When will the pushforward of a structure sheaf still be a structure sheaf? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T09:51:50Z http://mathoverflow.net/feeds/question/63301 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/63301/when-will-the-pushforward-of-a-structure-sheaf-still-be-a-structure-sheaf When will the pushforward of a structure sheaf still be a structure sheaf? YOURS 2011-04-28T15:42:10Z 2011-04-30T04:42:32Z <p>Let $f:X\rightarrow Y$ be a morphism of schemes.</p> <ol> <li><p>When $PicY\rightarrow PicX$ is an embedding and $f_{*}\mathscr{O}_{X}$ is invertible, it is the structure sheaf of $Y$.</p></li> <li><p>In the proof of Zariski's Main Theorem, we have: If $f$ is birational, finite, integral, and $Y$ is normal, then $f_{*}\mathscr{O}_{X}$ is the structure sheaf of $Y$.</p></li> </ol> <p>My questions are</p> <p>1) What exactly prevent $f_{*}\mathscr{O}_{X}$ to be a structure sheaf?</p> <p>2) Is there any necessary and sufficient condition(s) guarantee that $f_{*}\mathscr{O}_{X}$ is a structure sheaf?</p> http://mathoverflow.net/questions/63301/when-will-the-pushforward-of-a-structure-sheaf-still-be-a-structure-sheaf/63308#63308 Answer by J.C. Ottem for When will the pushforward of a structure sheaf still be a structure sheaf? J.C. Ottem 2011-04-28T16:34:35Z 2011-04-28T17:10:17Z <p>If $f:X\to Y$ is a proper morphism of noetherian shemes, then $f_*O_X=O_Y$ says that the fibers of $f$ are connected. This follows from a general form of Zariski's main theorem (Hartshorne III.11.3). </p> <p>Conversely, if $Y$ is in addition normal, then $f_*O_X=O_X$ holds. Indeed, there is a Stein factorization of the form $$X \xrightarrow{f'} Z={\bf Spec} (f_* O_X) \xrightarrow{g} Y$$where $g$ is finite and $f'$ has connected fibers. Furthermore $g_*O_Z=O_Y$ and ${f'}_*O_X=O_Z$. If the fibers of $f$ are connected, then $g$ must be birational (by Hartshorne III.10.3) and is in fact an isomorphism if $Y$ is normal. It follows that $f_*O_X=O_Y$ if and only if $f$ has connected fibers. </p> http://mathoverflow.net/questions/63301/when-will-the-pushforward-of-a-structure-sheaf-still-be-a-structure-sheaf/63344#63344 Answer by Karl Schwede for When will the pushforward of a structure sheaf still be a structure sheaf? Karl Schwede 2011-04-28T21:14:42Z 2011-04-30T04:42:32Z <p>Let me try to write an informal explanation as to why (and why not) you might have $f_* \mathcal{O}_X = \mathcal{O}_Y$. This is basically what J.C. Ottem wrote, but I'm trying to explain the reason at a slightly more philosophical level.</p> <p>Now $O_X$ is the sheaf of regular functions on $X$. Given an open set $U \subseteq Y$, the sections $\Gamma(U, f_* \mathcal{O}_X)$ is just $\Gamma(f^{-1}(U), \mathcal{O}_X)$. For this to be viewed as even a subset of functions on $U$, you would expect it to be constant / well-defined at the points of $U$. So consider some (closed) point $z \in U$. Therefore, you need a section $\sigma \in \Gamma(f^{-1}(U), \mathcal{O}_X)$ to be constant on the fiber $f^{-1}(z)$. Since $f$ is proper, this fiber is also proper, and thus the only sections are constant. I just lied of course, the only sections are the functions that are constant on each <em>connected component</em> of the fiber.</p> <p>Thus if you have fibers with multiple connected components, then you will expect that some of the sections $\sigma$ might be able to distinguish those connected components, and thus those sections of $f_* \mathcal{O}_X$ can't be viewed as functions on $Y$.</p> <p>Why does normality come into play? Well, the picture isn't quite as simple as what I just described. If a scheme $Z$ is non-normal, and its normalization $Z' \to Z$ is injective/bijective (for example, the normalization of the cusp), then you should view that normalization map as the inclusion of all the algebraic functions'' which can be defined on the points.