A noetherian ring whose ideals are idempotent is artinian

Question

I have to prove the folowing:

If $R$ is a Noetherian ring, and for every ideal $I$ of $R$ we have $I = I^{2}$, then $R$ is Artinian.

My first thought was to try to prove that the nilradical of $R$ is equal to the Jacobson, but I got nothing with that. Then I tried to prove that any prime ideal of $R$ is maximal, but I couldn't, so I got stuck.

Thank you for any help!

Answer 1

If $I=I^2$ and $I$ is finitely generated, then there exists $e\in I$, $e^2=e$ such that $I=(e)$; see here. Now take a prime ideal $\mathfrak p$ and look at $R/\mathfrak p$: any ideal of this ring has the form $I/\mathfrak p$ with $I$ ideal of $R$ containing $\mathfrak p$. We also have $I/\mathfrak p=(I/\mathfrak p)^2$ and $I/\mathfrak p$ is finitely generated, so $I/\mathfrak p$ is generated by an idempotent of $R/\mathfrak p$. Since $R/\mathfrak p$ is an integral domain its only idempotents are $0$ and $1$. We thus get that $I/\mathfrak p=(0)$ or $I/\mathfrak p=R/\mathfrak p$. This means that $R/\mathfrak p$ has only two ideal, that is, $R/\mathfrak p$ is a field.

Answer 2

If $I^2=I$ and $I$ is finitely generated, it is generated by an idempotent $e$, and thus it is a direct summand of $R$ (since $R=eR\oplus (1-e)R$).

Since $R$ is Noetherian, all ideals are finitely generated and hence by the last sentence they are summands of $R$. Consequently $R$ is semisimple, hence Artinian.

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Rings for which $I^2=I$ for all ideals are sometimes called "fully semiprime rings," and probably a more popular name I'm unaware of. Another way to summarize the above argument is that commutative fully semiprime rings are von Neumann regular and a Noetherian von Neumann regular ring is semisimple.

(Bonus unrelated discussion: I started learning about fully semiprime rings when learning about noncommutative rings whose ideals are all prime ideals. If I remember correctly, all ideals are prime iff $R$ is fully semiprime and the ideals are linearly ordered.)