**Proof.**
It is clear that if $U \subset X$ is an open such that $\mathcal{J}|_ U$ is generated by a quasi-regular sequence $f_1, \ldots , f_ r \in \mathcal{O}_ X(U)$ then $\mathcal{J}|_ U$ is of finite type, $\mathcal{J}|_ U/\mathcal{J}^2|_ U$ is free with basis $f_1, \ldots , f_ r$, and the maps in (3) are isomorphisms because they are coordinate free formulation of the degree $n$ part of (31.20.0.3). Hence it is clear that being quasi-regular implies conditions (1), (2), and (3).

Conversely, suppose that (1), (2), and (3) hold. Pick a point $x \in \text{Supp}(\mathcal{O}_ X/\mathcal{J})$. Then there exists a neighbourhood $U \subset X$ of $x$ such that $\mathcal{J}|_ U/\mathcal{J}^2|_ U$ is free of rank $r$ over $\mathcal{O}_ U/\mathcal{J}|_ U$. After possibly shrinking $U$ we may assume there exist $f_1, \ldots , f_ r \in \mathcal{J}(U)$ which map to a basis of $\mathcal{J}|_ U/\mathcal{J}^2|_ U$ as an $\mathcal{O}_ U/\mathcal{J}|_ U$-module. In particular we see that the images of $f_1, \ldots , f_ r$ in $\mathcal{J}_ x/\mathcal{J}^2_ x$ generate. Hence by Nakayama's lemma (Algebra, Lemma 10.20.1) we see that $f_1, \ldots , f_ r$ generate the stalk $\mathcal{J}_ x$. Hence, since $\mathcal{J}$ is of finite type, by Modules, Lemma 17.9.4 after shrinking $U$ we may assume that $f_1, \ldots , f_ r$ generate $\mathcal{J}$. Finally, from (3) and the isomorphism $\mathcal{J}|_ U/\mathcal{J}^2|_ U = \bigoplus \mathcal{O}_ U/\mathcal{J}|_ U f_ i$ it is clear that $f_1, \ldots , f_ r \in \mathcal{O}_ X(U)$ is a quasi-regular sequence.
$\square$

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