Timeline for Free division rings?
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11 events
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| Apr 27, 2010 at 0:10 | comment | added | Zoran Skoda | For the other problem, on embedding of Cohn's free field in n variables into 2 variable case, the answer would be probably not (unlike the group case), but I do not know. One could possibly consult the results of Reutenauer of similar flavour. | |
| Apr 27, 2010 at 0:07 | comment | added | Zoran Skoda | While there are no free fields nor free division rings in the sense of a left adjoint to forgetful functor, there is some useful notion, namely there is a division ring extending naturally the free associative algebra on given number of generators. Such a construction has been given by Amitsur and then another by P.M. Cohn, under the name "free field" (though noncommutative; and free is not the free of the universal algebra type). One should be warned that in general noncommutative case not every domain embeds in a field, and if it can, then the minimal way to do it is not always unique. | |
| Apr 25, 2010 at 21:50 | comment | added | Uri Andrews | @Arturo: That might be exactly the right formulation. Alternatively, perhaps I'm just asking whether there exists an embedding from a non-finitely generated division ring into a finitely generated division ring. (finite generation still makes sense without talking about presentations, I believe). | |
| Apr 25, 2010 at 19:05 | comment | added | Arturo Magidin | @Spice: I don't think you can make it sensible in the sense of getting an adjoint to the forgetful functor to sets, which is the usual meaning of "free object". I was tempted to try with "one-to-one maps into $R^{\times}$, but you will run into trouble if there are any "algebraic" relations between them; for example, mapping $x$ to $y^2$, leading to a non-trivial map that has a nontrivial kernel ($x-y^2$ would map to zero). So I think the answer is that there is no reasonable way to restrict the universal property to get a nice notion here. | |
| Apr 25, 2010 at 19:02 | comment | added | Arturo Magidin | @Uri: Here's one possible way of asking that question: every division ring is a division algebra over its center (which is nontrivial, since it contains $0$ and $1$). So we can ask whether there exists a division ring $R$, and a subring $S$ that contains $C(R)$, also a division ring, such that $R$ is finitely generated as a division algebra over $C(R)$ but $S$ is not finitely generated as a division algebra over $C(R)$. Or same question but relative to $C(S)$, or some common subring that is a division algebra. I'll think about it a bit. | |
| Apr 25, 2010 at 10:08 | comment | added | Uri Andrews | Thanks for this answer; you certainly dispelled my high dreams. How about an answer to the now meaningless second question? Is it possible to have a finitely generated division ring with infinitely many "independent" elements? I'm not sure what independent means here, or how to formulate it exactly. The perfect characterization would be, of course, that they generate a free division ring, but that's now meaningless. | |
| Apr 25, 2010 at 7:45 | comment | added | LSpice | Sorry, never mind, that was silly. In your notation, in any putative such free division ring, taking an 'extension' of the map sending $x$ and $y$ both to $1$ would show that $x - y = 0$, preventing extension of a map sending $x$ to $0$ and $y$ to $1$. Is there any possibility of an interesting replacement for the natural universal condition that you mention, or is it clear that no sensible such condition exists? | |
| Apr 25, 2010 at 7:37 | comment | added | LSpice | Perhaps it is more interesting to require that the set-theoretic map have image in $R^\times$, to avoid the obstruction you mention? In that case, perhaps a quotient of $\mathbb Q[X, X^{-1}]$ by a maximal ideal would do the job. | |
| Apr 25, 2010 at 3:44 | vote | accept | Uri Andrews | ||
| Apr 25, 2010 at 0:45 | history | edited | Arturo Magidin | CC BY-SA 2.5 |
corrected grammar
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| Apr 25, 2010 at 0:33 | history | answered | Arturo Magidin | CC BY-SA 2.5 |