# Covering maps of Riemann surfaces vs covering maps of $k$-algebraic curves

In going from Riemann surface theory to the theory of algebraic curves over fields $k$ that are not necessarily $\mathbb{C}$, I would like to understand more about how the notion of a covering map carries over.

If I have a compact, connected Riemann surface $M$, a cover of $M$ by another such Riemann surface, say $N$, then I am taking this to mean a holomorphic map $f:N\rightarrow M$ of finite degree $m>0$ (that is, the generic fiber of $f$ consists of $m$ points). At a $p\in M$ that is a regular value of $f$ (i.e. $p$ is not a branch point), then there is a open neighborhood $U$ of $p$ such that $f^{-1}(U)$ is the disjoint union of $m$ copies of $U$, and 'open' refers to the topology determined by the complex analytic structure on $M$.

If I have $M$ and $N$ nonsingular algebraic curves over a field $k$, then what can be said about $f^{-1}(U)$ when $f:N\rightarrow M$ is a finite regular map? What I mean by this is when the topology is the Zariski topology, I assume that the statement "$f^{-1}(U)$ is the disjoint union of $m$ copies of $U$" translates to open sets in that topology. When we regard a compact Riemann surface as a nonsingular curve over $\mathbb{C}$, then will these notions coincide?

Sorry if this is a trivial/ill-posed question. (My experience so far is more with differential geometry and complex analytic geometry....)

Many thanks.

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Non empty Zariski open sets in an irreducible curve are dense, so there is no pair of them which is disjoint. –  Mariano Suárez-Alvarez Jan 16 '10 at 19:06
Ahh, yes of course. Thanks. –  mathematically speaking Jan 16 '10 at 21:29

No, the property of having small neighbourhoods whose preimage is a disjoint union of $n$ homeomorphic open sets does not hold in the Zariski topology (once $n > 1$, i.e. the cover is non-trivial). The reason is that non-empty Zariski open sets are always very big; in the case of a curve, their complement is always just a finite number of points. In particular, two different non-empty Zariski opens are never disjoint.

There are two ways that one rescues the situation: the first is to use the differential topology view-point on covers: they are proper submersions between manifolds of the same dimension. In the context of compact Riemann surfaces, both source and target have the same dimension, and maps are automatically proper, so it is just the submersion property that is left to think about. It is a property about how tangent spaces map, which can be translated into the algebraic context (e.g. using the notion of Zariski tangent spaces).

So if $f: X \rightarrow Y$ is a regular morphism (regular morphism is the algebraic geometry terminology for an everywhere defined map given locally by rational functions) of projective curves, we can say that $f$ is unramified at a point $p \in X$ if $f$ induces an isomorphism from the Zariski tangent space of $X$ at $p$ to the Zariksi tangent space to $Y$ at $f(P)$.

For historical reasons, if $f$ is unramified at every point in its domain, we say that $f$ is etale (rather than a cover), but this corresponds precisely to the notion of a covering map when we pass to Riemann surfaces.

This leads to the more sophisticated rescue: one considers all the etale maps from (not necessarily projective or connected) curves $X$ to $Y$, and considers them as forming a topology on $Y$, the so-called etale topology of $Y$. This leads to many important notions and results, since it allows one to transport many topological notions (in particular, fundamental groups and cohomology) to the algebraic context.

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Thanks, that's great. This not only answers my question, but also shows me the scope of the situation. –  mathematically speaking Jan 16 '10 at 21:34