I was working through Contemporary Abstract Algebra by Joseph A. Gallian when I came across the following corollary.
Corollary: Irreducibility of $p^{th}$ Cyclotomic Polynomial
For any prime $p$, the $p^{th}$ cyclotomic polynomial $$\phi_p(x)=\frac{x^{p}-1}{x-1}=x^{p-1}+x^{p-2}+\cdots+x+1$$ is irreducible over $\mathbb{Q}$.
In the proof the author shifts the polynomial from $\phi_p(x)\rightarrow\phi_p(x+1)$ and shows that since $\phi_p(x+1)$ is irreducible over $\mathbb{Q}$, then $\phi_p(x)$ is irreducible over $\mathbb{Q}$.
I am able to follow the proof however I have the following question.
$f(x)$ is irreducible over $F$ if and only if $f(x+a)$ is irreducible over $F$, for some field $F$, $f \in F[x]$ and $a \in F$.
If the above is true, I should be able to prove it.
Note: The above might be a known mathematical fact but I was not able to find much literature on it. This was a follow up question I had after reading the proof.
Proof:
"$\rightarrow$" Since $f(x)$ is irreducible we can write $f(x)=a(x)b(x)$. This is about all I have (not much) but I cannot see how I can relate $f(x+a)$ to $f(x)=a(x)b(x)$.
If someone could offer a hint to help my understanding of polynomial shifts, in this context, it would be greatly appreciated. Thank you.
Update:
Claim: Polynomial shifts are ring homomorphisms.
Proof: Let $\phi:F[x]\rightarrow F[x+c]$ defined by $\phi(x):x\rightarrow x+c$.
Let $p(x),q(x) \in F[x]$.
(i) $\phi(p(x))=\phi(a_nx^n+\cdots + a_1x+a_0)=a_n(x+c)^n+\cdots +a_1(x+c)+a_0 \in F[x+c]$. Thus $\phi$ is a well-defined function.
(ii) $\phi(p(x)+q(x))=\phi((a_n+b_m)x^n+\cdots + (a_0+b_0))=(a_n+b_m)(x+c)^n+\cdots + (a_0+b_0) \in F[x+c]$. Thus $\phi(p(x)+q(x))=\phi(p(x))+\phi(q(x))$.
(iii) $\phi(p(x)q(x))=\phi(c_{m+n}x^{m+n}+\cdots + c_1x+c_0)=c_{m+n}(x+c)^{m+n}+\cdots +c_1(x+c)+c_0 \in F[x+c]$ where $c_{i}=\sum_{j=0}^{i} a_{j}b_{i-j}$. Thus $\phi(p(x)q(x))=\phi(p(x))+\phi(q(x))$.
Therefore, $\phi:x \rightarrow x+c$ is a ring homomorphism.
Hint: The shift is clearly bijective, as it has an inverse given by shifting in the other direction. So all you have to do is show that the shift is a homomorphism, that is to say $\phi (f+g)(x) = \phi (f(x)) + \phi(g(x))$ and $\phi (fg)(x) = \phi(f(x))\phi(g(x))$.
Both of these things are not so hard to check, but they can be a little laborious to compute.