Embed an algerbra A into an ultraproduct of its finitely generated subalgebras.

Question

The question is to prove that any algebra $A$ is embeddable into a ultraproduct of its finitely generated subalgebras. In the case of boolean algberas, this seems rather intuitive.

I don't know what to use here, say the algebra is $A$, well then $V(A)$ is generated by the finitely generated subalgebras so it makes sense that $A$ will embed into $\prod_{i\in I} B_i/U, B_i \le A, \forall i\in I, U :$ ultrafilter.

Since these are the finitely generated subalgebras maybe construct the ultra filter somehow of the set $X$ where $X$ contains all the generating sets? or some enumeration thereof to construct the appropriate $I$.

Not sure how to construct a map that would use this.

Answer 1

Let $I$ index the set of finitely generated subalgebras of $A$. For each $a\in A$ let $$U_a=\{i\in I:a\in A_i\}\;,$$ and let $\mathscr{A}=\{U_a:a\in A\}$. Then $\mathscr{A}$ is a centred family of subsets of $I$ (i.e., intersections of finitely many members of $\mathscr{A}$ are non-empty). In fact it’s clear that $\mathscr{A}$ is a filter base on $I$. Let

$$\mathscr{F}=\{F\subseteq I:\exists U\in\mathscr{U}\,(U\subseteq F)\}\;,$$

the closure of $\mathscr{A}$ under taking supersets; $\mathscr{F}$ is a filter on $I$ containing $\mathscr{U}$. The ultrafilter lemma now says that there is an ultrafilter $\mathscr{U}$ on $I$ such that $\mathscr{F}\subseteq\mathscr{U}$.

If you don’t want to take the ultrafilter lemma for granted, you can prove this directly using Zorn’s lemma. Let $\mathfrak{F}$ be the set of all filters on $I$ that contain $\mathscr{F}$. Show that the union of any chain in $\mathfrak{F}$ is a member of $\mathfrak{F}$, so that every chain in the partial order $\langle\mathfrak{F},\subseteq\rangle$ has an upper bound in the partial order. Zorn’s lemma then says that the partial order has a maximal element $\mathscr{U}$, and it’s straightforward to verify that $\mathscr{U}$ is an ultrafilter.

Define $f:A\to\prod_{i\in I}A_i$ by letting $f(a)=\langle a_i:i\in I\rangle$, where

$$a_i=\begin{cases} a,&\text{if }a\in A_i\\ 0_A,&\text{otherwise}\;. \end{cases}$$

Then define

$$f^{\mathscr{U}}:A\to\left(\prod_{i\in I}A_i\right)_{\mathscr{U}}:a\mapsto[f(a)]_{\mathscr{U}}$$

where for any $\langle a_i:i\in I\rangle\in\prod_{i\in I}A_i$ we define

$$[\langle a_i:i\in I\rangle]_{\mathscr{U}}=\left\{\langle b_i:i\in I\rangle\in\prod_{i\in I}A_i:\{i\in I:a_i=b_i\}\in\mathscr{U}\right\}\;,$$

the equivalence class of $\langle a_i:i\in I\rangle$ modulo $\mathscr{U}$; this should be the desired embedding.