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Let $k$ be a field. Let $X$, $Y$ and $C$ be varieties over $k$ such that $X\times_k C $ is isomorphic to $Y\times_k C$. Assume that $C$ is a curve. (The varieties $X$, $Y$ and $C$ are assumed to be smooth projective and geometrically connected over $k$.)

Does it follow that $X$ is isomorphic to $Y$?

My guess is that $X\to X\times C\cong Y\times C\to Y$ gives an isomorphism. (The first map is the inclusion and the second map is the projection. For the first map to be defined you might want to assume $C(k)$ is non-empty.) Anyway, the inverse map should be given by the reverse construction: $Y\to Y\times C\cong X\times C\to X$.

Is this correct? (No if $X=Y=C$. But let's assume $\dim X = \dim Y \geq 2$.)

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Your argument is not correct. For instance, $X$ and $Y$ and $C$ can all be $\mathbb{P}^1$ and the isomorphism in the middle step can be the one that swaps the factors. Then your composite map is a constant map (it's true that $X$ and $Y$ are isomorphic in this case, but the map you wrote down isn't an isomorphism between them). – user29743 Dec 21 '12 at 18:51
What if I take $X$ and $Y$ to be of dimension at least $2$? Then this phenomenon can't happen. – Harry Dec 21 '12 at 19:05
My particular counterexample won't happen, but the proof is still not correct. The problem is that you have no control over the map from $X \times C$ to $Y \times C$. (I believe I have read that the result you are trying to establish is false, by the way, but I don't know a reference or how to look it up! Sorry not to be more helpful.) – user29743 Dec 21 '12 at 19:38
Thank you very much! Your example did convince me that this is not going to be true. A counterexample shouldn't be hard to find, but it will be cumbersome. – Harry Dec 21 '12 at 19:41
Related question:… – Quimey Dec 21 '12 at 20:12

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