TITLE: Show $p(X)$ (over a field) is irreducible iff $p(X+a)$ is irreducible
QUESTION [8 upvotes]: Let $A$ be a field and let $p(X)$ be a polynomial over $A$. Let $a\in A$. 
Want to show:

$p(X)$ is irreducible if and only if $p(X+a)$ is irreducible. 

I suspect that I should use the substitution principle somehow, but that's as far as I've come. Completely stumped.

REPLY [11 votes]: One way to prove this is to instead prove that $p(X)$ is reducible if and only if $p(X + a)$ is reducible. 
To this end, note that if $p(X) = q(X)r(X)$, then $p(X+a) = \hat{q}(X)\hat{r}(X)$ where $\hat{q}(X) = q(X + a)$ and $\hat{r}(X) = r(X + a)$. Going the other way, if $p(X+a) = q(X)r(X)$, then $p(X) = \tilde{q}(X)\tilde{r}(X)$ where $\tilde{q}(X) = q(X - a)$ and $\tilde{r}(X) = r(X - a)$.

REPLY [5 votes]: Hint: Show that $\phi_a:A[X]\to A[X]$ defined as $p(X)\mapsto p(X+a)$ is an automorphism.