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Let $k$ be a field, and consider the field $K=k((X))=\text{Frac}(k[[X]])$. Then there exist a transcendence basis $\mathscr{B}$, i.e., a subset of algebraically independent element of $K$ such that $K\supseteq k(\mathscr{B})$ is an algebraic extension. I was wondering if such a basis should just consist of $X$, but than I do not see how can be $k(X)\subseteq k((X))$ an algebraic extension. Anybody has a clue about this?

Geronimo
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    I’m pretty sure $k((X))$ is never algebraic over $k(X)$. To begin with, if $k$ is countable, then so is the algebraic closure of $k(X)$, whereas $|k((X))|=2^\omega$. – Emil Jeřábek Nov 20 '13 at 13:31
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    A power series in $\mathbb F_q[[x]]$ is algebraic over $\mathbb F_q(x)$ iff it is automatic, by Christol’s theorem. This overview paper has more interesting information: http://maths.mq.edu.au/~alf/www-centre/alfpapers/a103.pdf . – Emil Jeřábek Nov 20 '13 at 14:09
  • Q: in this question what does the inner [] brackets in Frac(k( [[X]]) indicate? –  Nov 20 '13 at 14:08

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Certainly not! It is easy to show that $e^X:=\sum \frac{X^n}{n!} $ (I am assuming $\mathrm{char}(k)=0$ here) is not algebraic over $k(X)$. In fact, $k((X))$ is not finitely generated over $k(X)$, so your transcendance basis $\mathscr{B}$ must be infinite.