Let $f: X \longrightarrow Y$ be a morphism of varieties over $\mathbb{C}$ and assume it is finite of degree 1, i.e. it is surjective and
$$
[K(Y):K(X)] = 1 \quad \quad (*)
$$
i.e. the function fields are isomorphic.
Since one can show $f$ to be flat, this means all its fibres consist of a single point, hence the morphism is unramified hence étale (see some exercise in Hartshorne in the chapter on smooth morphisms). It is also bijective.
Now, to show that $f$ is an isomorphism one can use the euclidian topology and the fact that étale coverings induce isomorphisms on the tangent spaces at all points, so $f$ is euclidian-locally an isomorphism. Since it is also bijective, it is an isomorphism globally.
Could anyone give me an elegant short proof that works in all characteristics? Or if this happens not to be true, an algebraic proof over $\mathbb{C}$?
If this proof works more generally for certain schemes, i'd be happy to know about it too.
Idea 1: Locally $f$ is given by an inclusion of rings $R \rightarrow S$ and i hope to somehow deduce from $(*)$ that $S$ is a $R$ module of rank 1, so indeed $R \cong S$ and we are done locally, hence also globally
Idea 2: By $(*)$, $X$ and $Y$ are birational, but i am not sure how to continue from here.
This is most probably just a simple application of commutative algebra knowledge that i lack.. Thanks!
Edit: This is incorrect! I worked with the assumption of smooth varieties but failed to mention so. Smoothness is necessary to show flatness (Hartshorne exercise III.9.3(a)) and without it the reasoning collapes, as Georges points out below. It seemed more natural to close this one and post a new question, this can be found at Nice proof for étale of degree 1 implies isomorphism.
Best Answer
The morphism $f$ need not be flat and the fibers do not all consist of one point.
As I explained in my answer to your preceding question, the normalization of the node $N=V(y^2-x^2-x^3)\subset \mathbb A^2_k$ is the finite morphism $$f\colon \mathbb A^1_k\to N\colon t\mapsto (t^2-1,t^3-t)$$ of degree $1$.
However the fiber over $P=(0,0)\in N$ does not consist of a single point, but of the two simple points $t=-1, t=+1$, so that $f$ is not flat.
The morphism $f$ is not étale and not bijective.
So it is not possible to give you an "elegant short proof" of these incorrect statements.