I'm working on an exercise from Dummit and Foote, and I've gotten down to the following lemma that makes everything I need work out, the only problem is that I'm not sure how to prove it (or whether or not it's even true!).
Here's the lemma. If $p(x)$ is a polynomial of degree $\ge 5$ in $\mathbb{Z}[x]$, then if $p(x) + 1$ is irreducible, $xp(x) + 1$ is irreducible. Here's where I've gone so far:
Since $p(x)+1$ is irreducible, $p(x+1)+1$ is irreducible (as per an example in D&F that says that implies $f(x)$ is irreducible iff $f(x+1)$ is irreducible), so consider $(x+1)p(x+1)+1$. Distributing, we get $xp(x+1) + p(x+1)+1$, and passing into the quotient ring [I'm trying to use another proposition that says that if $I$ is a proper ideal and $\overline{f(x)}$ is irreducible in $(R/I)[x]$ then $f$ is irreducible in $R[x]$) $Z[x]/(p(x+1)+1)$, we get $\overline{xp(x+1) + p(x+1)+1} = \overline{xp(x+1)}$. but in the quotient ring, $\overline{p(x+1)} = \overline{-1}$, so $\overline{xp(x+1)} = \overline{-x}$ (or, since $-1$ is a unit, just $\overline{x}$.)
Now I just need to show that in this quotient ring $\overline{-x}$ is irreducible, but I don't know how to go about that.
Any suggestions?
Unfortunately, the lemma is false. Let $p(x) = x^6 + x$. Then $p(x) + 1$ is irreducible, but $xp(x) + 1 = x^7 + x^2 + 1 = (x^2 + x + 1) (x^5 - x^4 + x^2 - x + 1)$.