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.
$\square$

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