Dimension of An Ultraproduct Field as a Vector Space over Another Ultraproduct Field 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 $\mathbb{X}$.
 A: Let $B$ be a basis for $K$ over $F$. Let $B^{*}$ denote the ultrapower $B^{X}/\mathcal{U}$. Then $F^{*}\subseteq B^{*}\subseteq K^{*}$. Since the statement "$B$ is linearly independent over $F$" is a conjunction of countably many first order statements, the set $B^{*}$ is also linearly independent. Therefore the degree of $[K^{*},F^{*}]$ is at least $B^{*}$.
If your ultrafilter is $\sigma$-complete, then the sentence "$B$ is a basis" is a countably long first order sentence. Since Los' theorem holds for infinitary logic on $\sigma$-complete ultrafilters, the set $B^{*}$ is also a basis. Therefore, for $\sigma$-complete ultrafilter $[K^{*},F^{*}]=|B^{*}|$, so in this case, the problem is reduced to the cardinality of ultrapowers.
In fact, the set $B^{*}$ is a basis if and only if $\mathcal{U}$ is $\sigma$-complete or $B$ is finite. If $\mathcal{U}$ is not $\sigma$-complete and $B$ is infinite, then let $b_{n}$ be a sequence of distinct elements in $B$ and let $A_{n}$ be a sequence of disjoint subsets of $X$ where $\bigcup_{n}A_{n}\in\mathcal{U}$, but $A_{n}\not\in\mathcal{U}$ for all $n$. Let $f:X\rightarrow K$ be a function where $f=b_{1}+...+b_{n}$ on $A_{n}$. Then $[f]\in K^{*}$, but $[f]$ is linearly independent from $B^{*}$.
