Timeline for Even XOR Odd Infinities?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Dec 26, 2014 at 21:52 | comment | added | Lord_Farin | @Emil, I've linked to this answer on maths.SE, here. I hope you don't mind, but thought I'd give you a heads up just in case. | |
| Aug 10, 2014 at 18:46 | history | edited | Emil Jeřábek | CC BY-SA 3.0 |
since the question was just bumped, fix MO 1.0 -> 2.0 formatting issues
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| Jan 24, 2013 at 19:10 | vote | accept | Russell Easterly | ||
| Jan 24, 2013 at 13:00 | comment | added | Emil Jeřábek | As I already mentioned, the only models of the second-order MA2 are the finite rings $\mathbb Z/n\mathbb Z$: if $M$ is any such model, let $f:\mathbb Z\to M$ be the unique ring homomorphism. Since $f(\mathbb N)$ contains $0$ and is closed under successor, it equals $M$ by induction, hence $f$ is surjective, and its kernel $I$ is nontrivial. Since $\mathbb Z$ is a PID, $I=n\mathbb Z$ for some $n>0$, and $M\simeq\mathbb Z/I$. | |
| Jan 24, 2013 at 12:49 | comment | added | Emil Jeřábek | Since it is not known whether MA proves the four-square theorem, this does not by itself imply that the f-s t holds in the 2-adics. Nevertheless, the f-s t does hold in $\mathbb Z_2$ (and in every $\mathbb Z_p$; moreover, if $p\equiv1\pmod4$, then $\mathbb Z_p$ satisfies a “two-square theorem”). $-1=2^2+1^2+1^2+(-7)$, and $-7$ has a 2-adic square root by Hensel’s lemma. (In general, a nonzero $a\in\mathbb Z_2$ is a square iff it can be written as $4^kb$, where $b\equiv1\pmod8$; for odd $p$, $a$ is a square iff $a=p^{2k}b$, where the Legendre symbol $\genfrac(){}{}{b\bmod p}p=1$.) | |
| Jan 24, 2013 at 1:03 | comment | added | Russell Easterly | Thanks Emil! If induction holds in the 2-adic integers does this mean the four square theorem is true in this model? If so, how would I represent -1 (...111) as four squares? Would this be a model for MA2? | |
| Jan 23, 2013 at 14:25 | comment | added | François G. Dorais | Very nice, Emil! | |
| Jan 23, 2013 at 14:22 | history | edited | Emil Jeřábek | CC BY-SA 3.0 |
added 4 characters in body
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| Jan 23, 2013 at 14:17 | history | answered | Emil Jeřábek | CC BY-SA 3.0 |