The Stacks project

OK, I think we can leave it to the reader to see the equivalences of 2, 3, and 4. Presumably this is already silently used in some places.

Thanks and fixed here.

No, it is fine as is.

In the statement, I think we should say "let $Z\subset X$ be a reduced closed subscheme" or something like that.

I think it could be interesting to expand the statement in the following way:

Let $X$, $Y$ be locally ringed spaces. Let $\mathcal{I} \subset \mathcal{O}_X$ be a sheaf of ideals locally generated by sections. Let $i : Z \to X$ be the associated closed subspace. Let $f : Y \to X$ be a morphism of locally ringed spaces. The following are equivalent: 1. The morphism $f$ factors through $Z$, 2. The map $f^*\mathcal{I} \to f^*\mathcal{O}_X = \mathcal{O}_Y$ is zero, 3. $\mathcal{I}$ is contained in the kernel of $\mathcal{O}_X\to f_*\mathcal{O}_Y$, and 4. $f^{-1}\mathcal{I}$ is contained in the kernel of $f^{-1}\mathcal{O}_X\to\mathcal{O}_Y$.

Moreover, if this is the case, then the morphism $g : Y \to Z$ such that $f = i \circ g$ is unique.

The already existing proof does 1$\Leftrightarrow$2 and, implicitly, 2$\Rightarrow$3. The proof of 3$\Rightarrow$4 is easy by the adjointness $f^{-1}\dashv f_*$ (the same strategy proves 4$\Rightarrow$3), so it is left to show 4$\Rightarrow$2, and this is straightforward.

The statement of (3) can be written in a slightly more general way without changing the proof: restate "(3) For any birational integral morphism $\alpha : X' \to X$ such that $X'$ has locally finitely many irreducible components [i.e., every quasi-compact open of $X'$ has finitely many irreducible components] there exists a factorization $X^\nu \to X' \to X$ and $X^\nu \to X'$ is the normalization of $X'$."

Regarding the proof of (4): I don't understand the sentence "as the factorization is unique it suffices to construct it locally on $Z$." When we construct the morphism $W\to \nu^{-1}(U)$, shouldn't we say something about the compatibility of these morphisms? (So they can be glued to give rise to a morphism $Z\to X^\nu$.) The following is the argument I devised: For $i=1,2$, let $\operatorname{Spec} A_i=U_i\subset X$ be open affine, and let $\operatorname{Spec} B_i=W_i\subset g^{-1}(U_i)$ be open affine. We want to see that the morphisms $W_i\to\nu^{-1}(U_i)\hookrightarrow X^\nu$ agree on $W_1\cap W_2$. Pick an open affine $\operatorname{Spec} A=U\subset U_1\cap U_2$ and an open affine $\operatorname{Spec} B=W\subset g^{-1}(U)\cap W_1\cap W_2$. We have canonically defined ring morphisms $A_i'\to B_i$ and $A'\to B$, where $A_i'$, $A'$ are, respectively, the integral closures of $A_i$, $A$ in $Q(A_i)$, $Q(A)$. We want to see that the diagram $$\require{AMScd} \begin{CD} A_i'@>>> B_1\\ @VVV@VVV\\ A'@>>> B\\ @AAA@AAA\\ A_2'@>>> B_2 \end{CD}$$ commutes. This can be done by studying this bigger diagram. In this latter diagram:

1. The horizontal squares commute by construction.

2. The back squares commute by functoriality of the total ring of fractions (with respect to ring morphisms which preserve nonzerodivisors). In particular, the open immersions $U\to U_i$, $W\to W_i$ preserve generic points of irreducible components, so the maps on global sections—which are morphisms of reduced rings—preserve nonzerodivisors.

3. The side squares commute because the map $Q(A_i)\to Q(A)$ (resp., the map $Q(B_i)\to B$) sends an integral dependence equation with coefficients in $A_i$ (resp., with coefficients in $B_i$) to an integral dependence equation with coefficients in $A$ (resp., in $B$).

4. The front squares commute because the rest of the diagram commutes and using that $B\to Q(B)$ is monic.

To Weixiao Lu: In a locally noetherian category, every inductive limit of injective objects is an injective object (Cor. 1, p.358 of Des Catégories Abéliennes by Gabriel).

*Sorry, I mean "finite type". Is it true also for general sheaves of Λ-modules?

Is finiteness needed here?

Oups, sorry for overlooking the word "quasi-coherent" in the 2nd line of the proof.

It may help to mention that the image of $\mathcal{G}_1\oplus \mathcal{G}_2\to \mathcal{F}$ is quasi-coherent.

