# Hartshorne's Exercise II.4.5(c). A third time.

It seems that I am the third one to pose this question. I think the proof in this question has a gap. He have not shown the lifting he constructed is compatible with $$\mathrm{Spec}k(p) \rightarrow X$$. And the proof given in wikiproof have not shown why Z satisfies the property as described in the exercise, which in my opinion is the most key part.

Any way to fill the gap is admired.

• Let me just add that this was asked about recently here with an incomplete answer. (I think you've probably seen this, since your title says "a third time" - I'm mostly adding the link to make it easy to resolve all of these questions if/when someone posts a full answer to one of them.) Nov 4 '20 at 8:36
• @KReiser I give a semi-proof. You may check out. Nov 7 '20 at 3:29


Warning: this post is somewhat long. If you have ideas on how to improve it or shorten it up while staying relatively true to the material presented in Hartshorne, please do not hesitate to say something in the comments.

First, a preliminary about valuation rings:

Lemma. If $$A\subset K$$ is a subring of a field, then $$A$$ is a valuation ring iff for every nonzero $$x\in K$$, at least one of $$x$$ and $$x^{-1}$$ belongs to $$A$$.

Proof. The forward direction is clear: $$x\cdot x^{-1}=1$$, so $$v(x)+v(x^{-1})=0$$ in the value group and therefore at least one of $$v(x),v(x^{-1})$$ is non-negative. For the reverse direction, if $$A\neq K$$, then $$A$$ has a nonzero maximal ideal $$\mathfrak{m}$$. If there's another maximal ideal $$\mathfrak{m}'$$, then we can find $$x\in\mathfrak{m}$$, $$y\in\mathfrak{m}'$$ with $$x\notin\mathfrak{m}'$$ and $$y\notin\mathfrak{m}$$. Then neither $$x/y$$ or $$y/x$$ can be in $$A$$, contradicting the asumption, so $$A$$ has a unique maximal ideal. Now suppose $$A'$$ is a local ring dominating $$A$$, and suppose $$x\in A'$$ - we have to show $$x\in A$$. If not, then $$x^{-1}\in A$$ and in fact $$x^{-1}\in\mathfrak{m}$$, so $$x,x^{-1}\in A'$$. But this means that $$x^{-1}$$ goes from being in the maximal ideal of $$A$$ to being a unit in $$A'$$, which is impossible because $$A'$$ dominates $$A$$. The claim is proven. $$\blacksquare$$

To solve the problem, we'll show that the conditions about centers of valuations on $$k(X)/k$$ give the appropriate conditions for the valuative criteria for separatedness and properness. We start by explaining how to connect the left side of the diagram in the valuative criteria to valuations of $$k(X)/k$$. Let $$R$$ be a valuation ring with field of fractions $$L$$ and suppose we have the standard commutative diagram:

$$\require{AMScd}$$ $$\begin{CD} \Spec L @>>> X\\ @VVV @VVV\\ \Spec R @>>> \Spec k \end{CD}$$

Let $$z$$ denote the unique point in the image of $$\Spec L\to X$$, and let $$Z$$ be the closure of $$z$$ equipped with the reduced induced subscheme structure. Then $$Z$$ is an integral subscheme of $$X$$ with generic point $$z$$ and function field $$k(Z)=\k(z)$$, the residue field at $$z$$. The map $$\Spec L \to X$$ gives us an inclusion $$\k(z)\subset L$$, and we let $$S=\k(z)\cap R$$. It is straightforward to check that $$S$$ is a valuation ring: for any element $$a\in \k(z)=\Frac(S)$$, we have that considering $$a$$ as an element of $$L$$, we have that either $$a\in R$$ or $$a^{-1}\in R$$, so either $$a\in S$$ or $$a^{-1}\in S$$. So our diagram can be rewritten as follows:

$$\require{AMScd}$$ $$\begin{CD} \Spec L @>>> \Spec \k(z) @>>> X\\ @VVV @VVV @VVV\\ \Spec R @>>> \Spec S @>>> \Spec k \end{CD}$$

and therefore it's enough to check the valuative criteria on valuation rings $$R$$ with fields of fractions $$L$$ the residue field of a point $$z$$ in $$X$$.

