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 $\mathbb{X}$.