Every proof I’ve seen that, for a commutative ring $A$, $$\newcommand{\Nil}{\operatorname{Nil}}\Nil(A)=\bigcap_{x\in \newcommand{\Spec}{\operatorname{Spec}}\Spec(A)}x$$ assumes Zorn’s lemma. So my question is,
Within ZF but not ZFC, does there exist a commutative unital ring $A$ such that $\Nil(A)\neq ∩_{x∈ \Spec(A)}x$? (Here, I define the nilradical to be the set of nilpotents of $A$.)
Perhaps it’s better to define $\Nil(A)$ to be the standard conclusion in commutative algebra, obfuscating this (potential?) issue, but I’m still curious.
Sure. For instance, let $A=\mathcal{P}(\mathbb{N})/\mathrm{fin}$ be the Boolean ring of subsets of $\mathbb{N}$ modulo the ideal of finite subsets. So explicitly, an element of $A$ is an equivalence class of subsets of $\mathbb{N}$ where two sets are equivalent if they differ by finitely many elements, addition is symmetric difference, and multiplication is intersection. A prime ideal in $A$ is equivalent to a nonprincipal ultrafilter on $\mathbb{N}$. There are models of ZF in which there are no nonprincial ultrafilters on $\mathbb{N}$, and so $A$ has no prime ideals. So the intersection of all primes ideals is all of $A$, but not all of $A$ is nilpotent (in fact, the only nilpotent element is $0$).