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Let $R$ be a (non-commutative) domain satisfying the (right) Ore condition; i.e. for all $a,b\in R$ one can find $\beta_1,\beta_2\in R$ such that $a\beta_1=b\beta_2$. In the well known construction of Ore, one considers pairs of elements $(a,b)$ (with $a,b\in R$, $b\neq 0$) together with the equivalence relation $(a,b)\sim(c,d)$ iff there exist $\beta_1,\beta_2\in R$ such that $a\beta_1=c\beta_2$ and $b\beta_1=d\beta_2$. The pair $(a,b)$ corresponds to an element of the type $ab^{-1}$. By introducing addition and multiplication on equivalence classes of pairs, one finds a division ring $F$ that contains $R$. An element $a\in R$ is embedded as $(a,1)$, and its inverse is simply $(1,a)$.

Now, assume that $R$ is also a $\ast$-algebra (over $\mathbb{C}$). Is there a canonical way of introducing a $\ast$-operation on $F$, such that $F$ becomes a $\ast$-algebra (over $\mathbb{C}$)?

I tried the following: Since $(ab^{-1})^\ast$ should equal ${b^\ast}^{-1} a^\ast$ one defines $(a,b)^\ast=(1,b^\ast)(a^\ast,1)$. It is easy to see that $((a,b)^\ast)^\ast=(a,b)$ and $(\lambda(a,b)+\mu(c,d))^\ast=\bar{\lambda}(a,b)^\ast+\bar{\mu}(c,d)^\ast$.

The problematic property is to show that $((a,b)(c,d))^\ast=(c,d)^\ast(a,b)^\ast$ (and right now I'm not even sure it is true). Does anyone have experience with (or can point me to a reference for) skew fraction fields of $\ast$-algebras?

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2 Answers

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Fun question! I think I can extend the $*$-operation less explicitly with a shortcut. But I hope someone double-checks the details here because I fear this may be a little too fast-and-loose.

Your $*$-operation is an anti-automorphism of $R$. In particular, because it is a right Ore domain, it's also a left Ore domain. Letting $S = R \setminus\{0\}$, we have $RS^{-1} = S^{-1}R$, and this is the "universal localization" of $R$ with respect to $S$ (it is universal among all rings with an $S$-inverting homomorphism $R \to T$); see Lam's Lectures in Modules and Rings, Corollary 10.14.

Let's think of $* = g \colon R \to R^{op}$ as an isomorphism of $R$ with its opposite ring that satisfies $g(g(x)^{op}) = x$ (I'm considering "double-$op$" to be the identity). In particular, $g(S)$ consists of nonzero elements, hence the composite $R \to R^{op} \to R^{op}(S^{op})^{-1} = (S^{-1}R)^{op} = (RS^{-1})^{op}$ is $S$-inverting. Thus it extends to a homomorphism $g' \colon RS^{-1} \to (RS^{-1})^{op}$. Now $g'(g'(x)^{op}) = g(g(x)^{op}) = x$ for all $x \in R$, so the same identity holds for the inverse of any element of $R$. And since every element of $R$ has the form $rs^{-1}$ with $r \in R$ and $s \in S$, we see that the identity holds for all elements of $RS^{-1}$, proving that $g'$ determines a new involution $*' \colon RS^{-1} \to RS^{-1}$.

I bet there's a proof using the explicit construction of $RS^{-1}$ that you give above, but I think the key is to use that $R$ is both left and right Ore and that $RS^{-1} = S^{-1}R$. I won't rob you of the joy of discovering such a proof for yourself!

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Thanks! I will have a closer look at your answer later today. You're confirming my suspicion that one needs to use both the left AND right Ore property to prove the fact that (ab)*=b*a*. As I wanted to learn these things throrougly, I have done all the calculations to check that the set of pairs is actually a division ring (which Ore does not explicitly do in his paper). It was tedious, and sometimes not completely trivial. It would then annoy me if I had to rely on a non-constructive proof to claim that it is a *-algebra :) –  Joakim Arnlind Nov 2 '12 at 7:03
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If I'm not mistaken, it follows immediately from abstract nonsense that the $*$-operation extends uniquely from $R$ to the field of fractions.

The notion of localization is more general than that of field of fractions, and hence it is sufficient to prove that if $S\subseteq R$ is any set with $S^*=S$, then $RS^{-1}$ inherits a unique $*$-operation.

But this is an immediate consequence of the universal property of $RS^{-1}$: the $*$-operation is a homomorphism $R^{op}\to R$, which can be composed with $R\to RS^{-1}$ to give $R^{op}\to RS^{-1}$. By $S^*=S$, this homomorphism maps every element of $S$ to an invertible element, and hence this induces $*:(RS^{-1})^{op}\to RS^{-1}$ which is compatible with the original $*:R\to R$. In particular, this compatibility guarantees also that the extended $*$-operation is antilinear.

The universal property should also guarantee that the extended $*$ is an involution as well.

Edit (see comments): The following is a proof sketch showing that $(RS^{-1})^{op}$ and $R^{op}S^{-1}$ are canonically isomorphic.

I think the easiest way to see this is to show that $(RS^{-1})^{op}$ has the universal property of $R^{op}S^{-1}$, i.e. that the canonical morphism $R^{op}\to (RS^{-1})^{op}$, which is defined simply as the opposite of $R\to RS^{-1}$, is the universal morphism with domain $R^{op}$ which maps $S$ to invertibles.

To see this, consider any $R^{op}\to T$ which maps $S$ to invertibles. Then its opposite $R\to T^{op}$ also maps $S$ to invertibles, and hence factors through $RS^{-1} \to T^{op}$. Taking opposites again gives the desired $(RS^{-1})^{op}\to T$.

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Thanks! How is it that *:R^{op}->RS^{-1} induces *:(RS^{-1})^{op} -> RS^{-1} in your argument? –  Joakim Arnlind Nov 2 '12 at 11:29
Do you know the universal property of localization? It states that if $T$ is any ring and $f:R\to T$ any ring homomorphism which sends every element of $S$ to an invertible element in $T$, then $f$ induces a unique homomorphism $R[S^{-1}]\to T$ which reproduces $f$ after composing with $R\to R[S^{-1}]$. In the case at hand, I have applied this to $R^{op}$ instead of $R$ itself, and also identified $R^{op}[S^{-1}]$ with $R[S^{-1}]^{op}$. (That these two are canonically isomorphic is again a consequence of the universal property.) –  Tobias Fritz Nov 2 '12 at 19:49
Yes, I do know about the universal property, and that an $S$-inverting homomorphism between to rings give rise to a (unique) homomorphism between the corresponding fraction fields. But, I can't straighten out the details of your last comment; i.e. why $R^{op}S^{-1}$ is isomorphic to $(RS^{-1})^{op}$. I guess that one argues that $(RS^{-1})^{op}$ is a fraction field of $R^{op}$ and then uses universality to equate it to $R^{op}S^{-1}$. But why is $(RS^{-1})^{op}$ is a fraction field of $R^{op}$? Explicit check? –  Joakim Arnlind Nov 3 '12 at 10:37
Abstract nonsense also gives you a canonical isomorphism between $R^{op}S^{-1}$ and $(RS^{-1})^{op}$. I have expanded my answer to include this, and also switched my notation to yours. –  Tobias Fritz Nov 3 '12 at 20:25
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