</p> <p>In fact, given any scheme $Z$ over an algebraically closed field of characteristic zero, the <em>seminormalization</em> $Z'$ of $Z$ can be exactly described as the scheme whose structure sheaf has all functions that make sense on the closed points of $Z$.''</p> <p>This is the point of view on seminormalization is described in: Leahy and Vitulli, <em>Seminormal rings and weakly normal varieties.</em> Nagoya Math. J. 82 (1981), 27–56</p> http://mathoverflow.net/questions/63301/when-will-the-pushforward-of-a-structure-sheaf-still-be-a-structure-sheaf/63371#63371 Answer by Sándor Kovács for When will the pushforward of a structure sheaf still be a structure sheaf? Sándor Kovács 2011-04-29T02:49:13Z 2011-04-29T02:49:13Z <p>Another issue that has not been addressed is what happens if $f$ is not proper. You may have intended to assume that it is, but it also an interesting question for not necessarily proper morphisms. For that matter, you could ask "if $f:X\hookrightarrow Y$ is an open embedding, when will $f_*\mathscr O_X$ be isomorphic to $\mathscr O_Y$?" You are also writing that "... if $f_*\mathscr O_X$ is a line bundle, then ...". It should be noted that this is actually a strong restriction. For instance if you have a generically finite morphism that satisfies this, then it has to be birational.</p> <p>For the question of an open embedding the answer is relatively simple. If the complement of $X$ in $Y$ has a non-empty codimension $1$ part, then $f_*\mathscr O_X$ is not even coherent, so little chance there. If the complement is of codimension at least $2$, then this is a condition on the singularities of $Y\setminus X$, and essentially equivalent to $Y$ being $S_2$ along $X\setminus Y$. </p> http://mathoverflow.net/questions/63301/when-will-the-pushforward-of-a-structure-sheaf-still-be-a-structure-sheaf/63374#63374 Answer by roy smith for When will the pushforward of a structure sheaf still be a structure sheaf? roy smith 2011-04-29T03:19:40Z 2011-04-29T06:20:27Z <p><strong>Q</strong>: Exactly what information is contained in $f_*\mathscr O_X$? Look at the definition. For any $U\subseteq Y$ open, $f_*\mathscr O_X(U) = \mathscr O_X(f^{-1}(U))$ = regular functions on $f^{-1}(U)$. So the information in $f_*\mathscr O_X$ is related to the sets in $X$ of form $f^{-1}(U)$.</p> <p>Cases where $f_*\mathscr O_X$ contains as little information about $X$ as possible.</p> <p>If $X$ is irreducible and projective and $f$ is constant, e.g. if $Y$ is affine, then the only non empty set of form $f^{-1}(U)$ in $X$ is $X$ itself. In this case $f_*\mathscr O_X$ is a skyscraper sheaf with stalk $k$ supported on the image point of $f$ in $Y$. There is very little information here about $X$, but perhaps we do see that $f$ is constant and that $X$ is connected. More generally, if $Z$ is a projective variety, $Y$ is any variety, and $X = Z\times Y$, and $f:Z\times Y\to Y$ is the projection, then $f^{-1}(U) = Z\times U$, so an element of $f_*\mathscr O_X(U)$, i.e. a regular function on $f^{-1}(U)$, is determined by its restriction to $\{p\}\times U$ for any $p\in X$, i.e., a regular function on $U$ in $Y$. Thus in this case we have $f_*\mathscr O_X = \mathscr O_Y$. Consequently in this case $f_*\mathscr O_X$ recovers $Y$, but contains no information at all about $X$.