The Stacks project

Proof. Since a regular immersion is a Koszul-regular immersion, see Divisors, Lemma 31.21.2, it suffices to prove the second statement. This translates into the following algebraic statement: Suppose that $I \subset A$ is an ideal generated by a Koszul-regular sequence $f_1, \ldots , f_ r$ of $A$. Then $A \to A/I$ is a perfect ring map. Since $A \to A/I$ is surjective this is a presentation of $A/I$ by a polynomial algebra over $A$. Hence it suffices to see that $A/I$ is pseudo-coherent as an $A$-module and has finite tor dimension. By definition of a Koszul sequence the Koszul complex $K(A, f_1, \ldots , f_ r)$ is a finite free resolution of $A/I$. Hence $A/I$ is a perfect complex of $A$-modules and we win. $\square$