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Suppose $f:X\to Y$ is a dominant morphism between varieties such that $k(Y)$ is a finite field extension of $k(X)$, $S\varsubsetneqq X$ is a proper closed subset (subvariety) of $X$. Is it always true that $\overline{f(S)}\ne Y$?

KReiser
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user498029
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  • The linked duplicate shows that the answer is yes: reduce to the case of $S$ irreducible, then consider the field extension $k(\overline{f(S)})\to k(S)$, which shows that $\overline{f(S)}$ has dimension at most $\dim S$, and hence cannot be equal to $Y$. – KReiser Jun 13 '21 at 06:39
  • @KReiser Why $\overline{f(S)}$ has dimension at most $\dim S$ shows that it is not equal to $Y$. – user498029 Jun 13 '21 at 06:45
  • This is because your morphism is also finite so dimX=dimY. If the image of your proper subvariety was all of Y, Then it would be of the same dimension as X which contradicts the inequality. – LBE Jun 13 '21 at 06:52
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    The fact that $k(X)\subset k(Y)$ is finite means that the transcendence degrees of $k(X)$ over $k$ and $k(Y)$ over $k$ are equal, hence $X$ and $Y$ have the same dimension. (It is not true that $k(X)\subset k(Y)$ finite means that $X\to Y$ is a finite morphism.) Therefore as $S\subsetneq X$ means $\dim S<\dim X$, we find that $\dim \overline{f(S)}\leq \dim S <\dim X=\dim Y$ and therefore $\overline{f(S)}$ is a proper subset of $Y$. – KReiser Jun 13 '21 at 06:59

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