In the case where $$z$$ is the generic point of $$X$$, this gives the answer immediately: $$R$$ is a valuation ring of $$k(X)$$, so by assumption it has at most one (respectively, a unique) center $$x\in X$$, which means that there exists at most one (respectively, a unique) lifting $$\Spec R\to X$$ making the relevant diagram commute by lemma II.4.4. To show the result in general, we will prove that if $$X$$ is an integral scheme of finite type over a field so that every valuation on $$k(X)/k$$ has at most one (respectively, a unique) center on $$X$$, then the same is true for every integral closed subscheme $$Z\subset X$$. We can reduce this further to proving that if $$X$$ satisfies the condition on valuations, then every closed integral subscheme $$Z\subset X$$ of codimension one does as well by downward induction. To show this, we start with a lemma.

Lemma. Let $$f:X\to Y$$ be a proper dominant (equivalently, proper surjective) morphism of integral schemes over $$k$$. Every valuation on $$k(X)/k$$ has at most one (respectively, a unique) center on $$X$$ iff the same is true for valuations of $$k(Y)/k$$ and $$Y$$.

Proof.

We have four things to prove:

• (i) If every valuation on $$k(X)/k$$ has at most one center, then every valuation on $$k(Y)/k$$ has at most one center;
• (ii) If every valuation on $$k(X)/k$$ has a center, then every valuation on $$k(Y)/k$$ has a center;
• (iii) If every valuation on $$k(Y)/k$$ has at most one center, then every valuation on $$k(X)/k$$ has at most one center;
• (iv) If every valuation on $$k(Y)/k$$ has a center, then every valuation on $$k(X)/k$$ has a center.

Let $$R$$ be a valuation ring for $$k(Y)/k$$. As $$f:X\to Y$$ is dominant, it maps the generic point of $$X$$ to the generic point of $$Y$$ and thus induces an injection of fields $$k(Y)\hookrightarrow k(X)$$ which we may assume to be an inclusion. Let $$R'$$ be a valuation ring of $$k(X)$$ dominating $$R\subset k(X)$$. (We note that this also implies that $$R'\cap k(Y)=R$$: if $$k(Y)\cap R'$$ contained an element $$e$$ not in $$R$$, then $$e^{-1}$$ is in $$R$$, therefore $$\mathfrak{m}_{R'}\cap R\neq \mathfrak{m}_R$$ which contradicts the fact that $$R'$$ dominates $$R$$.) By lemma II.4.4, a center $$y$$ for $$R$$ on $$Y$$ is equivalent to $$R$$ dominating $$\cO_{Y,y}$$, which implies $$R'$$ dominates $$\cO_{Y,y}$$ as subrings of $$k(X)$$. This means we have a valuative diagram

$$\require{AMScd}$$ $$\begin{CD} \Spec k(X) @>>> X\\ @VVV @VVV\\ \Spec R' @>>> Y \end{CD}$$

and by the valuative criteria for properness, we have a unique lifting $$\Spec R'\to X$$. Thus for every center $$y\in Y$$ of $$R$$, we get a unique center $$x\in X$$ of $$R'$$. So if there is at most one center in $$X$$ for all valuation rings $$R'$$ of $$k(X)/k$$, then there is at most one center in $$Y$$ for any valuation ring $$R$$ on $$Y$$, and (i) is proven.

