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?

• 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. – Macadam 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? – Macadam 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

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.