*A friend of mine and I were talking about computable algebra, and this question came up. The answer may already be known, but I couldn't find it with Google:*

Suppose I have a countable field, $k$. Then the Brauer group of $k$, $Br(k)$, is also countable. This means that I can talk about the computability-theoretic complexity of $Br(k)$. Specifically, we can define the *spectrum* of $Br(k)$, denoted $Sp(Br(k))$, to be the set of degrees computing copies of $Br(k)$: $$Sp(Br(k))=\{X: \exists G\le_T X(G\cong Br(k))\}.$$ My question is, how does the complexity of $Br(k)$ in this sense relate to the complexity of $k$? Specifically:

(1) If $k$ has a computable copy, does $Br(k)$ have a computable copy?

A second question, pointing in the opposite direction:

(2) Is there a field $k$ with a computable copy, such that $Sp(Br(k))$ has a least element which is nonzero?

*EDIT: I should point out that this is a* strengthening *of "has no computable copy": in particular, there are structures with no computable copy, but such that for every noncomputable set $X$, there is a copy $\mathcal{A}$ which does not compute $X$! (This is the "Slaman-Wehner theorem.")*

A positive answer to (2) would be a strong negative answer to (1): it would mean that not only does $Br(k)$ not always have to be computable, but we can code some specific non-computable set into $Br(k)$.

Okay, so why would anyone care?

Well, besides just being generally interested in computable algebra, I'm intrigued by the specific obstacles this question faces. Prima facie, the Brauer group $Br(k)$ is quite complicated: its elements are equivalence classes of central simple algebras. This raises a pair of questions, right off the bat:

How hard is it to tell that a given (finitely presented) algebra over $k$ is central simple?

How hard is it to tell that two central simple algebras are Brauer equivalent?

On the face of it, both of these questions are $\Sigma^1_1$: the former quantifies over ideals, and the latter over isomorphisms. This complexity seems to block any easy approach to getting a positive answer to (1). In fact, my suspicion is that both of these questions are $\Sigma^1_1$-complete, when phrased appropriately.

*EDIT: Actually, I overestimated the complexity of the first question: "centrality" is at worst $\Pi^0_2$ ("is there an element not in the field which commutes with everything?"), and "simplicity" is also at worst $\Pi^0_2$ ("is there a nonzero $a$ such that there is no witness to $(a)$ containin 1?"). So "is a (code for a) central simple algebra" is at worst $\Pi^0_2$. Still pretty bad, but vastly better. I suspect now that Brauer equivalence may also be substantially simpler than $\Sigma^1_1$. . .*

However, *even knowing* that each question is as complicated as possible, we would still not have a negative answer to (1); this is because a presentation of $Br(k)$ is just a presentation of $Br(k)$ as an *abstract group*, and doesn't tell us anything about which elements correspond to what algebras. So conceivably, identifying central simple algebras is $\Sigma^1_1$-complete, but every computable field has a computable Brauer group.

Given that, it seems the best way to get a negative answer to (1) would be to get a positive answer to (2), by coding some specific noncomputable set (let's face it, probably $0'$) into the Brauer group of a computable field. This is the part where, for me, things get really interesting: my limited understanding of algebra suggests that this approach would be extremely difficult. So, in addition to some computability theory and some descriptive set theory, this problem seems to really interact with algebra in a serious way. In particular, to the best of my knowledge computability theory hasn't really been "pointed at" cohomology in a serious way; and this problem looks like a good candidate for that to happen.

initial segmentof the naturals, to allow for finite structures. (And the term is the standard one in the field.) $\endgroup$ – Noah Schweber May 1 '14 at 20:466more comments