To show that the existence of a center for all valuation rings $$R'$$ of $$k(X)/k$$ implies the existence of a center for all valuation rings of $$k(Y)/k$$, let $$R$$ and $$R'$$ be as in the previous paragraph. As $$R'$$ has a center on $$X$$ by assumption, we get a map $$\Spec R'\to X$$ by lemma II.4.4, and composing with the map $$X\to Y$$, we obtain a map $$\Spec R'\to Y$$. I claim that $$\Spec R'\to Y$$ factors through $$\Spec R\to Y$$. This can be seen from examining the maps on local rings: letting $$x\in X$$ be the image of the closed point of $$\Spec R'$$ and $$y\in Y$$ the image of $$x$$, we have a sequence of local maps of local rings $$\cO_{Y,y}\to \cO_{X,x}\to R'$$ which are all injections because the maps on function fields are. But $$\cO_{Y,y}$$ lands inside $$k(Y)\subset k(X)$$, which means it lands inside $$k(Y)\cap R'$$, which is exactly $$R$$. Therefore $$R$$ dominates $$\cO_{Y,y}$$, so $$y$$ is a center for $$R$$ and (ii) is proven.

Now suppose $$R'$$ is a valuation ring for $$k(X)/k$$ with two centers $$x,x'$$. Then by the logic above, the images of $$x$$ and $$x'$$ must be centers for $$R=R'\cap k(Y)$$, and $$x$$ and $$x'$$ cannot map to the same point, otherwise this would violate our result that every center for $$R$$ uniquely determines a center for $$R'$$. Therefore we've proven (iii) by contrapositive.

Finally, if $$R'$$ is a valuation ring for $$k(X)/k$$, then $$R=R'\cap k(Y)$$ is a valuation ring for $$k(Y)/k$$, and so if $$R$$ has a center, we get a valuative diagram as above. As $$X\to Y$$ is proper, we get a lifting $$\Spec R\to X$$, and considering the composition $$\Spec R'\to \Spec R\to X$$, we see that $$R'$$ has a center on $$X$$ by lemma II.4.4, so we've proven (iv) and we're done. $$\blacksquare$$

Now recall the construction of the normalization from exercise II.3.8: for a $$X$$ an integral $$k$$-scheme of finite type, we let $$X'$$ denote the normalization, which comes with a natural dominant map $$\nu:X'\to X$$ which is finite in our case. By exercise II.4.1, a finite map is proper, so the map $$\nu:X'\to X$$ verifies the conditions of the lemma (we also note $$\nu$$ is surjective).

Let $$Z\subset X$$ be a codimension one integral closed subvariety. As finite and surjective morphisms are stable under base change, we have that $$Z\times_X X'\to Z$$ is finite and surjective. Since finite morphisms are closed, each irreducible component of $$Z\times_X X'$$ must map to a closed irreducible subset of $$Z$$, and therefore by surjectivity there must be an irreducible component of $$Z\times_X X'$$ which surjects on to $$Z$$. Let $$Z'\subset X'$$ be such an irreducible component equipped with the reduced induced subscheme structure. As $$Z'\to Z\times_X X'$$ is a closed immersion, it is finite, so the composite $$Z'\to Z$$ is a finite surjective morphism, and as finite morphisms are proper by exercise II.4.1, this satisfies the conditions of our lemma. So it suffices to show that if $$Z$$ is a closed codimension one integral subscheme of a normal integral scheme $$X$$ of finite type over a field, then the condition about valuations on $$X$$ implies the condition about valuations on $$Z$$.

Let $$Z\subset X$$ as described in the previous sentence, and let $$z\in Z$$ be the generic point of $$Z$$. Note that $$\cO_{X,z}$$ is a DVR by theorem I.6.2A: it's a noetherian local domain of dimension one which is integrally closed. Suppose $$R\subset k(Z)=\k(z)$$ is a valuation ring trivial on $$k$$, and let $$q:\cO_{X,z}\to \cO_{Z,z}=\k(z)$$ be the natural quotient map. Define $$S=q^{-1}(R)\subset\cO_{X,z}$$. I claim $$S$$ is a valuation ring. Let $$e\in k(X)$$ be an arbitrary nonzero element. As at least one of $$e,e^{-1}$$ are in $$\cO_{X,z}$$, we may assume $$e$$ is actually in $$\cO_{X,z}$$. If $$e\in\mathfrak{m}_z$$, then $$e\in S$$. If $$e\notin\mathfrak{m}_z$$, then $$e,e^{-1}\in\cO_{X,z}$$ and so at least one of $$q(e)$$ or $$q(e^{-1})=q(e)^{-1}$$ is in $$R$$, which implies that one of $$e$$ or $$e^{-1}$$ is in $$S$$, which implies it is a valuation ring.

