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Suppose I have two affine varieties $X_1, X_2 \in \mathbb{A}^n_{\mathbb{C}}$. Let $U$ be a Zariski-open set. Suppose I know that $$ \emptyset \not = X_1 \cap U = X_2 \cap U. $$ Does it then follow that $\dim X_1 = \dim X_2$? I think it does... Any comments are appreciated. Thank you very much!

Edit: Without the assumption of irreducibility the statement is not true, as shown in the comments. So I would like to suppose we know $X_1$ and $X_2$ are irreducible, and how to prove it in this case.

Johnny T.
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    Intersect $X_1=(zx=zy=0)$, a plane and a line, and $X_2=(x=y=0)$, just the line, with the complement of $z=0$. The intersection is just the line (without the origin) in both cases, but $X_1$ has dimension $2$, while $X_1$ has dimension $1$. –  Aug 03 '18 at 11:01
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    Some people require varieties to be irreducible, is that the case in your case? –  Aug 03 '18 at 11:05
  • Thank you for the explanation!! I am not actually sure but I think I do. Let me add it. – Johnny T. Aug 03 '18 at 11:10
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    Yes, it has been asked before here. –  Aug 03 '18 at 11:31
  • @spiralstotheleft I am confused now, why is it that even though it's true for irreducible ones, it doesn't work in general? I mean all varieties are union of irreducible ones after all... – Johnny T. Aug 03 '18 at 11:38
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    The thing is where the open set intersects them. In a reducible one the open set can be "very small" as in the example in my comment. In the irreducible case, if you for the open set to intersect the variety, then it intersects in a big chunk. –  Aug 03 '18 at 11:41
  • I think I got it now. Thank you! – Johnny T. Aug 03 '18 at 13:57

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