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Let $X$ be a projective algebraic curve and consider a `uniformizing' map $h:X \rightarrow \mathbb{P}^1$. Is there any connection between this notion of uniformizer and a uniformizer of the maximal ideal of the local ring at a point $P \in X$?

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There is no connection because, as far as I know, there is no such thing as a "uniformizing map $X\to \mathbb P^1$".

Assume $X$ is a smooth irreducible projective curve over an algebrically closed field $k$ of arbitrary characteristic.
Then a morphism $h:X\to \mathbb P^1$ (which is not the constant map with value $\infty$) is exactly the same as a rational function $f\in Rat(X)$.
Each point $P\in X$ has an associated local ring $k\subsetneq \mathcal O_P\subsetneq k(T)$, which is is a discrete valuation ring by smoothness of $X$.
The function $h$ is said to be a uniformizer of $X$ at $P$ if $h$ is a generator of the maximal ideal $\mathfrak m_P$ of $\mathcal O_P$.
This is equivalent to saying that $h$ is non-ramified (or in alternative terminology étale) at $P$.

The interpretation of "uniformizing map" in your question might be that $h$ be a uniformizer at each $P\in X$.
However such a situation only occurs if $h:X \xrightarrow {\simeq} \mathbb P^1$ is an isomorphism: this result follows from Riemann-Hurwitz and is the algebraic geometer's way of saying that $\mathbb P^1_k$ is simply connected.

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That makes sense - I should have put that the map was an isomorphism, in which case $h$ is the uniformizer at each $P$ as you explained, and so the two notions do match up. Thanks – bob Jul 12 '12 at 18:56

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