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True or False:

Let $A \supseteq R$ be an integral ring extension of $R$, with $A$ being a domain. If every non-zero prime ideal of $R$ is a maximal ideal, then every non-zero prime ideal of $A$ is also maximal.

This is a follow up to another problem that's exactly identical except that in the first problem $A$ is not necessarily a domain. I found a counterexample for that problem, using $\mathbb{Z} = R$ and $\mathbb{Z}[x]/(x^2) = A$.

I'm thinking that being a domain changes something important here, but I can't figure out what. By the Going Up/Maximality Theorems in my book, I know that every prime ideal in $A$ lying over a prime ideal in $R$ is maximal, so I really only need to show that A has no prime ideals that don't lie over a prime ideal in R or that prime ideals of A that aren't lying over a prime ideal in R are maximal, but I'm getting stuck here.

There's a hint in my book that says that if {$0$} is a prime ideal in $R$, then the only prime ideal of A lying over {$0$} is {$0$}, but I'm not sure how that's helpful since we're dealing specifically with non-zero prime ideals in the proof.

Insight is appreciated!

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Any non-zero prime ideal $\mathfrak{q}\subseteq A$ lies over some non-zero prime ideal $\mathfrak{p}=\mathfrak{q}\cap R \subseteq R$. Then you are done by those going up/maximality theorems that you mention.

Added: since the answer is rather short and you also asked for insight, I will add some more. In an integral ring extension $R\subseteq A$ we have $$ \dim_{\text{Krull}}(A)=\dim_{\text{Krull}}(R) $$ The exercise you are proving is a particular case of this (for $\dim_{\text{Krull}}(R)=1$, assuming it is not a field).

You probably knew a particular case of this already: an integral extension of a field is a field, meaning that the smallest ring containing your field and some algebraic element is again a field. But if you add a transcendental element $T$ you don't get a field anymore, you get a principal ideal domain.

This has a nice geometric interpretation in terms of actual dimension of topological spaces: fields are points, principal ideal domains are curves, ... (the ring being the functions on the geometric space). This doesn't apply on arbitrary rings and is only a particular case, of course, but I think it is a nice way to think of it.

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  • $\begingroup$ That makes perfect sense. I appreciate it! $\endgroup$ – Joe Apr 9 '18 at 21:31

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