Timeline for Are the positive multiplicative group and the additive group of the field of real algebraic numbers isomorphic?
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Dec 28, 2017 at 22:06 | history | edited | Qfwfq | CC BY-SA 3.0 |
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| Dec 28, 2017 at 21:34 | history | edited | YCor |
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| Dec 28, 2017 at 21:21 | history | edited | François G. Dorais |
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| Apr 11, 2016 at 20:15 | vote | accept | Asaf Shachar | ||
| Apr 11, 2016 at 18:45 | history | edited | Ben McKay | CC BY-SA 3.0 |
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| Apr 11, 2016 at 18:14 | answer | added | Arturo Magidin | timeline score: 10 | |
| Mar 24, 2016 at 23:37 | comment | added | Arturo Magidin | So... should this be written up as a formal answer? | |
| Mar 24, 2016 at 16:23 | comment | added | Emil Jeřábek | For the additive group, an argument easier to check is that for any $d\ge1$, there are real algebraic numbers $\alpha$ of degree $d$ (such as $2^{1/d}$), in which case $\{1,\alpha,\dots,\alpha^{d-1}\}$ is linearly independent. | |
| Mar 24, 2016 at 16:22 | history | edited | Arturo Magidin | CC BY-SA 3.0 |
it was unclear that the final paragraph was connected to the prior one; easy to miss the ":". better this way.
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| Mar 24, 2016 at 16:17 | comment | added | Arturo Magidin | @S.Carnahan: Yes, you're right. The additive group is of course infinite dimensional: for example, $\{\sqrt{p}\mid p\text{ is prime}\}$ is (additively) independent over $\mathbb{Q}$. As for the multiplicative group, the primes themselves are independent. | |
| Mar 24, 2016 at 16:11 | comment | added | Emil Jeřábek | Now wait a minute, I somehow lost “algebraic” along the way—the groups are in fact countable. So, there is actually some work to do to show they have the same dimension (namely, countably infinite). This should be easy. | |
| Mar 24, 2016 at 16:10 | comment | added | S. Carnahan♦ | @ArturoMagidin I think cardinality is not quite enough in the countable case. Perhaps the existence of an infinite set of independent elements would help. | |
| Mar 24, 2016 at 16:06 | comment | added | Emil Jeřábek | Wow, that was close. | |
| Mar 24, 2016 at 16:04 | comment | added | Arturo Magidin | @EmilJeřábek: snap! | |
| Mar 24, 2016 at 16:04 | comment | added | Arturo Magidin | They are both torsionfree abelian divisible groups, so they are both isomorphic to a direct sum of copies of $\mathbb{Q}$; since they have the same cardinality, they have to be isomorphic. | |
| Mar 24, 2016 at 16:04 | comment | added | Emil Jeřábek | Yes, since the theory of divisible torsion-free abelian groups (aka $\mathbb Q$-linear spaces) is uncountably categorical. | |
| Mar 24, 2016 at 15:52 | history | asked | Asaf Shachar | CC BY-SA 3.0 |