The argument seems to hold in the case of locally Noetherian scheme.

At the end of the proof, instead of "hence the local sections $f_U$ glue to a global section $f$ as desired," one could be a little bit more explanatory by saying "hence the local sections $f_U$ glue to a global section $g\in R$. In particular, $bg|_U=bf|_U$ for all open affine $U\subset X$; whence $bg=bf$, so $g=f$."

$j$the should be $j$th

In the third sentence, shouldn't one say "hence, by the proof of Lemma 10.22.1, it equals $V(I)$ where $I$ is generated by the idempotents..."?

Instead of "any pairwise intersection $V_\alpha\cap V_\beta$ occurs as a $V_\alpha$," one could write "any finite intersection $V_{\alpha_1}\cap\cdots\cap V_{\alpha_n}$ equals $V_\beta$, for some $\beta$" (the latter is what one actually uses to deduce $T\cap V_\alpha=\emptyset$ for some $\alpha$).

I don't understand why we can assume that $R$ is a subring of $R_\mathfrak{p}$ (in the sentence “as $R \subset R_{\mathfrak p}$ is flat we see that $R_{\mathfrak p} \subset Q(R) \otimes_R R_{\mathfrak p}$”). This is not the case for a non-domain $R$: the preimage of the prime $(0)\subset R_\mathfrak{p}$ along $R\to R_\mathfrak{p}$ is prime and thus different to $(0)\subset R$.

Here's the proof I've thought of:

The total ring of fractions is not a functor: in general a ring homomorphism $A\to B$ does not induce a ring homomorphism $Q(R)\to Q(R')$, let alone one that makes the corresponding square commute (this boils down to the observation that ring homos do not preserve nonzerodivisors in general). However, if $S\subset R$ is any multiplicative set, then the ring homo $\varphi:R\to S^{-1}R$ does preserve nonzerodivisors and it induces then a ring homomorphism $Q(\varphi):Q(R)\to Q(S^{-1}R)$ that makes the diagram $$\require{AMScd} \begin{CD} R@>{\varphi}>> S^{-1}R\\ @VVV@VVV\\ Q(R)@>>{Q(\varphi)}> Q(S^{-1}R) \end{CD}$$ commute.

In particular, if $S=R\setminus\mathfrak{p}$, where $\mathfrak{p}\subset R$ is prime, the commutative diagram is $$\begin{CD} R@>>> R_\mathfrak{p}\\ @VVV@VVV\\ Q(R)@>>>Q(R_\mathfrak{p}) \end{CD}$$ Suppose $x\in Q(R)$ satisfies an integral equation over $R$. Then we can map this equation to $Q(R_\mathfrak{p})$ to deduce that the image of $x$ in $Q(R_\mathfrak{p})$ is integral over $R$ and hence over $R_\mathfrak{p}$. Since we are assuming that $R_\mathfrak{p}$ is a normal domain, it follows that the image of $x$ in $Q(R_\mathfrak{p})$ lies in $R_\mathfrak{p}$.

Define $I=\{f\in R\mid fx\in R\}$, which is an ideal of $R$. To see that $I$ is the unit ideal, it suffices to see that $I$ in not contained in $\mathfrak{p}$. Write $x=g/s$ inside $Q(R)$, where $g,s\in R$ and $s$ is a nonzerodivisor. Then, inside $Q(R_\mathfrak{p})$, we can write $g/s=h/t$, where $h\in R$ and $t\in R\setminus\mathfrak{p}$, i.e., $gt=hs$ in $Q(R_\mathfrak{p})$ and therefore in $R_\mathfrak{p}$. Hence, there is $u\in R\setminus\mathfrak{p}$ with $gtu=hsu$ in $R$. Thus, inside $Q(R)$, we can write $\frac{g}{s}tu=hu\in R$, so $tu\in I\cap (R\setminus\mathfrak{p})$. In particular, $I\not\subset\mathfrak{p}$.

Example suggestion: Let $R$ be a local ring and $\mathfrak{p}$ a prime ideal which is not maximal. Then, the localization map $R \to R_{\mathfrak{p}}$ is not a local ring map.

Shouldn't it be $U(\psi\circ\varphi)=r_{\psi}^{-1}(U(\varphi))$? I'd expect to obtain $U(\psi\circ\varphi)$ by first forming the open $U(\varphi)$ in $Y$, and then pulling it back to $Z$ to form $r_{\psi}^{-1}(U(\varphi))= U(\psi)\times_{r_{\psi},Y}U(\varphi)$.

There's a small typo in the second line of the second paragraph of the proof of Theorem 05A9. Should be $M_{\alpha}$ I think?