Let's work over $\mathbb C$, Consider a finite morphism between two integral curves $f: X\rightarrow Y$, let $\eta_Y$ (resp $\eta_X$) be the generic points


  • What is $X_{\eta_Y}=f^{-1}(\eta_Y)$ ? is it just $\eta_X$?

  • If the answer is no, haw about the case where $f$ is an double cover?

  • How about the general case of two integral schemes (rather then curves)?



Recall that finite morphisms are closed, so that closed points of $X$ map to closed maps of $Y$.
On the other hand the generic point of $X$ (the only non closed point) cannot be sent to a closed point $y\in Y$, else all points of $X$ would be sent to $y$ too, by continuity of $f$, and $f$ would be constant and thus certainly not finite.
All in all we have proved that $$f^{-1}(\{\eta_Y\})=\{\eta_X\}$$

Since you are interested in a generalization, here is one:
Given a morphism of completely general integral schemes $f:X\to Y$ we have the equivalence $$ f \operatorname {is dominant}\: (i.e.\operatorname {has dense image}) \iff f(\eta_X)=\eta_Y$$ If moreover $f$ is proper (for example projective or finite) its image is closed and we thus deduce:$$ f (X)=Y\iff f(\eta_X)=\eta_Y $$ Be careful that this last equivalence does not say that $f^{-1}(\{\eta_Y\})=\{\eta_X\}$ as shown by the unique $k$-morphism $$f:X=\mathbb P^1_k\to Y=\operatorname {Spec}(k)$$ for which $f^{-1}(\{\eta_Y\})=\mathbb P^1_k$


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