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I have seen the following construction and I would be very happy if someone could explain its meaning to me.

We start from a smooth projective algebraic variety $X$ over a field of characteristic zero $k$ and a reduced effective divisor with simple normal crossings $D$. Let $V=Z−D$ and let $U \to V$ be an étale cover.

What's the meaning of taking $\pi: Y \to X$ the normalization of $X$ in the function field $k(U)$?

Does it mean that $U \hookrightarrow Y$ and that the complement is a divisor with normal crossings $E$ such that $\pi(E)=D$?

Thanks for your help

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2 Answers 2

up vote 4 down vote accepted

Let $X$ be a variety (a separated integral scheme) with function field $K = k(X)$, maybe assumed normal. Let $L$ be a finite separable extension of $K$. From this data, we can construct a variety $Y$ with $k(Y) = L$ together with a finite surjective map $\pi: Y\to X$, called the normalization of $X$ in $L$.

If $X$ is affine, equal to $\mathrm{Spec}(A)$, the construction is just $Y = \mathrm{Spec}(B)$ where $B$ is the integral closure of $A$ in $L$. By finiteness of integral closure, the resulting map $Y\to X$ is finite.

In the general case, cover $X$ by open affines, their intersections are affine again, so we can glue and obtain $Y$ this way. If you don't like gluing, you could define a sheaf of $\mathcal{O}_X$-algebras $\mathcal{A}$ by $\mathcal{A}(U) = $ integral closure of $\mathcal{O}_X(U)$ (and maybe sheafify) and then define $Y = \mathrm{Spec}_X \mathcal{A}$.

In your situation, you already have a part of $Y$ (you have a map $U\to X$ and define $L = k(U)$). This gives a map $Y\to X$ such that $U\to X$ factors through an open immersion $U\to Y$.

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Thanks for your answer! –  normali Dec 28 '12 at 22:18
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To make the globalization construction a bit tidier (in the same way that the viewpoint of representable functors tidies up global gluing constructions), one can also incorporate a universal property: any dominant map $X' \rightarrow X$ from a normal variety to $X$ equipped with a $k(X)$-inclusion $L \rightarrow k(X')$ uniquely factors through $\pi$. –  user30180 Dec 28 '12 at 23:53

$Y$ is a normal variety and $\pi$ is a finite morphism that extends $U\to V$. The branch locus of $\pi$ is a divisor contained in $D$. $Y$ can be singular over the singular points of $D$.

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Thanks for your answer –  normali Dec 28 '12 at 22:18
    
You're welcome! –  rita Dec 29 '12 at 8:06

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