Eisenstein’s Irreducibility Criterion

Theorem: Let

\[ f(x) = a_0 + a_1 x + ... + a_n x^n \]

be a polynomial with integer coefficients. Suppose a prime $p$ divides each of $a_0, a_1, ..., a_{n-1}$ (every coefficient except the leading coefficient), and that $p^2$ does not divide $a_0$. Then $p(x)$ has no factors with integer coefficients.

Proof: Suppose $f = g h$ for polynomials $g, h$ with integer coefficients. Look at this factorization modulo $p$: we get $f(x) = a_n x^n$, so $g(x) = b_d x^d$, $h(x) = c_e x^e$ for some constants $b_d, c_e$ and for some integers $d, e$ with $d e = n$. This implies the constant term of $g(x)$ is a multiple of $p$, and similarly for the constant term of $h(x)$, hence $p^2$ divides the constant term of $f(x)$, a contradiction.∎

Gauss' Lemma

We usually combine Eisenstein’s criterion with the following theorem for a stronger statement. (The name "Gauss' Lemma" has been given to several results in different areas of mathematics, including the following.)

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.