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Given two morphisms of schemes $X\to S$ and $Y\to S$, suppose $U$ is an open subscheme of $X$ and $U$ is dense in $X$, is $U\times_SY$ dense in $X\times_SY$?

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    $\begingroup$ Not necessarily, let $S=X=\mathbb{A}^1$, let $U=\mathbb{A}^1\setminus\{0\}$ and $Y=\{0\}$, then $X\times_S Y=Y$ and $U\times_S Y=\emptyset$. $\endgroup$
    – Roland
    Commented Feb 7, 2019 at 15:16
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    $\begingroup$ @Roland That seems like a perfectly fine answer to me. $\endgroup$
    – Servaes
    Commented Feb 7, 2019 at 15:23
  • $\begingroup$ @Arthur Thank you! I know few examples about algebraic geometry, I scan a textbook and find you are right. $\endgroup$ Commented Feb 7, 2019 at 15:57
  • $\begingroup$ @Roland If $X$ is irreducible, $X\to S$ is proper, $Y\to S$ is projective, is $U\times_SY$ dense in $X\times_SY$? $\endgroup$ Commented Feb 7, 2019 at 16:05
  • $\begingroup$ @Borntobeproud My example satisfies all these extra assumptions. $\endgroup$
    – Roland
    Commented Feb 7, 2019 at 16:13

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