Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have the following definition for normalization of scheme: Let $X$ a integral scheme and $L\supseteq K(X)$ an algebraic extension. So $\pi:X'\to X$ is a normalization of $X$ in $L$ if $X'$ is normal, $K(X')=L$, $\pi$ is integral and $\pi$ extend the canonical map $\mathrm{Spec}(L)\to X$.

My first problem is to prove the uniqueness. My idea was to get it with a universal property: $\pi:X'\to X$ is the normalisation of $X$ in $L$ iff for all $Y$ normal with $K(Y)=L$, $Y\to X$ integral there a unique $Y\to X'$ so that the diagram we think is commutative. Is it that? I'm not sure because I can't verify it because of my second problem. If my approach via univeral property were false, how get the uniqueness?

My second problem was the affine case. I suspect that if $X$ is affine with $X=\mathrm{Spec}(A)$ we have $X'=\mathrm{Spec}(A')$ with $A'$ the integral closure of $A$ in $L$. Whatever the definition (of normality in $L$) that I take I have to check the integrity of $f:X'\to X$ that is: for all $U\subseteq X$ open we have $f^{-1}(U)$ affine and $\mathcal{O}_{X'}(f^{-1}(U))$ integral over $\mathcal{O}_X(U)$. My problem is to check the first part: why for all $U\subseteq X$ open we have $f^{-1}(U)$ affine?

share|improve this question
The property of a morphism of schemes being affine (in the sense that $f^{-1}(U)$ is affine whenever $U$ is an affine open of the target of $f$) is affine local on the target in the sense that, if there is an affine open covering $\bigcup_iU_i$ of the target such that $f^{-1}(U_i)$ is affine for all $i$, then $f$ is affine. This implies that a morphism between affine schemes is affine. –  Keenan Kidwell Feb 20 '13 at 15:08
Thanks, it seems to solve my problem. But it seems also that the locality of affiness need the notion of quasi-coherent sheave (to be proved), notion that I do not yet study. I can admit that for now. –  Gabriel Soranzo Feb 20 '13 at 15:36
I don't understand what you mean by "the notion of quasi-coherent sheave (to be proved)," but you don't need anything about quasi-coherent sheaves. You just need to prove that if $X$ is a scheme with $f_1,\ldots,f_n\in\mathscr{O}_X(X)$ such that $X=\bigcup_{i=1}^nX_{f_i}$ and each $X_{f_i}$ is affine, then $X$ is affine. Here $X_f=\{x\in X:f_x\notin\mathfrak{m}_x\}$ is the locus where $f$ ``doesn't vanish." This is the analogue for a general scheme $X$ of a standard open (and when $X=\mathrm{Spec}(A)$, $f\in A$, $X_f=D(f)$ is the standard open associated to $f$). –  Keenan Kidwell Feb 20 '13 at 16:40
Well... I undertstand that if $X=\cup_i U_i$ where the $U_i$ are affine then one can then write $X=\cup_i\cup_{j=1}^n (U_i)_{f_{i,j}}$ but I don't understand how one can deduce that the $(U_i)_{f_{i,j}}$ become "global" (I mean that the $f_{i,j}\notin\mathcal{O}_X(X)$) and how the union become finite. And of course I don't see how to deduce the affiness of $X$... Maybe the better were that someone give me a reference that contain the details of that? –  Gabriel Soranzo Feb 20 '13 at 20:29
What I should have said above was that $(f_1,\ldots,f_n)=\mathscr{O}_X(X)$, which implies that $X=\bigcup_i X_{f_i}$. I've given the details in an answer. –  Keenan Kidwell Feb 20 '13 at 20:51
add comment

1 Answer

up vote 0 down vote accepted

Claim: If $X$ is a scheme with global sections $f_1,\ldots,f_n$ which generate $\mathscr{O}_X(X)$ and such that each $X_{f_i}$ is affine, then $X$ is affine.

Proof: First of all, $X$ is quasi-compact since it can be covered by finitely many affine opens. Next, since $X_{f_i}$ is affine, for any $g\in\mathscr{O}_(X)$, $X_g\cap X_{f_i}=X_{gf_i}$ is a standard open in $X_{f_i}$ (the standard open associated to the image of $g$ in $\mathscr{O}_X(X_{f_i})$). In particular, $X_{gf_i}$ is affine. Applying this with $g=f_j$, we get that $X_{f_ifj}=X_{f_i}\cap X_{f_j}$ is affine for all $i,j$, so $X$ is quasi-separated. For any qcqs scheme $X$, the natural map $\mathscr{O}_X(X)_f\rightarrow\mathscr{O}_X(X_f)$ is an isomorphism for all $f\in\mathscr{O}_X(X)$. The pullback of the natural morphism $X\rightarrow\mathrm{Spec}(\mathscr{O}_X(X))$ along the open subscheme $D(f_i)$ is a morphism $X_{f_i}\rightarrow D(f_i)$. (If $f:S\rightarrow S^\prime$ is a morphism of schemes and $V\subseteq S^\prime$ is an open subscheme, then the pullback of $f$ along $U$ is by definition the induced morphism $f^{-1}(V)\rightarrow V$, which can be identified with the base change of $f$ along $V\hookrightarrow S^\prime$, thus the term ``pullback.") We have $D(f_i)=\mathrm{Spec}(\mathscr{O}_X(X)_{f_i})$, and this is canonically identified with $\mathrm{Spec}(\mathscr{O}_X(X_{f_i}))$ by what I've said above. The morphism $X_{f_i}\rightarrow D(f_i)$ can then be identified with the natural morphism $X_{f_i}\rightarrow\mathrm{Spec}(\mathscr{O}_X(X_{f_i}))$, which is an isomorphism by assumption (this is what it means for $X_{f_i}$ to be affine). Thus $X_{f_i}\rightarrow D(f_i)$ is an isomorphism. The condition that $(f_1,\ldots,f_n)=\mathscr{O}_X(X)$ is equivalent to $\mathrm{Spec}(\mathscr{O}_X(X))=\bigcup_{i=1}^n D(f_i)$. So we have a morphism $X\rightarrow\mathrm{Spec}(\mathscr{O}_X(X))$ and an open cover of the target such that the pullback along each open in the cover is an isomorphism. It follows that the morphism is itself an isomorphism, so $X$ is affine.

This is the key fact that is needed to prove that affineness of a morphism is affine local on the target in the sense I described in the comments to the question. If you would like more details, see the exercises to the second section in the second chapter of Harthshorne. Also this is almost certainly proved in great detail in the Stacks Project, probably in the chapter on morphisms (wherever affine morphisms are treated).

share|improve this answer
Thanks. My second problem is solved. I repost my first problem to separate my questions. –  Gabriel Soranzo Feb 21 '13 at 14:35
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.