</p> <p>In general, if $f:X\to Y$ is a projective morphism with every fiber connected, and $Y$ is any normal variety, then $f_*\mathscr O_X = \mathscr O_Y$, so again $f_*\mathscr O_X$ contains little information about $X$. Recall that if $X$ is a projective variety then every morphism out of $X$ is a projective morphism, and more generally a projective morphism $X\to Y$ is one that factors via an isomorphism of X with a closed subvariety of $\mathbb P^n\times Y$, followed by the projection $\mathbb P^n\times Y\to Y$. Suppose that $f:X\to Y$ is any projective morphism. Then the fibers $f^{-1}(y)$ over points $y \in Y$ are all finite unions of projective varieties. Therefore for any open set $U\subseteq Y$ containing the point $y$, the only regular functions in $\mathscr O_X(f^{-1}(U)) = f_*\mathscr O_X(U)$ are constant on every connected component of the fiber $f^{-1}(y)$. Thus $f_*\mathscr O_X$ can contain little information about $X$ and $f$, other than at most the connected components of the fibers. We shall see below that it contains exactly this information.</p> <p>Cases where $f_*\mathscr O_X$ contains as much information about $X$ as possible.</p> <p>If $f:X\to Y$ is a map of affine varieties, then the global sections of $f_*\mathscr O_X$ determine $X$ completely, since then $H^0(Y,f_*\mathscr O_X) = H^0(X,\mathscr O_X)$, and then $X = \mathrm{Spec}h^0(X,\mathscr O_X)$, is the unique affine variety with coordinate ring $H^0(X,\mathscr O_X)$. The generalization of this case is that of any affine map $f:X\to Y$, since then $X$ can be recovered by patching together the analogous construction from $H^0(U,f_*\mathscr O_X)$ for affine open sets $U\subseteq Y$. Thus $X$ is completely determined by $f_*\mathscr O_X$ for any affine map $f:X\to Y$, and this is essentially the only case. I.e. in general $f_*\mathscr O_X$ is always a quasi coherent $\mathscr O_Y$ algebra, and if we want it to determine a variety, as opposed to a "scheme", it is reasonable to assume for all $U\subseteq Y$ affine open, that $f_*\mathscr O_X(U)$ is a finitely generated k algebra, as well as an $\mathscr O_Y(U)$ algebra. We may call temporarily such an $\mathscr O_Y$ algebra "of finite type". Thus if $f:X\to Y$ is any morphism such that $f_*\mathscr O_X$ is of finite type, then the patching construction above yields not necessarily $X$, but a variety $Z$ and an affine map $h:Z\to Y$ which factors via a map $g:X\to Z$, where $f = h\circ g$, and where $g_*(\mathscr O_X) = \mathscr O_Z$. In particular then, we have <code>$f_*\mathscr O_X = (h\circ g)_*(\mathscr O_X) = h_*(g_*(\mathscr O_X))= h_*(\mathscr O_Z)$</code>. So since $h$ is affine, $f_*\mathscr O_X = h_*(\mathscr O_Z)$ determines not $X$, but $Z$. (Kempf, section 6.5.)</p> <p>The case of an arbitrary projective morphism.</p> <p>Now when $f:X\to Y$ is any projective morphism, then $f_*\mathscr O_X$ is a coherent $\mathscr O_Y$-module, hence we get a factorization of $f$ as $h\circ g:X\to Z\to Y$, where $h:Z\to Y$ is affine, and where also $h_*(\mathscr O_Z) = f_*\mathscr O_X$. Then $h$ is not only an affine map, but since $h_*(\mathscr O_Z)$ is a coherent $\mathscr O_Y$-module, $h$ is also a finite map. Moreover $g:X\to Z$ is also projective and since $g_*(\mathscr O_X) = \mathscr O_Z$, it can be shown that the fibers of $g$ are connected. Hence an arbitrary projective map $f$ factors through a projective map g with connected fibers, followed by a finite map $h$. Thus in this case, the algebra $f_*\mathscr O_X$ determines exactly the finite part $h:Z\to Y$ of $f$, whose points over $y$ are precisely the connected components of the fiber $f^{-1}(y)$.</p> <p>One corollary of this is "Zariski's connectedness theorem". If $f:X\to Y$ is projective and birational, and $Y$ is normal then $f_*\mathscr O_X= \mathscr O_Y$, and all fibers of $f$ are connected, since in this case $Z = Y$ in the Stein factorization described above. If we assume in addition that $f$ is quasi finite, i.e. has finite fibers, then $f$ is an isomorphism. More generally, if $Y$ is normal and $f:X\to Y$ is any birational, quasi - finite, morphism, then $f$ is an embedding onto an open subset of $Y$ ("Zariski's 'main theorem' "). More generally still, any quasi finite morphism factors through an open embedding and a finite morphism.</p>