Universal skew field of fractions

Question

Let $R$ be a non-commutative ring with unity. A skew field $K$ containing $R$ is called a universal skew field of fractions of $R$ if

it is generated by $R$ as a skew field (i.e. there is no proper subskewfield of $K$ containing $R$)
for any skew field $L$ and any morphism of rings $\varphi : R \to L$ there exists a subring $K_0$ of $K$ with $R \subseteq K_0$ and a morphism of rings $\theta : K_0 \to L$ extending $\varphi$ and with the property that $$ \forall x \in K_0 (x^{-1} \in K_0 \Leftrightarrow \theta(x) \neq 0). $$

This is the definition I saw in a noncommutative geometry course. We went on to examine a few examples, but a lot of things are still not clear to me:

Why would the 'naive' generalisation of the universal property in the commutative case not give the right notion in the non-commutative case?
In what category is the above definition of a universal skew field of fractions actually an initial object? Don't we need uniqueness of $K_0$ and $\varphi$?
Because why else would the universal skew field of fractions be unique up to isomorphism?

Note that uniqueness does not follow from the fact that $R$ generates $K$. An example from my course (sketch):

Let $k$ be a commutative field, $k<x_1, x_2>$ the polynomial ring in two non-commuting variables. For every $n \in \mathbb{N}$, $n \geq 2$ consider the skew polynomial ring $k[t][X, n]$, in which $tX = Xt^n$. This is a right Ore ring, hence can be embedded into a skew field. Also one has an embedding $$ \phi_n : k < x_1, x_2 > \to k[t][X, n] $$ by setting $\phi_n(x_1) = X$ and $\phi_n(x_2) = Xt$, but the fields generated by $\phi_n(k < x_1, x_2 >)$ and $\phi_m(k < x_1, x_2 >)$ are not isomorphic if $n \neq m$, as one has $$ \phi_n(x_1)^{-1}\phi_n(x_2)\phi_n(x_1) = X^{-1}XtX = tX = Xt^n = Xtt^{n-1} = Xt(X^{-1}Xt)^{n-1} = \phi_n(x_2)(\phi_n(x_1)\phi_n(x_2))^{n-1} $$ and this depends on the choice of $n$!

I don't know how I should give you a source though, it's just a course and my notes are all I have. — Bib-lost, May 10 '17 at 19:13
"Why would the 'naive' generalisation of the universal property in the commutative case not give the right notion in the non-commutative case?" Do you mean "Can't we just do the same thing we do for commutative domains for noncommutative domains and get a divison ring?" or are you aware of the difficulties already? — rschwieb, May 10 '17 at 19:26
I know that in general there does not even need to be a skew field extending a non-commutative domain. By naive definition of the universal property I meant: for any skew field $L$, any homomorphism $R \to L$ extends to a unique homomorphism $K \to L$. One can define this of course, but I'm guessing that it is a too strong condition if one wants to obtain enough examples. — Bib-lost, May 10 '17 at 20:10

score 4 · Accepted Answer · answered May 11 '17 at 09:16

4

The example you give shows why the 'naive' notion doesn't work for non-commutative rings in general. The free algebra $R=k\langle x_1,x_2\rangle$ has several different skew fields of fractions. But any map between skew fields of fractions of $R$ must be an isomorphism, so there could only be a "universal" (in the naive sense) skew field of fractions if there were a unique skew field of fractions.

The right category in which to take the initial object is pretty much described by the second part of your definition of the universal skew field of fractions:

If $R$ is a ring then the skew fields of fractions of $R$ form a category where a morphism from $R\hookrightarrow K$ to $R\hookrightarrow L$ is a "specialization"; i.e., a (necessarily surjective) ring homomorphism $\beta: K_0\to L$ from a subring $K_0$ of $K$ with $R\subseteq K_0$ with the property that $\beta$ is the identity on $R$ and $\ker(\beta)$ is precisely the set of non-units of $K_0$. The universal skew field of fractions is the initial object in this category.

answered May 11 '17 at 09:16

Jeremy Rickard

28,899

But then are $K_0$ and $\beta$ uniquely defined for a map from the universal skew field of fractions to another skew field of fractions, or does one still need to consider some kind of equivalence relation on morphisms to obtain the uniqueness?
Thanks for the first part of your answer by the way, it's very enlightening!
– Bib-lost May 11 '17 at 15:30
@Bib-lost If there is a specialization from $K$ to $L$ then it is unique. Every element of $K$ and $L$ can be wtitten as a finite expression involving elements of $R$, addition, multiplication and inversion, but different such expressions will make sense in different skew fields, as you need to avoid inverting zero. $K_0$ is characterized as the set of evaluations in $K$ of all expressions that make sense in $L$. – Jeremy Rickard May 12 '17 at 14:41
Thanks. I guess there is still some things to prove formally, but at least it makes intuitive sense now. – Bib-lost May 13 '17 at 10:11

Duchamp Gérard H. E. · Answer 2 · 2017-12-14T14:00:59.723

Complement The category of Jeremy is ``unfolded'' w.r.t. that of P. M. Cohn (in Free Ideal Rings and Localization in General Rings) and it is also my way (because associativity is, IMHO, easier to write down completely). You recover Cohn's category (in the book above) by passing to quotient (the relation between two `unfolded specializations'' $\beta_i, i=1,2$ being that there is a common subring $K_{0i}, i=1,2$ where they coincide). It is an easy exercise to prove that, $F$ being the quotient functor, the initial points of Cohn are precisely the images of Jeremy's (hence essentially the same) by $F$.

Universal skew field of fractions

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