Number Theory - Eisenstein's Irreducibility Criterion

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 $f \in ℤ [x]$ . Then $f$ is irreducible over $ℤ [x]$ if and only if $f$ is irreducible over $ℚ [x]$ .

(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 $m n x^{r + s} \neq 0 (mod p)$ , 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.