What Are the Conditions for Eisenstein’s Criterion?

Eisenstein’s criterion is a quick test for proving that a polynomial cannot be factored into lower-degree polynomials over the rational numbers. The key is to find one prime that satisfies three divisibility conditions on the coefficients. The leading coefficient is deliberately excluded from the first condition.

The conditions

Let

$$f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$$

be a polynomial in $\mathbb{Z}[x]$, with $n\geq 1$. If there is a prime number $p$ such that

• $p$ does not divide $a_n$, the leading coefficient;
• $p$ divides each of $a_{n-1},a_{n-2},\ldots,a_0$;
• $p^2$ does not divide $a_0$, the constant term;

then $f$ is irreducible over $\mathbb{Q}$.

The last condition means that the constant term is divisible by $p$, but not by $p^2$. For instance, if $p=5$, a constant term of $10$ passes this test, while $25$ does not.

A worked example

Consider

$$f(x)=x^4+10x^3+5x^2+15x+5.$$

Take $p=5$. The leading coefficient is $1$, which is not divisible by $5$. Each other coefficient—$10$, $5$, $15$, and $5$—is divisible by $5$. Finally, the constant term is $5$, which is not divisible by $25$.

All three conditions hold, so $f(x)$ is irreducible over $\mathbb{Q}$. The criterion proves this without requiring you to try possible factorizations.

Why the test works

Suppose, for a contradiction, that such a polynomial factors over $\mathbb{Q}$. Let $c$ be the positive greatest common divisor of its coefficients, and set $F=f/c$. Since $p$ does not divide the leading coefficient of $f$, it does not divide $c. Dividing by$c$, which is coprime to$p$, preserves the three divisibility conditions, so the primitive polynomial$F$still satisfies Eisenstein’s conditions for$p$. By Gauss’s lemma, if$F$factors over$\mathbb{Q}$, it factors into positive-degree polynomials in$\mathbb{Z}[x]$. Write$F=gh$.

Reduce the coefficients modulo $p$. All nonleading coefficients of $F$ become zero, while its leading coefficient remains nonzero. Thus, modulo $p$, the polynomial is just its leading coefficient times $x^n$. Since the leading coefficient of $F$ is not divisible by $p$, neither factor loses its degree when reduced modulo $p$. The only way two positive-degree polynomials over a field can multiply to a nonzero constant times $x^n$ is for each to be a constant times a positive power of $x$. Therefore both factors have constant term divisible by $p$ before reduction. Their product has constant term divisible by $p^2$, contradicting the criterion’s third condition for $F$. Hence no such factorization exists. Since $f=cF$ and $c$ is a nonzero rational number, $f$ is irreducible over $\mathbb{Q}$ as well.

What “irreducible” means here

The conclusion is irreducibility over $\mathbb{Q}$: there is no factorization into two positive-degree polynomials with rational coefficients. For a primitive polynomial with integer coefficients, this is equivalent to irreducibility over $\mathbb{Z}$, by Gauss’s lemma. The criterion does not say that the polynomial has no roots or that it cannot factor over other fields, such as the complex numbers.

Common mistakes

• Including the leading coefficient in the divisibility requirement. It must be excluded: the prime must not divide the leading coefficient.
• Checking only that the constant term is divisible by $p$. It must not be divisible by $p^2$.
• Treating the test as necessary. Passing the test proves irreducibility, but failing to find a suitable prime does not prove reducibility. Some irreducible polynomials do not satisfy the criterion in their given form.
• Forgetting the coefficient ring. The usual statement applies to integer-coefficient polynomials and concludes irreducibility over $\mathbb{Q}$. A polynomial with rational coefficients can be cleared of denominators and its integer coefficients then tested.

When the coefficients do not pass directly

A change of variable can sometimes make the criterion applicable. For example, $x^4+1$ does not pass Eisenstein’s test in its original form. Substitute $x+1$ for $x$:

$$(x+1)^4+1=x^4+4x^3+6x^2+4x+2.$$

For this new polynomial, use $p=2$: every nonleading coefficient is divisible by $2$, the leading coefficient is not, and $2^2$ does not divide the constant term $2$. It is therefore irreducible over $\mathbb{Q}$. Since replacing $x$ by $x+1$ is reversible (replace $x$ by $x-1$), a factorization of $x^4+1$ would give a factorization of the transformed polynomial, so $x^4+1$ is irreducible as well.