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Let $U$ be an open set of the Riemann sphere, $z_i$ be $n$ distinct points of $U$, and $E$ the vector space of meromorphic functions on $U$ with poles of order no more than 2.

Let $F$ be the subspace of $E$ whose elements are holomorphic in a neighborhood of the $z_i$.

Does $E/F$ have finite dimension ? If so, what is it ?

It is clear that it has dimension at least $2n$, since the $\frac{1}{(z-z_i)^k}$, $k=1,2$, form a free family. However, I couldn't determine if there was more (intuition suggests not).

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    Why not $k\geq 3$? You can examine the simplest case $n=1$.2012-07-30
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    oh right, I'm sorry I forgot something in the question. I'll edit it.2012-07-30
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    also, yeah, the case $n=1$ is trivial. I'm interested in the case where $n$ is large2012-07-30
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    If $f$ is an element in $E$, you can subtract all principal parts of $f$ at $z_i$, then you'll get a function which is holomorphic near every $z_i$.2012-07-30
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    @jerrysciencemath : thanks ! I don't know why, but I thought that a principal part at some $z_i$ could have a pole at some other ^$z_j$, which is of course not the case. so this works and the dimension is n, right ?2012-07-30
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    No, it's $2n$ since the space of rational functions arising as principal parts is $2n$ dimensional.2012-07-30
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    yes of course, typo2012-07-30
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    if jerrysciencemath wants to make an answer, I'll validate it. otherwise, should I : 1. answer it myself ? 2. close it ?2012-07-30
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    @Glougloubarbaki I'll make an answer, thanks~2012-07-30

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If $f$ is an element in $E$, you can subtract all principal parts of $f$ at $z_i$, then you'll get a function which is holomorphic near every $z_i$. Since every principal part is a linear combination of $\frac{1}{z-z_i}$ and $\frac{1}{(z-z_i)^2}$, we can deduce that $E/F$ is a complex vector space of dimension $2n$.