Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$, $Y$ and $Z$ be Noetherian schemes.

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?

If not, which results can we obtain?

(In his textbook on Algebraic Geometry, Liu asks to prove that the answer is always "yes"...)

EDIT. So, indeed, the problem statement in the book is wrong...

share|improve this question
    
each point of $Y$'' should read each point of $X$'', I guess. –  Emerton Feb 9 '10 at 19:39
    
Yes, sorry. I'll correct it. –  Wanderer Feb 9 '10 at 19:40
add comment

3 Answers

up vote 6 down vote accepted

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.

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 $y \times Z.$

(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.)

share|improve this answer
    
Thanks. I wondered whether I was missing something; this is exercise 3.11 (page 145) of Liu's book, in which he asks to prove that the result is true - i.e. that $Y \times_X Z \to X$ is always smooth/étale/unramified. –  Wanderer Feb 9 '10 at 19:48
    
So the exercise is false, or am I missing something? –  Georges Elencwajg Feb 9 '10 at 20:34
    
Yes, that's the point. –  Wanderer Feb 9 '10 at 20:57
1  
Scrogneugneu (that's for me). Thanks Arne and Matt. The exercise is completely wrong as explains clearly Matt. If I remember well, originally I though about the case of two smooth curves Y, Z which are finite covers of a smooth curve X. Then $Y\times_X Z\to X$ is <b>ramified</b> at every point $(y,z)$ in the product such that one is étale and the other one is ramified. I don't know how this s.. heu, incorrect statement came. –  Qing Liu Feb 9 '10 at 21:04
    
Tiens,les djeunz disent encore scrogneugneu? (that's for Qing Liu) –  Georges Elencwajg Feb 9 '10 at 21:22
show 1 more comment

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.

[EDITED] According to Matt's remark below, we must suppose $Y\to X$ surjective.

share|improve this answer
2  
Do you need to assume that $Y \to X$ is surjective? Otherwise, if $Y$ was an open subset with non-empty complement, and $Z\to X$ was any map with image landing in the complement, the fibre product would be empty, hence have all possible properties, but $Z \to X$ could be quite bad. –  Emerton Feb 9 '10 at 22:02
    
Yes, this is clear. thanks ! –  Qing Liu Feb 9 '10 at 22:24
add comment

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}.

share|improve this answer
    
This follows formally from the fact that {smooth, unramified, etale} morphisms are stable under base change by the tensor product in the opposite category of commutative rings. The etale topology on Sch is merely the extension of the etale topology from CommRing^op. This follows from SGA 4.1.ii.2.5 and 4.1.ii.5 –  Harry Gindi Feb 9 '10 at 21:42
    
Without this fact, etale morphisms don't form a grothendieck topology. –  Harry Gindi Feb 9 '10 at 21:44
    
Your answer addresses part of the problem of showing that if $Y \to X$ smooth/etale/unram., then $Z\to X$ is smooth/etale/unram. if and only if $Y\times_X Z \to X$ is smooth/etale/unram., namely it addresses a part of the only if direction. (This is the revised exercise suggested in Qing Liu's answer.) It doesn't fully address that revised question, though (e.g. the converse is slightly more subtle, I think) and it doesn't address the original question at all. (Without wanting to speak for whoever downvoted your answer, this may go some way to explaining the downvote). –  Emerton Feb 9 '10 at 21:57
    
Incidentally, Qing Liu (the author of the book) has already stated that the exercise was in error. –  Emerton Feb 9 '10 at 22:21
1  
sure, but I was under the impression that you're really only supposed to vote down answers that are wrong. –  Harry Gindi Feb 9 '10 at 22:22
add comment

Your Answer

 
discard

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.