$f(x)$ is irreducible and primitive over $\Bbb Q$, then $f(x)$ is irreducible over $\Bbb Z$

Question

show that $f(x) (\in \Bbb Z[x])$ is irreducible and primitive over $\Bbb Q$, then $f(x)$ is irreducible over $\Bbb Z$.

How pf it?

I tried it.

MY pf)

Suppose that $f(x)$ is reducible over $\Bbb Z$.

then, $f(x)=Q(x)L(x)$ for some $Q(x),L(x) \in \Bbb Z[x]$ .

since $f(x)$ is primitive, $c(f(x))=1$.

Let $c(Q(x))= a$, $c(L(x))=b$. ($a,b$ is constant)

then $f(x)=aQ'(x)bL'(x)$. (Note $Q(x)=aQ'(x)$ , L(x)=bL'(x))

=> ${1\over ab}f(x)=Q'(x)L'(x)$.

by gauss theorem,

=> ${1\over ab}=1$ (since the product of primitive polynomial is also primitive)

so, $ab=1$ => $a=b=1$ or $a=b=-1$.

$f(x)=Q(x)L(x)$ is reducible over $\Bbb Z$ and $Q(x),L(x)$ are primitive.

it means that $f(x)$ is also reducible over $\Bbb Q$.

-contradiction-

is it all right? I know my pf is insufficient....plz edit! thanks :D

Answer 1

Since $f(x)$ is irreducible over $\mathbb{Q}$, it must be irreducible over $\mathbb{Z}$. Recall the idea of field extensions and this should be immediately clear.

Answer 2

Suppose $f(x)$ is not irreducible in $\Bbb Q[x]$, but $f(x)$ is irreducible in $\Bbb Z[x]$. We know that a irreducible polynomial in $\Bbb Z[x]$ is primitive.

Again $\Bbb Z$ is a UFD and $f(x)$ is non-constant primitive polynomial in $\Bbb Z[x]$. Then $f(x)$ is irreducible in $\Bbb R[x]$ if $f(x)$ is irreducible in $\Bbb Q(\Bbb R)[x]$.

So $f(x)$ is irreducible in $\Bbb Q[x]$. A contradiction. So $f(x)$ is not irreducible in $\Bbb Z[x]$.