Let $X$ be a quasi-projective scheme over $\mathbb{C}$ and $Y$ be a proper scheme over $\mathbb{C}$, and $f:X\to Y$ be a bijective, etale morphism.

Is this necessary for $f$ to be an isomorphism? If the answer is not, furthermore assume $X$ is also proper(hence projective), one can show that $f$ is actually finite. Then is $f$ an isomorphism under this additional assumption?

When they are both varieties and $X$ is also proper(hence projective), one can show that $f$ is actually finite, and hence $f$ is an isomorphism (see for example, Harris' Algebraic Geometry: A First Course, p. 179.). But I'm a little bit confused about non-reduced case.

Thanks for any help.


1 Answer 1


This must have already been discussed on here, but I cannot find the correct post.

Fact: Let $f:Y\to X$ be a bijective etale morphism of finite type $\mathbf{C}$-schemes. Then, $f$ is an isomorphism.

Proof: Note that $\Delta:Y\to Y\times_X Y$ is surjective. To see this, note that since $\Delta(Y)$ is a locally closed subset of $Y\times_X Y$its complement if non-empty must contain a $\mathbf{C}$-point. So, it suffices to prove that $\Delta(Y)$ contains all $\mathbf{C}$-points of $Y\times_X Y$. But, this is clear since if $(y_1,y_2)$ is a $\mathbf{C}$-point of $Y\times_X Y$ then $f(y_1)=f(y_2)$ and since $f$ is injective this implies that $y_1=y_2$ and so $(y_1,y_2)\in\Delta(Y)$. By Tag 01S4 this implies that $f$ is universally injective. Since $f$ is also etale, Tag 02LC implies that $f$ is an open embedding. Since $f$ is a bijection, it is thus an isomorphism. $\blacksquare$

Exercise: What properties of $\mathbf{C}$ did I implicitly use? What generality can one extend it to?


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