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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.

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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.

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  • $\begingroup$ 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? $\endgroup$
    – KReiser
    Commented May 19, 2023 at 19:47
  • $\begingroup$ @KReiser I meant $\ker \varphi$ as an ideal in $A$ (and its minimal primes) $\endgroup$
    – math54321
    Commented May 19, 2023 at 19:48
  • $\begingroup$ Ah, okay, I see what you mean. $\endgroup$
    – KReiser
    Commented May 19, 2023 at 19:53
  • $\begingroup$ Thank you! Way simpler than I thought. $\endgroup$
    – freeRmodule
    Commented May 20, 2023 at 13:40

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