Basis of $\mathbb{F}[[x]]$ over $\mathbb{F}$ without AC

Asked May 01 '15 at 15:54

Active May 02 '15 at 18:39

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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'.

edited May 02 '15 at 18:39

asked May 01 '15 at 15:54

Emolga

3,527

So... it's the algebraic dual of the ring of polynomials? – Asaf Karagila May 01 '15 at 16:35
@AsafKaragila I think OP means the formal power series over $\Bbb F$. Or perhaps you're saying that the two are the same. – Ben Grossmann May 01 '15 at 16:39
@Omnomnomnom: I am asking about the second thing. – Asaf Karagila May 01 '15 at 16:40
@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 $\mathbb{F}^{\omega}$. – Matt Samuel May 01 '15 at 16:46
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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). – Asaf Karagila May 01 '15 at 17:31

Basis of $\mathbb{F}[[x]]$ over $\mathbb{F}$ without AC

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