## Gauss's Lemma

Gauss's LemmaTheorem: 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 \[m n x^{r+s} \ne 0 \pmod {p}\]

, a contradiction.)In other words,

\[p\]

divides each coefficient of \[a g(x)\]

or \[b h(x)\]

, a contradiction. Hence \[a b = 1\]

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 Pascalâ€™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.