This question is inspired by Pace Nielsen's recent question [Does a left basis imply a right basis, without AC?](https://mathoverflow.net/questions/189090/does-a-left-basis-imply-a-right-basis-without-ac).


For any field $k$, the field $k(x)$ of rational functions in one variable has an explicit $k$-basis given by partial fractions: the set 
$$B(x,k)=\left\{x^i:i\geq 0\right\}\cup \left\{\frac{x^i}{m(x)^j}:\mbox{$m(x)$ monic irreducible, } 0<j, 0\leq i<\operatorname{deg}(m)\right\}$$
is a basis.


By induction, $k(x\_1,\dots,x\_n)$ has a $k$-basis for any $n$, without needing to assume the axiom of choice. The obvious way of doing this gives a basis that depends on the choice of an order for the variables.



> 
> In ZF set theory without choice, must $k(X)$ have a $k$-basis for an arbitrary set $X$?
> 
> 
> 


The existence of such a basis certainly doesn't need the full strength of AC. The statement that every set has a total order is known to be independent of ZF but not to imply AC (see for example the various references in [Are all sets totally ordered ?](https://mathoverflow.net/questions/37272/are-all-sets-totally-ordered)). If $X$ has a total order, and for $y\in X$ we let $$X\_{<y}=\left\{x\in X: x<y\right\},$$
then the set of finite products $t\_1t\_2\dots t\_n$, where 
$$1\neq t\_i\in B\left(y\_i,k(X\_{<y\_i})\right)$$
for some $y\_1<\dots < y\_n$,
is a $k$-basis of $k(X)$.



Does k(X) have a k-basis for every set X, without AC?
I will show that, if $k$ is countable and has characteristic $0$, then $k(X)$ has a basis. (This whole proof takes place without choice.)


For any $x \in X$, and any $f \in X$, we define an element $r\_x(f)$ of $k(X \setminus \{ x \})$ as follows: We have $k(X) \cong k(X \setminus \{ x \})(x)$, which embeds into the field of [formal Laurent series](http://en.wikipedia.org/wiki/Formal_power_series#Formal_Laurent_series) $k(X \setminus \{ x \})((x))$. Let $r\_x(f)$ be the coefficient of $x^0$ in this Laurent series. For a finite subset $I$ of $X$, let $r(I)$ be
$$\{ f \in k(I) : r\_x(f) =0\ \forall\_{x \in I} \}.$$


For any finite set $I$, we have $k(I) = \bigoplus\_{J \subseteq I} r(J)$ so, since $k(X)$ is the union of the $k(I)$ for $I$ finite, we have $k(X) = \bigoplus\_{I \subseteq X,\ \# I < \infty} r(I)$. So it is enough to show $\bigoplus\_{I \subseteq X,\ \# I < \infty} r(I)$ has a basis. For this purpose, it is enough to show that $r(\{ x\_1, x\_2, \ldots, x\_n \})$ has an $S\_n$ invariant basis. One can then transport this basis to each of the $r(I)$ with $\#I = n$. (More precisely, look at all $n!$ bijections between $\{ 1, \ldots, n \}$; each of them gives a transported basis in $r(I)$. Using the $S\_n$ invariance, their union is also a basis in $r(I)$. Thus, we get a basis in $r(I)$ without having to choose an ordering of $I$.)


We fix particular representatives $Sp\_{\lambda}$ for the irreducible representations of $S\_n$. This can be done in a choice free way by, for example, [writing down formulas with Young tableaux](http://en.wikipedia.org/wiki/Specht_module). For any $S\_n$-representation $W$, let $W\_{\lambda} = \mathrm{Hom}\_{S\_n}(Sp\_{\lambda}, W)$. Then there is an obvious map $\bigoplus\_{|\lambda| = n} W\_{\lambda} \otimes Sp\_{\lambda} \longrightarrow W$ and representation theory tells us this map is an isomorphism.
We will show that $r(\{ x\_1, \ldots, x\_n \})\_{\lambda}$ has a countably infinite basis for each $\lambda$, and we will pick a particular one without choice. Thus 
$$r(\{x\_1, \ldots, x\_n \}) \cong \bigoplus\_{|\lambda| = n} k^{\oplus \omega} \otimes Sp\_{\lambda} \cong (k S\_n)^{\bigoplus \omega}$$
 as $S\_n$ representations, and the latter has an obvious permutation basis.


**Lemma:** Any subspace $V$ of $k^{\oplus \omega}$ has a countable or finite basis, which can be found with no choices.


**Proof:** Let $e\_i$ be the standard basis vectors of $k^{\oplus \omega}$. Let $k^n$ be the subspace of $k^{\oplus \omega}$ spanned by $e\_1$, $\dots$, $e\_n$. Let $S \subseteq \omega$ be the set of $n$ for which $V \cap k^n \neq V \cap k^{n-1}$. For each $s \in S$, there is a unique $v\_s \in V$ such that $v\_s \in k^n \cap V$, the coefficient of $e\_s$ in $v\_s$ is $1$, and $v\_s$ does not contain the basis vectors $e\_{s'}$ for $s' \in S$, $s < n$. These $v\_s$ form a basis. (In other words, we use row reduced echelon form.) $\square$


Now, $k(x\_1, \ldots, x\_n)$ has a countable basis. So $r(\{ x\_1, \ldots, x\_n \})$ has a countable or finite basis by the lemma. The Specht module $Sp\_{\lambda}$ has an explicit finite basis which can be written down using standard Young tableaux, so $\mathrm{Hom}(Sp\_{\lambda}, r(\{ x\_1, \ldots, x\_n \}))$ has a basis (note that this is $\mathrm{Hom}$ of vector spaces). The $S\_n$ equivariant maps are a subspace of this so, using the lemma again, $\mathrm{Hom}\_{S\_n}(Sp\_{\lambda}, r(\{ x\_1, \ldots, x\_n \}))$ has an explicit finite or countable basis, as desired.


Finally, we must check that $\mathrm{Hom}\_{S\_n}(Sp\_{\lambda}, r(\{ x\_1, \ldots, x\_n \}))$ is infinite dimensional. The easiest way is probably to note that, for any distinct positive integers $(a\_1, \ldots, a\_n)$, the span of $x\_1^{a\_{\sigma(1)}} \cdots x\_n^{a\_{\sigma(n)}}$ forms a copy of $k S\_n$ inside $r(\{ x\_1, \ldots, x\_n \})$, and these are all linearly independent.


