Timeline for Is the class of additive groups of rings axiomatizable?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Aug 6, 2015 at 19:03 | vote | accept | Michał Masny | ||
| Aug 4, 2015 at 22:26 | answer | added | Keith Kearnes | timeline score: 16 | |
| Mar 29, 2012 at 15:00 | comment | added | Sean Eberhard | Thanks for the examples. This question is less trivial than I thought. :) | |
| Mar 29, 2012 at 13:39 | comment | added | Michał Masny | @TomGoodwillie Why is it important that we include the trivial group? | |
| Mar 29, 2012 at 13:26 | comment | added | Tom Goodwillie | @Sean: Here is a torsion-free example. Consider a nontrivial subgroup of $\mathbb Q$. If it admits a nonzero multiplication, then it is isomorphic (scaling by some rational number) to a unitary subring of $\mathbb Q$. This excludes examples like the group of all rational numbers with square-free denominator. | |
| Mar 29, 2012 at 13:18 | comment | added | Tom Goodwillie | The trivial group, with the only possible ring structure, is unitary, so by excluding zero multiplication you are in some sense excluding too much. | |
| Mar 29, 2012 at 13:15 | comment | added | user6976 | continued: Since $k=0$ or $1$, $k=1$, $\alpha*\alpha=\alpha$. Taking $s=\beta$, we get $m\beta*\beta=\beta$ where $m$ is $0,1$ or $2$. Therefore $m\ne 0$. Similarly all other coefficients of the unit are not zeroes which is a contradiction since the direct sum consists of sums with finite supports. Thus the ring $\mathbb{Z}_2+\mathbb{Z}_3+...$ is not the additive group of any unitary ring. | |
| Mar 29, 2012 at 13:11 | comment | added | user6976 | @Sean: I assume you mean unitary rings. How about $\mathbb{Z}_2 + \mathbb{Z}_3+...+\mathbb{Z}_p+...$? Suppose that $\alpha$ is the generator of $\mathbb{Z}_2$, $\beta$ is the generator of $\mathbb{Z}_3$,... and $k\alpha+m\beta+...$ is the unit in the ring, $k$ is $0$ or $1$, $m$ is $0,1$ or $2$,... . Consider $\alpha*\beta$. Note that $2\alpha*\beta=3\alpha*\beta=0$, hence $\alpha*\beta=\beta*\alpha=0$. Also $(k\alpha+m\beta)*s=s$, $k\alpha*s+m\beta*s=s$ - for every $s$. Taking $s=\alpha$, we get $k\alpha*\alpha+m\beta*\alpha=\alpha$. Hence $k\alpha*\alpha=\alpha$. | |
| Mar 29, 2012 at 13:08 | comment | added | Michał Masny | @SeanEberhard Yes, as far as I know there is only zero multiplication on every divisible torsion group. | |
| Mar 29, 2012 at 12:52 | history | edited | Michał Masny | CC BY-SA 3.0 |
added 111 characters in body
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| Mar 29, 2012 at 12:52 | comment | added | Michał Masny | I realized I forgot to exclude zero multiplication right after I posted but I didn't have time to edit. Sorry! I'm editing now. | |
| Mar 29, 2012 at 12:26 | comment | added | Sean Eberhard | ... and every finitely generated abelian group is the additive group of some associative unitary commutative ring, by looking at the classification: en.wikipedia.org/wiki/…. Is there any (necessarily not finitely generated) abelian group which is not the underlying additive group of some commutative ring? | |
| Mar 29, 2012 at 12:11 | comment | added | Andreas Blass | When a subset of your adjectives doesn't contain "unitary," the answer is trivial, because you can take any abelian group and make it a ring by defining all products to be zero. | |
| Mar 29, 2012 at 11:58 | history | asked | Michał Masny | CC BY-SA 3.0 |