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The Stacks project

Lemma 10.63.20. Let k be a field. Let S be a finite type k algebra. If \dim (S) > 0, then there exists an element f \in S which is a nonzerodivisor and a nonunit.

Proof. By Lemma 10.63.5 the ring S has finitely many associated prime ideals. By Lemma 10.61.3 the ring S has infinitely many maximal ideals. Hence we can choose a maximal ideal \mathfrak m \subset S which is not an associated prime of S. By prime avoidance (Lemma 10.15.2), we can choose a nonzero f \in \mathfrak m which is not contained in any of the associated primes of S. By Lemma 10.63.9 the element f is a nonzerodivisor and as f \in \mathfrak m we see that f is not a unit. \square

Comments (7)

Comment #7753 by Kentaro Inoue on

I think this lemma can be proved more easily. Since has infinite maximal ideals, there exists a maximal ideal which is not equal to any associated prime ideal. We can take an element which is not contained in any associated prime ideal. Then is a desired one.

Comment #8319 by Zichen Lu on

Why does the ring S have infinitely many maximal ideals? Lemma 0ALW only implies that S has infinitely many prime ideals, but not maximal ideals.

Comment #8320 by on

Thanks for pointing out this insufficient argument. Of course this is easily fixed by saying that an infinite Jacobson space has infinitely many closed points; I will do this when I next go through all the comments again. Does anybody have another suggestion for fixing this?

Comment #8324 by Rijul Saini on

I just wanted to remark that by Noether normalization, it is enough to see that the affine line over k has infinitely many closed points.

There are also:

  • 16 comment(s) on Section 10.63: Associated primes

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