Equivalence of Definitions of Nilradical of Ring

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Theorem

Let $A$ be a commutative ring.


The following definitions of its nilradical are equivalent:


Definition 1

The nilradical of $A$ is the subset consisting of all nilpotent elements of $A$.

Definition 2

Let $\operatorname{Spec} \left({A}\right)$ be the prime spectrum of $A$.


Then the Nilradical of $A$ is:

$\displaystyle \operatorname{Nil} \left({A}\right) = \bigcap_{\mathfrak p \mathop \in \operatorname{Spec} \left({A}\right)}\mathfrak p$

That is, it is the intersection of all prime ideals of $A$.


Proof

By Nilpotent Element is Contained in Prime Ideals, $\operatorname{Nil}(A)$ is contained in the intersection of all prime ideals.

It remains to prove the other inclusion.


Let $f \in A$ be not nilpotent.

Let $S$ be the set of ideals of $A$ that are disjoint from $\{f^n : n \in \N\}$.

By Zorn's Lemma, $S$ has a maximal element $P$.

In particular, $f\notin P$.

We want to show that $P$ is prime.

Let $a,b\in A$ with $a,b\notin P$.

Then the sums of ideals $(a)+P$ and $(b)+P$ contain $P$ strictly.

By the maximality of $P$, there exist $n,m\in\N$ with $f^n\in (a)+P$ and $f^m\in (b)+P$.

Then $f^{m+n}\in (ab)+P$.

Thus $(ab)+P\notin S$; in particular $ab\notin P$.

Thus $P$ is prime, and $f\notin \bigcap_{\mathfrak p \in \operatorname{Spec} \left({A}\right)}\mathfrak p$.

$\blacksquare$


Axiom of Choice

This theorem depends on the Axiom of Choice.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.


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