Suppose $$z'\in Z$$ is a center for $$R$$ on $$Z$$. Then $$\cO_{X,z'}=q^{-1}(\cO_{Z,z'})$$ is dominated by $$S$$, so $$z'$$ is a center for $$S$$ on $$X$$. As $$Z\to X$$ is injective on underlying sets, this implies that if $$R$$ has two distinct centers on $$Z$$, $$S$$ must have two distinct centers on $$X$$. This shows that if every valuation on $$k(X)/k$$ has at most one center on $$X$$, then every valuation on $$k(Z)/k$$ has at most one center on $$Z$$. Now suppose $$z'\in X$$ is a center for $$S$$: then $$\cO_{X,z'}\subset S\subset \cO_{X,z}$$, so $$z'\in \ol{z}=Z$$ and taking the quotient by the maximal ideal of $$\cO_{X,z}$$ we see that $$\cO_{Z,z'}$$ is dominated by $$R$$. We are done.

• It might be worthwile to note that $\mathcal O_{Z, z} = k(Z) = \kappa(z)$ though. That just took me some minutes to see :D Sep 9 at 8:18
• Sure, that's a reasonable suggestion. Glad the answer is helping you out. Sep 9 at 8:20

Let me give an outline of proof, which is from my teacher. Indeed my knowledge is insufficient to fill the proof. I can just tell the geometry meaning of the proof.

First we can choose a completion of $$X$$, denoting by $$\tilde{X}$$. Then $$K(X) = K(\tilde{X})$$ (or they are similar. In fact, I don't know what exactly happens when taking completion. Maybe something like blowing up at $$Z$$ below. But I don't know blowing up neither.). Now we have a diagram $$\require{AMScd} \begin{CD} \mathrm{Spec}(L) @>>> X;\\ @VVV @VVV \\ \mathrm{Spec}(S) @>>> \mathrm{Spec}(k). \end{CD}$$ The image of $$\mathrm{Spec}(L)$$ gives an irreducible closed subset $$Z$$ of $$X$$, whose generic point is $$z$$. We can pull back $$S$$ to be a valuation ring $$S'$$ of $$k(z)$$, and subsequently pull back $$S'$$ to be a ring in $$\mathcal{O}_{X,z}\subset K(X)$$, namely $$S''$$. Taking the maximum with respect to domination, we have a valuation ring $$R$$ of $$K/k$$, whose restriction to $$\mathcal{O}_{X,z}$$ is $$S''$$. Now $$R$$ dominates some $$\mathcal{O}_{X,x}$$ by hypothesis.

The problem is $$x$$ may not in $$Z$$. But for the commutative diagram $$\require{AMScd} \begin{CD} \mathrm{Spec}(L) @>>> \tilde{X};\\ @VVV @VVV \\ \mathrm{Spec}(S) @>>> \mathrm{Spec}(k), \end{CD}$$ and because of the properness of $$\tilde{X}$$, we always have a lifting. By doing the same thing above, we have $$\tilde{x} \in \tilde{Z}$$, and $$\mathcal{O}_{\tilde{X},\tilde{x}}$$ is dominated by $$R$$. So as the counterpart corresponding to $$\tilde{x}$$, $$x$$ is in $$Z$$.

• I think that as currently written this has too many gaps/typos/holes to really resolve the question. It's clearly supposed to be along the lines of the proof given in EGA II 7.3.10, but it needs improvement. Perhaps you can ask for more details and improve the post. Nov 7 '20 at 4:19