MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\mathbb{X}$.

share|cite|improve this question

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 $B^{*}$.

share|cite|improve this answer
As far as I can see, the statement "$B$ is linearly independent over $F$" isn't a first-order statement, but it is a (countable) conjunction of first-order statements. The point is that each possible length for an alleged linear dependence relation needs to be treated separately. Fortunately, such conjunctions are also preserved by ultraproducts, so your conclusion, that $B^*$ is linearly independent (over $F^*$) remains correct. – Andreas Blass Dec 25 '12 at 23:46
Andreas, thanks for pointing that out. – Joseph Van Name Dec 26 '12 at 0:27
So $B^{\ast}$ is linearly independent but not a basis, assuming there is no measurable cardinal. What then, would the cardinality of a basis for $K^{\ast}$ be? – Shankman Dec 31 '12 at 20:16
If $[K:F]=|K|$, then by a simple cardinality argument, we have $[K^{*}:F^{*}]=|K^{*}|=|B^{*}|$. I conjecture that $[K^{*},F^{*}]=|B^{*}|$ for all ultrapowers, but I have no proof :(. – Joseph Van Name Jan 8 '13 at 0:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.