Does the ring of formal power series $\mathbb{F}[[x]]$ as a vector space over $\mathbb{F}$ admit a basis without assuming the Axiom of choice, at least in some special cases of $\mathbb{F}$?

I'm trying to find a basis explicitly.

In the case $\mathbb{F}[x]$ we can simply take $B=\{ x^0,x^1,... \}$. But even $\mathbb{F}_2[[x]]$ does not seem to have any obvious basis.

For every infinite-dimensional vector space $\; |V| = \max(|F|, \dim V)$, hence we know that the basis is of size $|V|=\max(2^{|\mathbb{F}|},{\aleph})$, and thus intuitively it seems 'too large to be explicit'.

  • $\begingroup$ So... it's the algebraic dual of the ring of polynomials? $\endgroup$ – Asaf Karagila May 1 '15 at 16:35
  • $\begingroup$ @AsafKaragila I think OP means the formal power series over $\Bbb F$. Or perhaps you're saying that the two are the same. $\endgroup$ – Omnomnomnom May 1 '15 at 16:39
  • $\begingroup$ @Omnomnomnom: I am asking about the second thing. $\endgroup$ – Asaf Karagila May 1 '15 at 16:40
  • $\begingroup$ @AsafKaragila There is a vector space isomorphism with the linear dual of the polynomial ring. A formal power series is determined by the coefficients of each $x^i$, which are arbitrary. This essentially describes the linear functionals on the polynomial ring. Really what we would want is a basis of $\mathbb{F}^{\omega}$. $\endgroup$ – Matt Samuel May 1 '15 at 16:46
  • 3
    $\begingroup$ This seems like a reasonable start for finding an answer. In particular when $\Bbb F$ is $\Bbb R$ or $\Bbb C$, we can easily show that the answer is negative (namely, consistently there is no basis). $\endgroup$ – Asaf Karagila May 1 '15 at 17:31

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