$f(x+a)$ irreducibility means $f(x)$ irreducibility Let $a~\in~\mathbb{Z}$ and let $f(x)~\in~\mathbb{Z}\left[x\right]$. Suppose that $f(x+a)$ is irreducible over $\mathbb{Z}$. Prove that $f(x)$ is irreducible over $\mathbb{Z}$.
My idea is:
$f(x)=u(x)*v(x) \ldots \tag{1}$
$f(x+a)=u(x+a)*v(x+a) \ldots \tag{2}$
Since $(2)$ above is irreducible so either $u(x+a)$ or $v(x+a)$ is constant. If $u(x+a)$ is constant than so is $u(x)$ and same for $v(x)$. Hence $f(x)$ is irreducible too.
Is it OK?
 A: The most important property of $R[x]$, where $R$ is a (commutative) ring, is that given any ring $S$, a ring homomorphism $\alpha\colon R\to S$ and an element $s\in S$, there exists a unique ring homomorphism $\alpha_s\colon R[x]\to S$ such that


*

*$\alpha_s(r)=\alpha(r)$, for $r\in R$

*$\alpha_s(x)=s$


This is known as the universal property of the ring of polynomials.
The ring homomorphism just does
$$
\alpha_s(r_0+r_1x+\dots+r_nx^n)=\alpha(r_0)+\alpha(r_1)s+\dots+\alpha(r_n)s^n
$$
You can consider $R=\mathbb{Z}$, $S=\mathbb{Z}[x]$ and $\alpha$ the standard embedding; then you can first choose $s=x+a$, but also $s=x-a$. Since
$$
\alpha_{x-a}(\alpha_{x+a}(x))=x,
\qquad
\alpha_{x+a}(\alpha_{x-a}(x))=x,
$$
the same universal property of the ring of polynomials says that $\alpha_{x+a}$ is a ring isomorphism. In this case, we have
$$
\alpha_{x+a}(f(x))=f(x+a)
$$
by just applying the statements above.
Clearly a ring isomorphism preserves irreducibility of elements.

If you want the same, but expressed in less rigorous terms, if $f(x+a)=u(x)v(x)$, then
$$
f(x)=u(x-a)v(x-a)
$$
This is the intuition behind the more rigorous argument that, if $\alpha_{x+a}(f(x))=u(x)v(x)$, then
$$
f(x)=\alpha_{x-a}(\alpha_{x+a}(f(x)))=
\alpha_{x-a}(u(x)v(x))=
\alpha_{x-a}(u(x))\alpha_{x-a}(v(x))=
u(x-a)v(x-a)
$$
A: Yes, your proof is correct. However, I'd like to show you an alternate proof:
$$f(x)=u(x)*v(x) \ldots \tag{1}$$
$$f(x+a)=u(x+a)*v(x+a) \ldots \tag{2}$$
If we have $f(x)$ reducible like in $(1)$, then by $(2)$, we have $f(x+a)$ is reducible. Therefore:
$$f(x) \text{ is reducible} \implies f(x+a) \text{ is reducible}$$
Take the contrapositive:
$$f(x+a) \text{ is irreducible} \implies f(x) \text{ is irreducible}$$

I like this proof better because it makes more sense to me, but you should do whatever you are more comfortable with.
