I have a question about this post https://math.stackexchange.com/questions/1729747/ solving Exercise 8.4.G of Vakil's THE RISING SEA Foundations of Algebraic Geometry:
8.4.G EXERCISE (THE CONDITION OF A LOCALLY CLOSED EMBEDDING BEING A REGULAR EMBEDDING IS OPEN) Show that if a locally closed embedding $\pi:X\rightarrow Y$ of locally Noetherian schemes is a regular embedding at $p$, then it is a regular embedding in some neighborhood of $p$ in $X$.
Question. In Vakil's definition of a $B$-regular sequence $x_1,\dots,x_r \subset B$, it is required that $(x_1,\dots,x_r)\subsetneq B$, which is a "closed condition" as it corresponds to the support of the finitely generated $B$-module $B/(x_1,\dots,x_r)$ on $\mathrm{Spec} B$. Then I think that in the slogan "the condition of a locally closed embedding being a regular embedding is open", "open" should be "locally closed".
The author of the cited post didn't explicitly address this point. Could anybody explain why the closed condition that $ (B/(x_1,\dots,x_r))_\mathfrak{q}\neq 0$ doesn't make the condition of a locally closed embedding being a regular embedding non-open? Thanks in advance.
Update. I think I see the reason why it is an open condition: on some principal affine open $D(b)$, $X$ is cut out by $I=(x_1,\dots,x_r)$ for $x_1,\dots,x_r\in B_b$. Since the annihilator of $B_b/(x_1,\dots,x_r)$ as a $B_b$-module is $(x_1,\dots,x_r)$, $X$ is the support of $B_b/(x_1,\dots,x_r)$. So the closed condition is just saying that $X$ is a closed subscheme...
