Does every prime ideal of a ring contain a minimal prime?

Question

Let $R$ be a ring and $\mathfrak p \subset R$ a prime ideal. I think that it is easy to show that $\mathfrak p$ contains a minimal prime ideal of $R$, by the following argument: pass to the localization $\varphi : R \to R_\mathfrak p$, fetch a minimal prime $\mathfrak q \subset R_\mathfrak p$ and contract it back to $R$ to get a minimal prime $\varphi^{-1}(\mathfrak q) \subset \mathfrak p$. However, the commutative algebra wiki suggests that only in a Noetherian ring do prime ideals always contain a minimal prime.

Is there anything wrong with my reasoning, and, if so, what is it?

Why can you "fetch a minimal prime $\mathfrak{q} \subset R_\mathfrak{p}$"? Localization preserves ideals contained in $\mathfrak{p}$. — Joshua P. Swanson, Sep 20 '20 at 06:01
@JoshuaP.Swanson: Because $R_\mathfrak p$ is a local ring, hence not the zero ring, hence it contains minimal prime ideals with respect to inclusion. — isekaijin, Sep 20 '20 at 06:05
Wow, I must be tired tonight. Anyway, yes, of course. Stacks Lemma 10.16.2(4) says it explicitly. — Joshua P. Swanson, Sep 20 '20 at 06:16

score 0 · Accepted Answer · answered Sep 20 '20 at 06:46

0

You do not need to use localizations.

Hint: The intersection of any decreasing chain of prime ideals is a prime ideal. Hence by Zorn's lemma every prime ideal contains a minimal prime ideal.

answered Sep 20 '20 at 06:46

markvs

19,896

It is really the same argument in disguise, right? – isekaijin Sep 20 '20 at 07:18
I am not sure it is the same. – markvs Sep 20 '20 at 14:41

Does every prime ideal of a ring contain a minimal prime?

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