Let $K$ be a field and $X\to Spec(K)$ and $Y\to Spec(K)$ and $Z\to Spec(K)$ three smooth morphisms of schemes.

I have no intuition of relative smoothness. What is an example of such schemes and (not necessary smooth) morphisms $Y\to X$ and $Z\to X$ such that $Y\times_X Z$ is not smooth over $Spec(K)$?

Are there conditions (apart from ''it is an isomorphism'') on $X \to Spec(K)$ such that there are no examples?


Here is an example where $Y\times_X Z$ is not smooth over $Spec(K) $.

Take $X=Spec(K[T])=\mathbb A^1_K, Y =Spec(K)=\lbrace *\rbrace $, and for $Z$ take the surface $Z=V(Y^2-X^3-T)\subset Spec(K[X,Y,T])=\mathbb A^3_K$.
The morphisms are $Z\to X:(x,y,t)\mapsto t$ and $Y\to X:*\mapsto 0 $.
The three schemes $X,Y,Z$ are smooth over $K$.
However the fiber product $Y\times_X Z$ is the cusp $Y^2-X^3=0$ in the plane $Spec(K[X,Y])$ and is thus not smooth over $K$, just as you required.

A smoothness criterion
No reasonable condition on $X$ alone will ensure that $Y\times_X Z$ is smooth: after all in the example above $X=\mathbb A^1_K$ and it is impossible to conceive of a smoother, better behaved $K$-scheme than the affine line!
The correct condition you want is that the morphism $Y\to X$ be smooth and that $Z$ be smooth over $K$.
Then the morphism $Y\times_X Z\to Z $ will be smooth too because smoothness is preserved by base change.
And the morphism $f: Y\times_X Z\to Spec(K) $ will be smooth, just as you wished: indeed the composition of smooth morphisms is smooth and this morphism $f$ is the composition of the smooth morphisms $Y\times_X Z\to Z $ and $Z\to Spec(K)$.
Notice carefully that you don't have to require that $X$ nor $Y$ be smooth over $K$ for the criterion to apply!

A criterion for the criterion
Yes, but how do we ensure that the morphism $f:Y\to X$ is smooth?
Answer: a necessary and sufficient condition for $f$ to be smooth is that $f$ be flat and that all fibers $f^{-1}(y)\subset X$ of closed points $y\in Y$ be smooth over $K$.

  • $\begingroup$ Can it be that $X$ is etale over $Spec(K)$ and there is still an example? $\endgroup$ – Daniel Dreiberg Sep 3 '12 at 7:55
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    $\begingroup$ Dear Daniel, if $X$ is étale over $K$ and if $Y$ is smooth over $K$, then $Y\to X$ is automatically smooth (EGA $IV_4$, 17.3.4), so that the "smoothness criterion" applies: in other words there is no (counter)example. $\endgroup$ – Georges Elencwajg Sep 3 '12 at 11:27
  • $\begingroup$ Thank you, your answers are always very enlightening! Is there a typo in the ''smoothness criterion'' confusing $X$ and $Y$? $\endgroup$ – Daniel Dreiberg Sep 3 '12 at 13:23
  • $\begingroup$ If $f$ goes from $Y$ to $X$, how can you consider the fiber $f^{-1}(y)$ of an element in $Y$ and how is this a subset of the target $X$? Maybe, I am just lost. $\endgroup$ – Daniel Dreiberg Sep 3 '12 at 19:57
  • $\begingroup$ Dear Daniel, you are absolutely right, there was a typo that I have just corrected: it should indeed be $f^{-1}(x)\subset Y$. The misunderstanding arose because you wrote that the typo was in "A smoothness criterion", whereas in fact it was in "A criterion for the criterion". Anyway, I am the only one responsible for this sillyness and I thank you for calling my attention to it. $\endgroup$ – Georges Elencwajg Sep 3 '12 at 20:25

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