Nilpotent elements in a quotient of a polynomial ring.

Question

I'm trying to prove the following statement. I'm not entirely sure if it's true though I suspect it is:

Consider the ring $A = \mathbb Z[x]/(f_1,\ldots,f_n)$. Then the element $x \in A$ is nilpotent if and only if $x$ is nilpotent in $A \otimes_\mathbb{Z} \mathbb{F}_p$ for all primes $p$ and $x$ is nilpotent in $A \otimes_\mathbb{Z} \mathbb{Q}$.

One direction is obvious. For the other direct I've been trying to fiddle around with the correspondence theorem to obtain that $x$ is the the nilradical of $A$ to no avail.

Answer 1

Let $I:=(f_1,\dots,f_n)$. Since $x$ is nilpotent in $I \iff x\in \underset{\underset{\mathfrak{p}\supset I}{\mathfrak{p}\text{: prime ideal}}}{\bigcap}\mathfrak{p}$, if $x$ is not nilpotent there exists a prime ideal $\mathfrak{p}$ that does not contain $x$. If $\mathbb{Z}\cap \mathfrak{p}=(p),p$ a prime number, $x$ is not nilpotent in $A/\mathfrak{p}=A/\mathfrak{p}\otimes_\mathbb{Z} \mathbb{F}_p$ and not in $A\otimes \mathbb{F}_p$. If $\mathbb{Z}\cap \mathfrak{p}=0$, $x$ is not nilpotent in $A/\mathfrak{p}$ and not in $A\otimes_\mathbb{Z} \mathbb{Q} \subset \mathrm{Frac}(A/\mathfrak{p})$.

Answer 2

In fact I came up with a nice argument for the general fact. The general result follows from the specific case above by considering the map $\mathbb Z[x] \to A$ picking out the element $a$, but let's prove it in general just for fun.

Let us prove: Let $A$ be a commutative ring. Then $a \in A$ is nilpotent if and only if $a$ is nilpotent in $A \otimes \mathbb Q$ and $A \otimes \mathbb F_p$ for each prime $p$.

Let $a \in \Gamma(\mathrm{Spec} A)$. We want to prove that $a$ is in every prime ideal. Pick a prime ideal $\mathfrak p$ and consider $A_\mathfrak{p}/\mathfrak p$, the residue field at $\mathfrak p$. The map $A \to A_\mathfrak{p}/\mathfrak p$ factors through $A \otimes \mathbb F_p$ or $A \otimes \mathbb Q$ (depending on the characteristic of the residue field). In any case, there exists an $n(\mathfrak p)$ with $a^{n(\mathfrak p)} \otimes 1$ vanishing in one of these. But this means that $a^{n(\mathfrak p)}$ vanishes in $A_\mathfrak{p}/\mathfrak p$ which is equivalent to $a^{n(\mathfrak p)} \in \mathfrak p$ which is equivalent to $a \in \mathfrak p$. Since $\mathfrak p$ is arbitrary we are done.