. This contradicts the primeness of
p. We deduce that there is no such p and c is
a unit. In other words, fg is primitive.
Let R be a UFD and let f,g in R[x]. Then
Proof: Write f=cf1 and g=dg1
where c,d are, respectively, the contents of f,g.
Thus, fg=cdf1g1. Since
f1g1 is primitive, by the Gauss Lemma,
we are done.
Now for our UFD R let k= be the field of fractions
of R. For example, if R=Z then k=Q.
The next result has the very first theorem we stated as the special case
Theorem Let R be a UFD with field of fractions
Let f be a polynomial in R[x].
Then f is primitive and irreducible in k[x]
iff it is irreducible in R[x].
Proof: One direction is trivial. If f is primitive
and irreducible in k[x] then how could it factorize
over R[x]? By primitivity a factorization ch
where c is a nonunit of R is impossible. By irreduciblity
in k[x] such a factorization where both g,h have
smaller degree than f is impossible.
So let's prove the other implications. Suppose that f is irreducible
in R[x] (hence clearly primitive). Suppose that
f=gh where g,h in k[x] have degree <
deg f. clearing denominators we can rewrite this as
a,b are in R and g1,h1
are in R[x]. Taking contents and using the corollary
we get that
Subsituting in (*) we have a nontrivial factorization of f in
R[x], a contradiction.