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Suppose that the additive group of the ring $R$ is a finitely generated abelian group. If $P$ is a maximal ideal of $R$, show that $R/P$ is a finite field. Show that every prime ideal of $R$ that is not maximal is a minimal prime ideal.

I managed to prove the first part, but I'm not sure about the second. And I have no idea how to connect prime ideals of $R$ with the condition on the additive group of $R$.

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  • $\begingroup$ In other words, you want to show it is has Krull dimension $\leq 1$. And it should be obvious that since $(R,+)$ satisfies the ACC on subgroups, $(R,+,\cdot)$ is a Noetherian ring. Now a $0$ dimensional Noetherian ring is Artinian, so the prime ideals are maximal, and the set of nonmaximal primes is empty, so that case is easy. I see a lot of stuff online about $1$-dimensional Noetherian rings, so it seems like these are good clues to begin with when searching for a reason the additive group will cause the ring to be $1$-dimensional. $\endgroup$ – rschwieb Apr 25 at 13:34
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It is equivalent to show that every non-minimal ideal is maximal. If $P\subsetneq Q\subset R$ are prime ideals, we can view $Q$ as an ideal of the integral domain $R/P$. The additive group of $R/P$ is also finitely generated, so it suffices to show that if $R$ is an integral domain with finitely generated additive group, then every non-zero prime ideal of $R$ is maximal.

Suppose $R$ is an integral domain whose additive group is finitely generated, and let $n$ be the rank of the additive group. Suppose $P\subset R$ is a non-zero prime ideal, and take $x\in P\backslash\{0\}$. Since multiplication by $x$ is a bijection of $R$ onto $xR$, the rank of $xR$ is $n$, so the rank of $P$ is also $n$. The quotient of a finitely generated abelian group by a subgroup with the same rank is finite, so $R/P$ is a finite integral domain, hence a field. It follows that $P$ is maximal.

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  • $\begingroup$ $Q$ is a subset of $R$, not of $R/P$, how can it be viewed as an ideal of $R/P$? Also, I'm not quite following why it's suffices to show that in an integral domain with finitely generated additive group every non-zero prime ideal is maximal. $\endgroup$ – user419669 Apr 27 at 2:00
  • $\begingroup$ Really I mean $Q/P$ is a prime ideal of $R/P$. Once we know that non-zero prime ideals of integral domains with finitely generated additive group are maximal, we get that $Q/P$ is a maximal ideal of $R/P$, and it follows that $Q$ is a maximal ideal of $R$. $\endgroup$ – Julian Rosen Apr 27 at 3:23

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