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Apr 1, 2012 at 20:14 vote accept jlk
Apr 1, 2012 at 20:14 answer added jlk timeline score: 3
Mar 31, 2012 at 12:31 answer added Georges Elencwajg timeline score: 18
Mar 31, 2012 at 11:33 comment added François Brunault Darij's argument show in fact that for any field $k$ and any $k$-vector space $V$, the natural map $k^{\mathbf{N}} \otimes V \to V^{\mathbf{N}}$ is injective (it is bijective if and only if $V$ is finite dimensional).
Mar 31, 2012 at 6:12 comment added Filippo Alberto Edoardo I gave a wrong answer sometime ago, using injectivity. As it seems ok, let me try again:starting with infinitely many power series $\sum_ia_n^{i}x^i$ with $a_n\in\mathbb{C}$ all algebraically independent, the ideal generated by all of these is not generated by finitely many of them, I guess, for an argument of transcendence degree. But I still lack a neat proof, so I do not post it as an answer.
Mar 31, 2012 at 3:18 comment added darij grinberg ... as sequences of elements of $k$. But due to the linear independence of the $f_i$, this yields that $\left(s_i\right)_j = 0$ for all $i$ and $j$, and thus $T=0$.
Mar 31, 2012 at 3:17 comment added darij grinberg The natural map is injective, by a simple argument: Let $T$ be a tensor in $k\left[\left[x\right]\right]\otimes_k F$ which gets mapped to $0$ by this map. Then, we can write $T$ as $\sum\limits_{i\in I} s_i \otimes f_i$ for some finite set $I$, some $s_i\in k\left[\left[x\right]\right]$ and some linearly independent $f_i\in F$. Now, the condition that $T$ gets mapped to $0$ by the natural map rewrites as $\sum\limits_{i\in I} f_is_i=0$. Hence, every $j\in\mathbb N$ satisfies $\sum\limits_{i\in I} f_i\left(s_i\right)_j=0$, where we treat power series in $k\left[\left[x\right]\right]$ ...
Mar 30, 2012 at 21:50 history asked jlk CC BY-SA 3.0