Consider the set $I$ of polynomials $p(x)$ in $\mathbb Z[x]$ such that $p(1)$ is even. Prove that this is a non-principal ideal.
That this is an ideal is clear. I was wondering whether my proof that $I$ is non-principal correct?
Assume $I=(f)$. Since $2\in I$, $2=f(x)g(x)$ for $f,g\in \mathbb Z[x]$, and this implies that $f$ must be constant. This constant can only be equal to $2$ because otherwise $2\notin I$. But on the other hand, $x^2+1\in I$, so $x^2+1=2g(x)$. This is impossible because the LHS is not divisible by $2$.
Your proof is correct, apart from a very minor detail ($f(x)$ could also be $-2$, but then $2$ would be a generator as well).
A slightly different approach: the map $\varphi\colon\mathbb{Z}[x]\to\mathbb{Z}[x]$ defined by $\varphi\colon p(x)\mapsto p(x+1)$ is an isomorphism (with inverse $p(x)\mapsto p(x-1)$). Note that $\varphi(I)$ is the ideal consisting of the polynomials $p$ such that $p(0)$ is even.
It's quite clear that $\varphi(I)=(2,x)$, which is not principal, because a possible generator $f(x)$ must have even constant term, but then $x=f(x)g(x)$ is not possible, unless the constant term is $0$.