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$\def\codim{\operatorname{codim}} \def\sO{\mathscr{O}} \def\sI{\mathscr{I}} \def\sF{\mathscr{F}} \def\sG{\mathscr{G}}$In p. 306 of Görtz, Wedhorn, Algebraic Geometry I, 2nd ed., we find defined:

[Let $X$ be any scheme.] Assume that $Y$ is a closed subscheme of $X$ such that there exists an open affine covering $(U_i)$ of $X$ and regular elements $f_i\in\Gamma(U_i,\mathscr{O}_X)$ such that $Y \cap U_i = V (f_i)$ for all $i$ (such subschemes will be studied in more detail in Volume II where they will be called regularly immersed of codimension 1).

My first question is:

(Q1). Does $Y$ really have codimension 1 in $X$?

(For any closed subset $Z\subset X$, the codimension is defined $\codim_XZ=\underset{x\in Z}{\inf}\dim\sO_{X,x}$; one can show $\codim_XZ=\underset{F\subset Z}{\inf}\codim_XF$, where $F$ moves in the irreducible components of $Z$, cf. [ibid., sect. (5.8)].)

A few pages later, in the statement of Proposition 11.39, we find the implicit claim:

Let $X$ be a locally Noetherian scheme. Let $C\subset X$ be a closed integral subscheme of codimension 1, and denote $\sI$ to the ideal of sections on $X$ vanishing on $C$. Then $C$ is regularly immersed in $X$ if and only if, for all $x\in C$, $\sI_{x}=f\sO_{X,x}$, where $f\in\sO_{X,x}$ is a nonzerodivisor.

(I have translated the way they phrase it in the book.) It is clear that the condition is necessary. My second question is:

(Q2). Why is it sufficient?

I think they hint it in the proof of Theorem 11.40 (2): there they invoke the result “if $X$ is any ringed space, $\sF,\sG$ are $\sO_X$-modules of finite presentation, and $x\in X$, then any isomorphism $\sF_x\to\sG_x$ extends to an isomorphism $\sF|_U\to\sG|_U$, for some open $U\subset X$” (Proposition 7.27 (2)) to infer the existence of an open $U\subset X$ such that $\sI|_U=f\sO_{U}$, where $f\in\sO_X(U)$ is a regular section. My question is: how can we do this? First, we would need to show that $\sI$ is of finite presentation (it suffices to see it is $\sO_X$-coherent; in turn, it suffices to see it is of $\sO_X$-finite), but why so? Second, the invoked result only gives an isomorphism $\sI_U\cong f\sO_U$, but how one can obtain an equality from this?

Elías Guisado Villalgordo
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1 Answers1

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Q1. By Krull's principal ideal theorem (see wiki) the ideal generated by a non zero divisor has height $1$. So the answer is yes.

Q2. To say that an ideal $I$ in a ring $R$ is generated by a non-zero divisor is equivalent to saying that $I$ is free of rank one as a $R$-module. So the statement follows from the fact that if $F$ is a locally free coherent sheaf (ie all the stalks of $F$ are free) then there is an open covering $U_i$, such that $F_{U_i}$ is free (supposing that we work on a locally noetherian scheme). For this, see https://stacks.math.columbia.edu/tag/00NX

Damian Rössler
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  • In Q1, I suppose that the ring is noetherian. – Damian Rössler Oct 26 '23 at 15:09
  • The thing is that in Q1 $X$ is really arbitrary. That is how the book states the definition; no local Noetherianity assumption. For my second question, I just realized that the quasi-coherent ideal sheaf $\mathscr{I}$ on $X$ is $\mathscr{O}_X$-finite since $X$ is locally Noetherian. Applying this, we get an open neighborhood $U\subset X$ of $x$ where $f$ is defined and such that $f$ is an $\mathscr{O}_U$-basis of $\mathscr{I}|_U$; in turn, this implies that $f$ is regular on $U$. – Elías Guisado Villalgordo Oct 27 '23 at 07:24
  • @Elias Guisado Villalgordo. I suppose that the authors use the terminology "regularly immersed in cod $0$" by extension of what happens in the noetherian case. The notion of codimension is only well behaved in the noetherian situation anyway. – Damian Rössler Oct 27 '23 at 11:00