Eisenstein criterion on f(x+1) I need to show that the polynomial $f(x)=x^6+x^5+x^4+x^3+x^2+x+1$ is irreducible in $\mathbb{Z}[X]$ and in $\mathbb{F}_2[X]$. As we can't find a prime number p satisfying the conditions for the Eisenstein criterion, I did not know how to solve it. I looked into the solutions and they apply the Eisenstein criterion to $f(x+1)$ instead of $f(x)$. I don't understand why we can do this.
Could somebody explain this to me? And is proving irreducibility for $f(x+1)$ enough?
 A: 
I looked into the solutions and they apply the Eisenstein criterion to $f(x+1)$ instead of $f(x)$. I don't understand why we can do this.

It suffices to show irreducibility of $f(x+1)$. To see this, assume that $f(x)$ is reducible, then $f(x) = g(x)h(x)$ for some proper factors $g(x)$ and $h(x)$. In that case, you get that
$$f(x+1) = g(x+1)h(x+1)$$
where the right factors are still proper factors. Thus, reducibility of $f(x)$ implies that of $f(x+1)$. The contrapositive shows that it's sufficient to prove the irreducibility of $f(x)$.
A: Very simple: for any $a$ in a ring $R$, the map $\;\begin {aligned}[t]R[X]&\longrightarrow R[X]\\ f(X)&\longmapsto f(X+a)\end{aligned}$ is a ring isomorphism ( the inverse isomotrphism maps $f(X)$ onto $f(X-a)$).
Therefore it maps irreducible polynomials onto irreducible polynomials and vice-versa.
A: Irreducibility is the same as lack of rational (or perhaps real) roots. The graph of $f(x+1)$ is simply the graph of $f(x)$, shifted horizontally by one unit. This move preserves the lack (or presence) of rational (or real) roots.
