In order to prove that cyclotomic polynomial $\Phi_p(x)=x^{p-1}+\dots+x+1$ is irreducible in $\mathbb{Q}[x]$ we need to consider "it's shift", namely $\Phi_p(x+1)$ and in this case we can apply Eisenstein's criterion easily.
This is motivated because the mapping $\phi_a:\mathbb{Q}[x]\to \mathbb{Q}[x]$ defined by $\phi_a(P(x))=P(x+a)$ is bijective homomorphism (or automorphism).
Claim: $f(x)$ is irreducible iff $f(x+1)$ is irreducible.
But in the proof of this claim we do not use that this mapping is bijective (we use only that this is homomorphism).
Am I right? Would be very grateful for reply!
The proof goes like this: Suppose $f(x)$ is reducible (i.e. $f(x) = g_1(x)g_2(x)$ where $g_1(x)$ and $g_2(x)$ are not units). Then $\phi_1(f(x))=\phi_1(g_1(x))\phi_1(g_2(x))$ Since $\phi_1(g_1(x))$ and $\phi_1(g_2(x))$ are not units, $\phi_1(f(x))$ is reducible. Applying a similar argument to $\phi_{-1}=\phi_1^{-1}$ gives the other direction. (Since we showed that we have iff for reducibility, we have it for irreducibility as well)
Here, we were able to avoid the bijection by using $\phi_{-1}$. However, just knowing that the maps were homomorphisms was not sufficient since we also needed to know that the map took non-units to non-units.