The product of two projective varieties over $X,Y$ is the fibered product $X\times_{\mathbb{Z}} Y$. I want to show that the projections $X\times_{\mathbb{Z}} Y \to X$ and $X\times_{\mathbb{Z}} Y \to Y$ are smooth if $X,Y$ are smooth.This somehow eludes me. Can anyone help?

EDIT: First, I mean smooth over a field $k$. Second,for $S$-Objects $X\to S$ and $Y\to S$, the product is the object $X\times_{S} Y$. Hence for smooth $k$-schemes the product is $X\times_{k} Y$ and not $X\times_{\mathbb{Z}} Y$. So the result follows via base change as pointed out in the answer below.

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    $\begingroup$ Dear Carsten, You write about "varieties", which normally implies that you are working over a base-ring that is a field, but then you take the fibre product over $\mathbb Z$. So do you mean that $X$ and $Y$ are smooth schemes over $\mathbb Z$? Please clarify. Regards, $\endgroup$ – Matt E Jan 19 '12 at 21:39
  • $\begingroup$ Dear Matt, I mean smooth over $k$. But you are right on point: I am unsure what is normally meant by $X\times Y$. In Hartshorne it is $X\times_{\mathbb{Z}} Y$. So "product" means fibered product over $k$ for $k$-schemes? Thanks. $\endgroup$ – Carsten Jan 19 '12 at 21:47
  • $\begingroup$ Dear Carsten, Yes, for $k$-schemes, the product is taken over Spec $k$. (You can take the product over Spec $\mathbb Z$ if you want, but this is a different beast, and is not the usual product of varieties. E.g. Spec $k \times_{\mathbb Z} $ Spec $k$ is the Spec of the ring $k\otimes_{\mathbb Z} k.$ A moments reflection should show that e.g. $\mathbb C \otimes_{\mathbb Z} \mathbb C$ is a rather large ring. On the other hand, Spec $k \times_k $ Spec $k$ is precisely Spec $k$, and that is what you want in the category of varieties: a point cross a point should be a point.) Regards, $\endgroup$ – Matt E Jan 20 '12 at 0:25

Smoothness is preserved by base change.
So if $Y\to Spec(\mathbb Z)$ is smooth (in other words if $Y$ is smooth), so is $X\times_{\mathbb Z} Y\to X$.
(Projectiveness and being varieties are irrelevant.)

  • $\begingroup$ Since you now comment that you mean smooth over a field, I'll let you adapt my answer to that case. $\endgroup$ – Georges Elencwajg Jan 19 '12 at 21:51
  • $\begingroup$ thanks! Matt E's comment made me realize my mistake. $\endgroup$ – Carsten Jan 19 '12 at 21:59

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