TAOCP 4.6.1 Exercise 8

Let $f(x)$ be a polynomial with integer coefficients.

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Section 4.6.1: Division of Polynomials

Exercise 8. ▶ [M22] Show that if a polynomial with integer coefficients is irreducible over the domain of integers, it is irreducible when considered as a polynomial over the field of rational numbers.

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Solution

Let $f(x)$ be a polynomial with integer coefficients. Assume that $f(x)$ is irreducible over the domain of integers. We shall prove that $f(x)$ is irreducible over the field of rational numbers.

Suppose, on the contrary, that $f(x)$ is reducible over $\mathbf Q$. Then there exist nonconstant polynomials $g(x),h(x)\in\mathbf Q[x]$ such that

$$ f(x)=g(x)h(x). $$

Write

$$ g(x)=\operatorname{cont}(g),\operatorname{pp}(g(x)), \qquad h(x)=\operatorname{cont}(h),\operatorname{pp}(h(x)), $$

where the contents and primitive parts are taken in the sense of equation (3), with coefficients regarded as rational numbers. Since $\mathbf Q$ is a field, we may choose the primitive parts to be monic. Hence $\operatorname{pp}(g(x))$ and $\operatorname{pp}(h(x))$ are primitive polynomials with rational coefficients.

Let $a$ and $b$ be the least positive integers such that

$$ G(x)=a,\operatorname{pp}(g(x)), \qquad H(x)=b,\operatorname{pp}(h(x)) $$

have integer coefficients. By construction, $G(x)$ and $H(x)$ are primitive polynomials over the integers.

Since

$$ f(x) =\operatorname{cont}(g)\operatorname{cont}(h), \operatorname{pp}(g(x))\operatorname{pp}(h(x)), $$

we obtain

$$ ab,f(x) = ab,\operatorname{cont}(g)\operatorname{cont}(h), \operatorname{pp}(g(x))\operatorname{pp}(h(x)) = c,G(x)H(x), $$

where

$$ c=ab,\operatorname{cont}(g)\operatorname{cont}(h)\in\mathbf Q. $$

The polynomial $ab,f(x)$ has integer coefficients. Since $G(x)$ and $H(x)$ are primitive, Lemma G shows that $G(x)H(x)$ is primitive. Therefore every coefficient of $c,G(x)H(x)$ is an integer only if $c$ itself is an integer. Consequently,

$$ ab,f(x)=c,G(x)H(x) $$

is a factorization in $\mathbf Z[x]$.

Because $G(x)$ and $H(x)$ are primitive and nonconstant, neither is a unit in $\mathbf Z[x]$. Thus $ab,f(x)$ is reducible in $\mathbf Z[x]$.

Since $ab$ is a nonzero integer, it is the content of $ab,f(x)$ and $f(x)$ is primitive up to multiplication by a unit. By Lemma H, any factorization of $ab,f(x)$ into nonunit factors yields a factorization of $f(x)$ into nonunit factors in $\mathbf Z[x]$. Hence $f(x)$ is reducible over the integers, contradicting the hypothesis.

Therefore no factorization of $f(x)$ into nonconstant factors over $\mathbf Q$ can exist. Hence $f(x)$ is irreducible over $\mathbf Q$.

This completes the proof.

∎

Notes

The converse is immediate: if a polynomial is reducible over the integers, the same factorization is valid over the rationals. Therefore a polynomial with integer coefficients is irreducible over $\mathbf Z$ if and only if it is irreducible over $\mathbf Q$. This result is commonly called Gauss's theorem on irreducibility.