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Suppose $F \subseteq K$ are fields with $G$ an ultrafilter on an infinite set $X$. If $F^{\ast}$ and $K^{\ast}$ represent the ultraproducts respectively of $F$ and $K$, it is easy to see that $[K : F]$ is finite if and only if $[K^{\ast} : F^{\ast}]$ is finite. Denote a typical element of $K^{\ast}$ by $a^{\ast}$, where $a$ is a member of $\prod_{i \in X}K$. Also if $a \in K$, denote by $a^{\ast}$ the equivalence class in $K^{\ast}$ of the constant $a$-sequence in $\prod_{i \in X}K$.

If $[K : F]$ is finite, let $v_1, ... , v_n$ be a basis for $K$ over $F$. It then follows that $v_1^{\ast}, ... , v_n^{\ast}$ is a basis for $K^{\ast}$. For if $a^{\ast} \in K^{\ast}$, then for every $i \in X$, $a(i) \in K$ so there exist scalars $c_{i1}, ... , c_{in} \in F$ such that $a(i) = c_{i1}v_1 + ... + c_{in}v_n$. For each $1 \leq j \leq n$ we define $c_j \in \prod_{ i \in X}F$ by $c_j(i) = c_{ij}$. It follows that, for every $i \in X$, $a(i) = c_1(i)v_1 + ... + c_n(i)v_n$, which implies $a^{\ast} = c_1^{\ast}v_1^{\ast} + ... + c_n^{\ast}v_n^{\ast}$. Thus $v_1^{\ast}, ... , v_n^{\ast}$ span $K^{\ast}$. To show linear independence, suppose $c_1^{\ast}, ... , c_n^{\ast} \in F^{\ast}$ are such that $c_1^{\ast}v_1^{\ast} + ... + c_n^{\ast}v_n^{\ast} = 0$. Then $Q = \{ i \in X : c_1(i)v_1 + ... + c_n(i)v_n = 0 \} \in G$. But for any such $i \in Q$, by the linear independence of $v_1, ... , v_n$ we get that $c_1(i), ... , c_n(i) = 0$. In other words for every $1 \leq j \leq n$, $Q \subseteq \{ i \in X : c_j(i) = 0\}$. But this just means that $c_1^{\ast}, ... , c_n^{\ast} = 0$. This means that $v_1^{\ast}, ... , v_n^{\ast}$ form a basis for $K^{\ast}$ over $F^{\ast}$.

Conversely if $[K : F]$ is infinite, let $B$ be a basis for $K$ over $F$. Let $B'$ be the set with the elements $v \in B$ replaced by the equivalence class of their constant sequence in $\prod_{i \in X}K$, as $v^{\ast}$. Then $B'$ is a linearly independent set, i.e. any finite subset thereof is linearly independent. For if $v_1^{\ast}, ... , v_n^{\ast}$ are finitely many members of $B'$ with $c_1^{\ast}v_1^{\ast} + ... + c_n^{\ast}v_n^{\ast} = 0$ for some $c_1^{\ast}, c_2^{\ast}$ etc. $\in F^{\ast}$, then we can apply the same argument in the previous paragraph to show that $c_1^{\ast}, ... , c_n^{\ast}$ are all $0$. Thus $B'$ is a linearly independent subset. But any basis of $K^{\ast}$ would have cardinality at least as great as that of $B'$. But there are as many elements in $B'$ as there are in $B$, which by supposition is infinite. Therefore $[K^{\ast} : F^{\ast}]$ must be infinite.

Now my question is, what can be said about the cardinality of $[K^{\ast} : F^{\ast}]$ in the infinite dimensional case, besides the fact that it is at least as great as $[K : F]$? For every set containing $B'$ I have tried, for example $\pi(\prod_{i \in X}B)$ (where $\pi: \prod_{i \in X}K \rightarrow K^{\ast}$ is the canonical homomorphism), linear independence or span are each too much to hope for. An explicit basis for $K^{\ast}$ may be just about as easy to find as an explicit ultrafilter on $\mathbb{N}$.

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Do you need the external version? $B^*$, the *-transfer of $B$, is an internal basis in the sense that every element of $K^*$ can be written uniquely as a hyper-finite linear combination of elements of $F^*$; the main points being that the terms are indexed by a hyperinteger -- an element of $\mathbb{N}^*$ -- and form an internal sequence. (It's been a while since I've thought this way, but $B^*$, I think, consists of the equivalence classes represented by the elements of $\prod_{i \in X} B$) – Hurkyl Dec 25 '12 at 14:02
$B^{\ast}$ would be the "star" of all the elements in $B$, that is for every $v \in B$, $v^{\ast}$ would be the equivalence class of the constant $v$-sequence in $\prod_{i \in X}K$. However $\pi(\prod_{i \in X}B)$ contains more than just that. In the $i$th slot (for $i \in X$) you can have any member of $B$ you want. So in $\pi(\prod_{i \in X}B)$ you'd be looking at all possible combinations of basis elements in each slot and then modding out. – D_S Dec 25 '12 at 14:54
When I said $B^*$, I had in mind the application of the transfer principle to $B$. I'll call it ${}^\star B$ to distinguish it from what you think $B^*$ should be used for, and I believe ${}^\star B = \pi(\prod_X B)$. (I believe ${}^\star F = F^*$ and ${}^\star K = K^*$). By the transfer principle, if we internalize the notion of "basis for a vector space", then ${}^\star B$ is an internal basis for ${}^\star K$ over ${}^\star F$. An internal basis is not a basis in the external sense, though (unless it's finite); I wanted to check you don't want the internal notion of basis and dimension. – Hurkyl Dec 25 '12 at 15:11
I'm sorry, what do you mean by internal basis? I am only familiar with a vector space basis as it's defined in an undergrad linear algebra class--that is a set $C$ which spans the entire vector space (that is to say every member of $K^{\ast}$ is a linear combination of some finite collection of members of $C$) of which every finite subset is linearly independent. – D_S Dec 25 '12 at 15:44
It's an idea from non-standard analysis: any property of the "standard" model can be transferred to be a property of the non-standard model (e.g. to ultra-product). Unfortunately, I mainly understand these things in a synthetic way, and can't translate them to ultraproducts well. One characterization of internal basis is that every element of ${}^\star K$ can be written uniquely in the form $$\sum_{i=0}^H c_i b_i $$ where $H$ is an element of ${}^\star \mathbb{N} = (\prod_X N)/G$, the $c_i$'s are an internal sequence (indexed by the interval $[0,H]$ of elements of ${}^\star \mathbb{N}$)... – Hurkyl Dec 25 '12 at 16:28

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