Irreducible components of certain (potentially non-Noetherian) spectrum

Question

Let $\Bbbk$ be an algebraically closed field (not necessarily characteristic zero), and let $R$ be a finitely generated $\Bbbk$-algebra. Let $S\subseteq R$ be a $\Bbbk$-subalgebra, and let $X=\operatorname{Spec}S$.

Of course $S$ need not be finitely generated---but my question is, does $X$ still have irreducible components? Can we write $X=X_1\cup \dots \cup X_r$ such that each $X_i$ is irreducible?

Even simpler question: is it known whether $X$ must be a Noetherian topological space? I'm guessing the answer is no, since I can't find such a statement.

math54321 · Accepted Answer · 2023-05-19T20:08:40.670

2

For irreducible components: if $\varphi : A \to B$ is any ring map, then every minimal prime of $\ker \varphi$ in $A$ is the contraction of a minimal prime of $B$ (see here for proof). In particular, any subring of a ring with finitely many minimal primes also has finitely many minimal primes.

edited May 19 '23 at 20:08

answered May 19 '23 at 19:37

math54321

952

The phrasing in the first sentence has a $\ker$ in it which seems like it might not need to be there? Or am I misreading something? – KReiser May 19 '23 at 19:47
@KReiser I meant $\ker \varphi$ as an ideal in $A$ (and its minimal primes) – math54321 May 19 '23 at 19:48
Ah, okay, I see what you mean. – KReiser May 19 '23 at 19:53
Thank you! Way simpler than I thought. – freeRmodule May 20 '23 at 13:40

Irreducible components of certain (potentially non-Noetherian) spectrum

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