## Gauss's Lemma

Gauss's Lemma
Theorem: Let
$f \in \mathbb{Z}[x]$
. Then
$f$
is irreducible over
$\mathbb{Z}[x]$
if and only if
$f$
is irreducible over
$\mathbb{Q}[x]$
.
(In other words, Let
$f(x)$
be a polynomial with integer coefficients. If
$f(x)$
has no factors with integer coefficients, then
$f(x)$
has no factors with rational coefficients.)
Proof: Let
$f(x) = g(x)h(x)$
be a factorization of
$f$
into polynomials with rational coefficients. Then for some rational
$a$
the polynomial
$a g(x)$
has integer coefficients with no common factor. Similary we can find a rational
$b$
so that
$b h(x)$
has the same properties. (Take the lcm of the denominators of the coefficients in each case, and then divide by any common factors.)
Suppose a prime
$p$
divides
$a b$
. Since
$a b f(x) = (a g(x))(b h(x))$

becomes
$0 = (a g(x)) (b h(x))$
modulo
$p$
, we see
$a g(x)$
or
$b h(x)$
is the zero polynomial modulo
$p$
. (If not, then let the term of highest degree in
$a g(x)$
be
$m x^r$
, and the term of highest degree in
$b h(x)$
be
$n x^s$
. Then the product contains the term
$m n x^{r+s} \ne 0 \pmod {p}$
In other words,
$p$
divides each coefficient of
$a g(x)$
or
$b h(x)$
$a b = 1$
and we have a factorization over the integers.
Example: Let
$p$
be a prime. Consider the polynomial
$f(x) = 1 + x + ... + x^{p-1} .$

We cannot yet apply the criterion, so make the variable subsitution
$x = y + 1$
. Then we have
$g(y) = 1 + (y+1) + ... + (y+1)^{p-1} .$

Note
$f(x)$
is irreducible if and only if
$g(y)$
is irreducible.
The coefficient of
$y^k$
in
$g(y)$
is
$\sum_{m=k}^{p-1} {\binom{p-1}{k}} = {\binom{p}{k+1}} .$

The last equality can be shown via repeated applications of Pascal’s identity:
${\binom{n+1}{k}} = {\binom{n}{k}} + {\binom{n}{k-1}} .$

Alternatively, use the fact
$g(y) = \frac{(y+1)^p - 1}{(y+1) - 1}$

Thus
$p$
divides each coefficient except the leading coefficient, and
$p^2$
does not divide the constant term
$p$
, hence
$f(x)$
is irreducible over the rationals.