products and smooth/étale/unramified morphisms - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T09:39:49Z http://mathoverflow.net/feeds/question/14799 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/14799/products-and-smooth-etale-unramified-morphisms products and smooth/étale/unramified morphisms Wanderer 2010-02-09T19:22:54Z 2010-04-24T02:31:40Z <p>Let $X$, $Y$ and $Z$ be Noetherian schemes. </p> <p>If $f: Y \to X$ and $g: Z \to X$ are morphisms of finite type, such that at each point of $X$, at least one of the two morphisms is smooth/étale/unramified (at all points of its inverse image), can we conclude that the induced morphism $Y \times_X Z \to X$ is smooth/étale/unramified everywhere? </p> <p>If not, which results can we obtain?</p> <p>(In his textbook on Algebraic Geometry, Liu asks to prove that the answer is always "yes"...)</p> <p>EDIT. So, indeed, the problem statement in the book is wrong...</p> http://mathoverflow.net/questions/14799/products-and-smooth-etale-unramified-morphisms/14802#14802 Answer by Emerton for products and smooth/étale/unramified morphisms Emerton 2010-02-09T19:44:29Z 2010-04-24T02:31:40Z <p>No. As an extreme example, suppose that $g$ is the identity (which is etale everywhere), and that $f$ is not etale at some point. Then the fibre product is just $f$ again.</p> <p>But in fact, this is essentially the general case. If $g$ is etale (or smooth) at a point, then it is etale (resp. smooth) in a n.h. of that point, so we may replace $Z$ by the n.h. and so assume that $g$ is etale everywhere. Then if $f$ is not etale (or smooth) at a point $y \in Y$, the product will not be etale in a n.h. of $y \times Z.$</p> <p>(Imagine that $Y$ was e.g. a nodal curve with a node at $y$, and that $Z$ is a smooth curve. (Here $X$ is Spec of the ground field.) Then $Y\times Z$ is the product of a nodal curve and a smooth curve, which just looks like a cylinder over the nodal curve; it is singular all along the "cylinder" over the node.)</p> http://mathoverflow.net/questions/14799/products-and-smooth-etale-unramified-morphisms/14810#14810 Answer by Qing Liu for products and smooth/étale/unramified morphisms Qing Liu 2010-02-09T21:23:36Z 2010-02-09T22:57:34Z <p>A correct statement would be : suppose Y is unramified/étale/smooth over X, then $Y\times_X Z\to X$ is unramfied/étale/smooth iff $Z\to X$ is. </p> <p>[EDITED] According to Matt's remark below, we must suppose $Y\to X$ surjective. </p> http://mathoverflow.net/questions/14799/products-and-smooth-etale-unramified-morphisms/14812#14812 Answer by Harry Gindi for products and smooth/étale/unramified morphisms Harry Gindi 2010-02-09T21:33:11Z 2010-02-09T22:24:15Z <p>Smooth, unramified, and etale morphisms are stable under base change, so $Y\times_X Z \to Z$ is {smooth, unramified, etale} if $Y\to X$ is {smooth, unramified, etale}.</p>