I read on a webpage that the ultraproduct ${\prod_{p\in \mathcal{P} } \mathbb{F}_p / }_\mathcal{U}$ (where $ \mathcal{P} $ is the set of prime numbers and $\mathcal{U}$ a non principal ultrafilter over $\mathcal{P}$ ) is algebraic over $\mathbb{Q}$ but I can't figure out why this is true... Can we generalize it to say that any non principal ultrafilter of finite field is algebraic over it's prime field?
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1They are usually uncountable, no? – Mariano Suárez-Álvarez Oct 24 '16 at 15:54
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The claim is definitely not true - the ultraproduct has size continuum, which means it cannot be an algebraic extension of $\mathbb{Q}$.
Why does the ultraproduct have size continuum? This is a consequence of a neat fact. Ignore the field structure, and just focus on the underlying sets. Then we have:
If $\mathcal{U}$ is a nonprincipal ultrafilter on $\mathbb{N}$ and $A_i$ is a finite set for each $i\in\mathbb{N}$, then either $\{\vert A_i\vert: i\in\mathbb{N}\}$ is bounded or the ultraproduct $\prod A_i/\mathcal{U}$ has size continuum.
To see this, it suffices to construct a family of continuum-many sequences $f_r$ ($r\in 2^\omega$) with $f_r(i)\in A_i$ such that for $r\not=s$, $\{i: f_r(i)=f_s(i)\}$ is finite (since then the $f_i$s represent distinct elements of the ultraproduct). This is a good exercise. HINT: given an infinite binary sequence $r\in 2^\omega$, let $F_r: i\mapsto r\upharpoonright i$. Then $F_r$ and $F_s$ agree at most finitely often for $r\not=s$, and we can view $F_r(i)$ as an element of $2^i$ in a natural way. Now, do you see how to turn the $F_r$s into the desired $f_r$s?
Noah Schweber
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I think I can imagine how to create such a sequence, I willthink more about it, thanks! – WrabbitW Oct 24 '16 at 16:24
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@WrabbitW: If you get stuck, you’ll find a couple of methods here and more at the question of which that was a duplicate. – Brian M. Scott Oct 24 '16 at 18:49