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 \[m n x^{r+s} \ne 0 \pmod {p}\]
, 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 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.