# Prove that every prime ideal that isn't maximal is a minimal prime ideal

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 $$R$$.

• 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. – rschwieb Apr 25 '19 at 13:34

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.
• $Q$ 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. – user419669 Apr 27 '19 at 2:00
• 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 $R$. – Julian Rosen Apr 27 '19 at 3:23