## Compute the nilradical of F[x]/(p(x))

Let $F$ be a field and $p(x) \in F[x]$. Compute the nilradical of $F[x]/(p(x))$.

We begin with a lemma. (I think I already proved this, but I can’t find it.)

Lemma: Let $R$ be a unique factorization domain and $x = \prod p_i^{k_i}$ an element of $R$, where the $p_i$ are distinct irreducibles. Then $\mathcal{N}(R/(x)) = (\prod p_i)/(x)$. Proof: Suppose $y + (x) \in \mathcal{N}(R/(x))$. Then $y^n \in (x)$ for some $n$. Certainly then each $p_i$ must divide $y$, as otherwise no power of $y$ is divisible by $p_i$. Hence $y \in (\prod p_i)$. Conversely, we have $(\prod p_i)^{\max k_i} \in (x)$, so that $(\prod p_i) + (x) \in \mathcal{N}(R)$. $\square$

Now $F[x]$ is a unique factorization domain, and the lemma applies to $F[x]/(p(x))$.