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. $\square$

Comment #7753 by Kentaro Inoue on

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

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

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