Timeline for What are some nice uses of ultraproducts/ultrapowers?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 7, 2023 at 13:21 | history | made wiki | Post Made Community Wiki by David Roberts♦ | ||
| Jul 28, 2023 at 17:58 | comment | added | Alex Kruckman | @Anonymous Great! FYI, your construction is a special case of the "reduced product", a generalization of the ultraproduct that works with a general filter instead of an ultrafilter. In your case, quotienting by the direct sum ideal corresponds to taking the reduced product by the cofinite filter, which does not require choice. The use of choice in the ultraproduct proof is exactly in extending the cofinite filter to an ultrafilter (in algebraic language, extending the direct sum ideal to a maximal ideal). | |
| Jul 25, 2023 at 16:14 | comment | added | Anonymous | Wonderful! (And thanks to Alex and Tom for educating me on ultraproducts as well.) | |
| Jul 25, 2023 at 12:34 | comment | added | Tom Leinster | @Anonymous Thanks for taking up the challenge and finding a simpler, choice-free, proof. I've put an expanded version of your proof on the n-Category Café. | |
| Jul 25, 2023 at 10:33 | comment | added | Tom Leinster | @Anonymous Your original argument does implicitly use an ultraproduct. You quotient $\prod \mathbb{Z}/p$ by a maximal ideal $I$. That quotient is precisely the ultraproduct $(\prod \mathbb{Z}/p)/\mathcal{U}$ with respect to the following ultrafilter $\mathcal{U}$ on the set of primes: a subset $A$ of the primes belongs to $\mathcal{U}$ iff $x_A \in I$, where $x_A \in \prod \mathbb{Z}/p$ has $p$th coordinate $0$ if $p \in A$ and $1$ otherwise. | |
| Jul 23, 2023 at 18:27 | comment | added | Anonymous | The argument uses the axiom of choice (used to choose a maximal ideal in a nonzero ring); is that equivalent to using ultraproducts? In any case, I realized that even this is not necessary. Indeed, the ring $S = (\prod_p \mathbf{Z}/p)/(\oplus_p \mathbf{Z}/p)$ is clearly nonzero. It is also a $\mathbf{Q}$-algebra as any nonzero integer $n$ is invertible in $\prod_{p > n} \mathbf{Z}/p$ and thus also in $S$. So $\mathbf{Q} \times S$ has two distinct maps to $S$ (via either projection) which obviously agree on $\mathbf{Z}$, and thus $\mathbf{Z} \to \mathbf{Q} \times S$ is not an epimorphism. | |
| Jul 23, 2023 at 17:29 | comment | added | Alex Kruckman | @Anonymous +1 because your proof is simpler in the sense that it doesn't use the fact that $-1$ is a square mod $p$ for infinitely many $p$. But it does implicitly use ultraproducts: an ultraproduct of the $\mathbb{Z}/p\mathbb{Z}$ is exactly the same thing as a quotient of the product ring by a maximal ideal. So this doesn't answer the challenge, at least the way I read it. | |
| Jul 23, 2023 at 2:29 | comment | added | Anonymous | (For the challenge) Note that $E' = \prod_p \mathbf{Z}/p$ has a quotient field $E$ of characteristic $0$ (take any maximal ideal containing $\oplus_p \mathbf{Z}/p$). So if $\mathbf{Z} \to \mathbf{Q} \times E'$ was an epimorphism, by base change and composition-stability of epimorphisms, we learn that $\mathbf{Q} \to \mathbf{Q} \times E$ is an epimorphism. But this is clearly false: the latter has two visibly distinct maps to $E$ (via either projection) that become the same on the source. | |
| Jul 22, 2023 at 23:59 | history | answered | Tom Leinster | CC BY-SA 4.0 |