Does it make sense to talk about, say, the free division ring on 2 generators? If so, does the free division ring on countably many generators embed into the free division ring on two generators?
$\begingroup$
$\endgroup$
1
-
8$\begingroup$ One way of formulating this question rigorously would perhaps be to ask whether the forgetful functor from the category of division rings to the category of sets had an adjoint. $\endgroup$– Kevin BuzzardCommented Apr 25, 2010 at 9:04
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
8
If you had a "free division ring" $F$ on a set $X$, and any division ring $R$, then any set theoretic map $X\to R$ would correspond to a unique division ring homomorphism $F\to R$. If $X$ has at least two elements $x\neq y$, let $R=\mathbb{Q}$. Then you cannot extend both a set theoretic map $f\colon X\to R$ that sends $x$ to $0$ and $y$ to $1$, and a set-theoretic map $g\colon X\to R$ that sends $x$ to $1$ and $y$ to $0$: division rings are simple, so any homomorphism must be either one-to-one or the zero map. (I put both maps, in case one wonders whether you can have $x$ correspond to the zero element of $F$). So, no, you cannot have "free division rings", much like you cannot have "free fields".
answered Apr 25, 2010 at 0:33
-
$\begingroup$ 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. $\endgroup$– LSpiceCommented Apr 25, 2010 at 7:37
-
$\begingroup$ 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? $\endgroup$– LSpiceCommented Apr 25, 2010 at 7:45
-
$\begingroup$ 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. $\endgroup$ Commented Apr 25, 2010 at 10:08
-
1$\begingroup$ @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. $\endgroup$ Commented Apr 25, 2010 at 19:05
-
2$\begingroup$ 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. $\endgroup$ Commented Apr 27, 2010 at 0:07
$\begingroup$
$\endgroup$
There is a notion of free division ring, due to Paul Cohn. I don't know a good online reference, but this AMS bulletin article talks about the construction and gives references. They go by the name of "free skew field" rather than "free division ring".
In the usual category of division rings and ring homomorphisms, there is no free division ring in the sense of category theory, but Cohn's free division rings are pretty close. If you consider the category of division rings with specializations as morphisms, then Cohn's construction gives you precisely the free objects.
Let $D, D'$ be division rings. Then a specialization is a homomorphism $R \to D'$, where $R$ is a subring which generates $D$ as a division ring. (In other words, the smallest division ring within $D$ containing $R$ is $D$ itself.) Specializations are a reasonably natural idea. For example, let $C$ be the complex numbers, and $C(x)$ be the field of rational functions over $C$. Then all specializations between $C(x)$ and $C$ are given by sending $p(x)$ in $C(x)$ to the value $p(a)$ for some $a$ in $C$. This is not defined for all rational functions, but only for the rational functions that don't have a pole at $a$.
If you restrict to the category of fields over a fixed base field $k$ and specializations, the free fields are just of the form $k(x_1, \ldots, x_n)$, where $x_1, \ldots, x_n$ are indeterminates. In the noncommutative case, you get something much more complicated, but free division rings can be realized as subrings of the noncommutative analogue of Laurent series.
