It's well known that the nilradical of a commutative ring with identity $A$ is the intersection of all the prime ideals of $A$.


Every proof I found (e.g. in the classical "Commutative Algebra" by Atiyah and Macdonald) uses Zorn's lemma to prove that $x \notin Nil(A) \Rightarrow x \notin \cap\_{\mathfrak{p}\in Spec(A)} \mathfrak{p}$ (the other way is immediate).
Does anybody know a proof that doesn't involve it?



Nilradicals without Zorn's lemma
Since you asked for a proof, let me complement Chris Phan's answer by outlining a proof that relies only on the [Compactness Theorem](http://en.wikipedia.org/wiki/Compactness_theorem) for propositional logic, which is yet another equivalent to the [Ultrafilter Theorem](http://en.wikipedia.org/wiki/Boolean_prime_ideal_theorem) over ZF. 


Let A be a commutative ring and let x ∉ Nil(A). To each element a ∈ A associate a propositional variable pa and let T be the theory whose axioms are


* p0, ¬p1, ¬px, ¬px2, ¬px3,...
* pa ∧ pb → pa+b for all a, b ∈ A.
* pa → pab for all a, b ∈ A.
* pab → pa ∨ pb for all a, b ∈ A.


Models of T correspond precisely to prime ideals that do not contain x. Indeed, if P is such an ideal, then setting pa to be true iff a ∈ P satisfies all of the above axioms, and conversely. So it suffices to show that T has a model.


Since xn ≠ 0 for all n, one can verify using ideals over finitely generated subrings of A that the theory T is finitely consistent, i.e. any finite subset of T has a model. (What I just swept under the rug here is a constructive proof of the theorem for quotients of Z[v1,...,vn].) The Compactness Theorem for propositional logic then ensures that T has a model.


