If $f(x)\in \mathbb{Z}[x]$ is irreducible (over $\mathbb{Q}$), is it always possible to find $a$ and $b$ in $\mathbb{Q}$ with $f(ax+b)$ Eisenstein?

Question

My initial thought is no, simply because it seems too easy if it is true.

The simplest example of a nontrivial irreducible polynomial I could think of was $f(x)=x^2+1$. Unfortunately, $f(x+1)$ is Eisenstein at $p=2$.

Then I considered $f(x)=x^2+4$, skipping $2$ and $3$ as those are Eisenstein, and I believe I have a proof.

If $f(ax+b)=a^2x^2+2abx+(b^2+4)$ is Eisenstein at $p$ (where $a,b \in \mathbb{Z}$), then $p\nmid a^2$ so $p\nmid a$ and thus (as $p \mid 2ab$), we must have $p \mid 2b$.

If $p=2$, then $b$ must be even as $2\mid (b^2+4)$. But then $b^2$ is divisble by $4$ so $2^2 \mid b^2+4$, which contradicts Eisenstein's criterion.

If $p$ is odd, then $p\mid b$, so clearly we have $p\mid b^2$. As $p\mid b^2+4$, we must have $p\mid 4$, which is a contradiction.

So no $a$ and $b$ exist.

My question is: is it sufficient to show that no integer $a$ and $b$ exist?

Answer 1

To check that your proof works for rational numbers as well, you may just look at $f((mx+n)/k)$ where $m,n,k$ are integers. After factoring out $1/k^2$, this becomes $$m^2x^2+2mnx+(n^2+4k^2).$$ So you could fall back on your original proof by considering $a=m$ and $b=n$, and then the only difference is that the constant $4$ is now $4k^2$.

The one place where this difference affects your proof is in the case where $p$ is odd and you would then deduce that $p\mid 4k^2$. It follows that $p\mid k$, and you know that $p\mid n$, so now $p^2\mid (n^2+4k^2)$, contradicting Eisenstein.