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S Dec 26, 2014 at 11:49 history bounty ended CommunityBot
S Dec 26, 2014 at 11:49 history notice removed CommunityBot
S Dec 18, 2014 at 10:10 history bounty started Jeremy Rickard
S Dec 18, 2014 at 10:10 history notice added Jeremy Rickard Draw attention
S Dec 17, 2014 at 18:03 history bounty ended CommunityBot
S Dec 17, 2014 at 18:03 history notice removed CommunityBot
Dec 15, 2014 at 16:40 answer added Jeremy Rickard timeline score: 3
Dec 13, 2014 at 15:05 comment added user44143 I suspect the answer is "no". This could be proved by deducing AC from the statement "every vector space with a left basis has a right basis". This would strengthen @AndreasBlass's result deducing AC from "every vector space has a basis". (math.lsa.umich.edu/~ablass/bases-AC.pdf) Perhaps he or someone else can look at his old proof and rise to the challenge?
Dec 9, 2014 at 21:19 comment added Asaf Karagila I have to admit, that on a normal day I'd have jumped into some books to better my understanding of the objects in question, many of which I haven't actually met on the mathematical playground. But nowadays I'm too darn busy with my own work to do that. So I apologize, perhaps some other day!
S Dec 9, 2014 at 17:01 history bounty started Pace Nielsen
S Dec 9, 2014 at 17:01 history notice added Pace Nielsen Authoritative reference needed
Dec 9, 2014 at 1:11 history edited Pace Nielsen CC BY-SA 3.0
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Dec 7, 2014 at 21:27 comment added Pace Nielsen @David Handelman: I once talked to George about those examples, but it has been a few years now. I'll take a look again. Thank you for the pointer.
Dec 7, 2014 at 19:44 answer added Manny Reyes timeline score: 1
Dec 7, 2014 at 18:19 comment added მამუკა ჯიბლაძე @JeremyRickard Yes of course, thanks! I should recall it myself.
Dec 7, 2014 at 18:09 comment added Jeremy Rickard @მამუკაჯიბლაძე If $k$ is a commutative field, then the finite-dimensional division algebras over $k$ are classified by the Brauer group of $k$, with "opposite algebra" corresponding to inverse in the Brauer group. So a division algebra corresponding to an element of order greater than two in the Brauer group will not be isomorphic to its opposite algebra. The Brauer group of the rationals has such elements, and "isomorphic as $\mathbb{Q}$-algebras" is the same as "isomorphic as rings". I suspect there are $9$-dimensional examples over $\mathbb{Q}$, but I don't offhand know a reference to one.
Dec 7, 2014 at 15:41 comment added David Handelman There are old (1970--80s) results of George Bergman on weird bimodules over division rings (e.g., different dimensions on right and left), typically in connection with tensor products; have you seen these?
Dec 7, 2014 at 15:12 comment added მამუკა ჯიბლაძე @JeremyRickard I see, thank you for the elucidating example! Then maybe one could take an abelian group isomorphism of vector spaces - if not the opposite thing. Could you please also give a reference for examples of division rings not isomorphic to their opposites?
Dec 7, 2014 at 11:00 comment added Jeremy Rickard @მამუკაჯიბლაძე $D$ is not isomorphic to its opposite ring in general. But even when it is, there's more going on than an isomorphism of vector spaces. For example, take $D=\mathbb{C}(t)$, and make $D$ into a $D$-$D$-bimodule with $f(t)\in D$ acting on the left by multiplication by $f(t)$, and on the right by multiplication by $f(t^2)$.
Dec 7, 2014 at 10:14 comment added მამუკა ჯიბლაძე Risking being silly - is not a $D$-$D$-bimodule structure equivalent to an isomorphism of $D$-vector spaces? I mean, are not the categories of left and right vector spaces equivalent? In fact, is not $D$ always isomorphic to its opposite?
Dec 7, 2014 at 7:34 answer added Qiaochu Yuan timeline score: 3
Dec 7, 2014 at 4:50 history asked Pace Nielsen CC BY-SA 3.0