Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 3 down vote accepted

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.

share|cite|improve this answer
Huh? I thought i showed earlier that it was flat, better review that.. Thanks! – Joachim Mar 25 '13 at 17:50
You were absolutely right Georges, thanks. See the edit i made to the question. – Joachim Mar 26 '13 at 12:48
Dear Joachim: you are displaying a fine sense of fair play by accepting an answer pointing out inaccuracies in your question. I find this a much more precious quality than anything related to mathematics. My sincere congratulations! – Georges Elencwajg Mar 26 '13 at 12:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.