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I assume the argument you have in mind is the folowing: suppose $f \in \cap \mathfrak{p}$; then to show that $f$ is nilpotent, it suffices to show that the localized ring $A_f$ is zero. And indeed, if $f$ is in every prime ideal of $A$, then $A_f$ has no prime ideals at all; since every nonzero ring has a maximal ideal by Zorn's Lemma, we must have $A_f = 0$.

This line of reasoning easily adapts to show that in fact, the statement that $\operatorname{Nil}(A) = \cap \mathfrak{p}$ implies that every nonzero ring has a prime ideal. Indeed, suppose that $A$ were nonzero with no prime ideals; then $\cap \mathfrak{p} = A$, so every element of $A$ is nilpotent. In particular, $1 = 1^n = 0$, so $A = 0$.

It would therefore appear that (following

Following Eric Rowell's answer) that , this theorem is very close to being equivalent to the axiom of choice (however, it does not obviously imply the existence of maximal ideals).

1

I assume the argument you have in mind is the folowing: suppose $f \in \cap \mathfrak{p}$; then to show that $f$ is nilpotent, it suffices to show that the localized ring $A_f$ is zero. And indeed, if $f$ is in every prime ideal of $A$, then $A_f$ has no prime ideals at all; since every nonzero ring has a maximal ideal by Zorn's Lemma, we must have $A_f = 0$.

This line of reasoning easily adapts to show that in fact, the statement that $\operatorname{Nil}(A) = \cap \mathfrak{p}$ implies that every nonzero ring has a prime ideal. Indeed, suppose that $A$ were nonzero with no prime ideals; then $\cap \mathfrak{p} = A$, so every element of $A$ is nilpotent. In particular, $1 = 1^n = 0$, so $A = 0$.

It would therefore appear that (following Eric Rowell's answer) that this theorem is equivalent to the axiom of choice.