Solution: Kaplansky’s Commutative Rings 2.2.13

If $R$ is a commutative ring and $A$ an $R$ module, let $Z(A)$ denote the zero divisors of $R$ on $A$. In other words, $Z(A) = \{ r\in R : \exists_a(ra = 0) \}$. If $I$ is the annihilator of $A$, then it is known that any prime minimal over $I$ is contained in $A$. In particular, any minimal prime of $R$ is contained in $Z(R)$. (Recall that a prime $P\subset R$ minimal if it does not properly contain any other prime, and $P$ is said to be minimal over $I$ if there is no inclusion $I\subseteq Q\subset P$ with $Q$ another prime.)

Problem. If $R$ is a ring with no nonzero nilpotent elements, then $Z(R) = \cup_P P$ where $P$ ranges over all the minimal primes of $R$.

Solution. Since $P$ is a minimal prime, we see (by the fact mentioned at the beginning of this post) that $\cup P \subseteq Z(R)$, so we just have to show the reverse inclusion.

The nilradical of $R$ (the ideal containing the nilpotents), is the intersection of all the primes. Since every prime contains a minimal one, we see that $\cap P = 0$ as $P$ runs over the minimal primes. Now let $a\in Z(R)$ with $a\not=0$ be a zero divisor. Let $b\in R$ be nonzero be such that $ab = 0$. Since $\cap P = 0$, there exists a minimal prime $P$ such that $b\not\in P$. But $ab = 0\in P$ so $a\in P$. Hence $Z(R)\subseteq \cup P$, where $P$ runs over the minimal primes.

Can you think of an counterexample when the condition “no nonzero nilpotents” is dropped?

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