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By $A$ we denote an arbitrary commutative ring. Suppose that for each $a\in A$ $(a)=(a^2)$ (i.e. $\forall a\in A$ $\exists b$ such that $ba^2=a$).

It is easy to see that this implies that each prime ideal in $A$ is maximal. For example, suppose $\rho$ is a prime ideal, i.e. $A/\rho$ is an integral domain. So, for each $\bar{a}\in A/\rho$ ($\bar{a}\ne \bar{0}$) there exists $\bar{b}\in A/\rho$ such that $\bar{b}\bar{a^2}=\bar{a}$. It follows that $\bar{a}\bar{b}=\bar{1}$ as $A/\rho$ is an integral domain.

Well, I'd like to understand if the converse is true. So, is it true that if each prime ideal is maximal then $\forall a$ $(a)=(a^2)$? I would like to add that I am assuming $R$ is reduced (no nilpotent elements).

rschwieb
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user106200
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1 Answers1

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Yes, what you are talking about is true. The rings you start off describing are called von Neumann regular rings. In fact:

For a commutative ring $R$, all prime ideals of $R$ are maximal iff the Jacobson radical $J(R)$ of $R$ is nil and $R/J(R)$ is von Neumann regular.

Since $J(R)=Nil(R)$ in this case, requiring that $R$ is reduced implies that $J(R)=\{0\}$, so you get that a commutative reduced ring for which all prime ideals are maximal is von Neumann regular.

rschwieb
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