Timeline for Is Q_r algebraically isomorphic to Q_s where r and s denote different primes? [closed]
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| May 20, 2023 at 15:57 | comment | added | Vik78 | Another way to see this: $\mathbb{Q}$ is a prime field, so any isomorphism must be a map of extensions of $\mathbb{Q}$, and therefore maps the rings of integers of the respective fields onto each other. But $p$ is a unit in the ring of integers of $\mathbb{Q}_l$, whereas it is not a unit in the ring of integers of $\mathbb{Q}_p$. | |
| May 20, 2023 at 15:44 | history | edited | Guntram | CC BY-SA 4.0 |
added 31 characters in body
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| Jan 4, 2012 at 13:49 | comment | added | Chandan Singh Dalawat | And any integer which is $\equiv1\mod8$ and $\equiv-1\mod3$ is a square in $\mathbf{Q}_2$ but not in $\mathbf{Q}_3$. | |
| Dec 23, 2011 at 18:25 | comment | added | sibilant | Isn't ${\mathbb Q}_{2}$ isomorphic to ${\mathbb Q}_{-2}$? | |
| Dec 23, 2011 at 15:56 | vote | accept | gottigen | ||
| Dec 23, 2011 at 15:00 | history | closed |
user9198 Felipe Voloch Chandan Singh Dalawat GH from MO Igor Rivin |
too localized | |
| Dec 23, 2011 at 9:06 | comment | added | Alex B. | This has also been asked on MSE, where it really belongs. I have voted to close. | |
| Dec 23, 2011 at 8:53 | answer | added | Guntram | timeline score: 13 | |
| Dec 23, 2011 at 8:48 | comment | added | KConrad | ... except for ${\mathbf Q}_2$ and ${\mathbf Q}_3$, which have the same number of roots of unity. But $-2$ is a square in ${\mathbf Q}_3$ but not in ${\mathbf Q}_2$. | |
| Dec 23, 2011 at 8:43 | comment | added | Chandan Singh Dalawat | One can more simply compare the groups of roots of unity contained in the two fields. | |
| Dec 23, 2011 at 7:48 | comment | added | Homology | Never: a Galois group $G=\mathrm{Gal} (F/\mathbb{Q}_b)$ is solvable, and in fact $[[G,G],[G,G]]$ is a $p$-group (contained in the wild inertia), and for any $p$ one can choose $F$ so that it is a non-trivial $p$-group. | |
| Dec 23, 2011 at 7:43 | comment | added | Martin Bright | No. For example, $\mathbb{Q}_3$ is not isomorphic to $\mathbb{Q}_5$, for the following reason: any field isomorphism would have to map $-1$ to $-1$; but $\mathbb{Q}_5$ contains a square root of $-1$, whereas $\mathbb{Q}_3$ does not. | |
| Dec 23, 2011 at 7:23 | history | asked | gottigen | CC BY-SA 3.0 |