# Does the image of the projection morphism of a fiber product contain an open set?

Let $f: X\times_S Y \to X$ be the projection morphism in the definition of fiber product.$U\subset X\times_S Y$ be an open set. Does $f(U)$ contain a non-empty open set of $X$? I know this can be reduced to the affine case.

If it is not true in general, can we save it by adding extra condition such as :$X,Y$ are noetherian, integral, etc.?

The problem comes from the attempt to prove: when $X$ is a noetherian integral separated scheme which is regular in codimension one, then $X\times_{\operatorname{Spec}\mathbb{Z}}\operatorname{Spec}(\mathbb{Z}[t])$ is also integral.

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Unfortunately, it is completely false that if $U\subset X\times_{S} Y$ is open, then $f(U)$ contains a non-empty open subset of $X$. Here is why.

Take $X=S$ and for $X\to S$ just take the identity $X=S \to S$.
Then the morphism $f: X\times_{S} Y \to X$ is just the initially given arbitrary morphism $u: Y\to S$, for which there is absolutely no reason that the image of an open subset should contain a non-empty open subset. For a completely concrete example, embed a point in a line:
$$u: Y=Spec(k) \to S=Spec(k[T])=\mathbb A^1_k : (0) \mapsto (T)$$.

The good news
Fortunately your initial question is not affected at all by what precedes:
If a scheme $X$ is integral, so is $X\times_{Spec(\mathbb Z)} Spec(\mathbb Z[T])$
This result boils down to the fact that if $D$ is a domain, then the polynomial ring $D[T]$ is a domain too: indeed, constructing the product can be done locally on open affines of $X$ . Note joyfully that absolutely no hypothesis of noetherianness, regularity, ... is required of $X$.

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Yes, I got you, thank you! –  Li Zhan Sep 13 '11 at 19:15