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I have a (noncommutative) division algebra D which is finite dimensional over its center F. I know that every subfield of D which contains F properly is a maximal subfield of D. What can we say about D?

Is there any characterization of such division algebras?

Does anybody know any book or paper that discusses this?

By the way, the set of such division algebras is obviously not empty because, for example, the (real) quaternion algebra (or any division algebra of prime degree) is in that set.

(The degree of D is the square root of the dimension of D over F.)

I hope my question is not too trivial! Thanks.

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    $\begingroup$ Does your example have composite degree? That would be very interesting as such an example can't exist when $F$ is a number field. If $F$ is a number field, then $D$ has a cyclic splitting field, which has a proper intermediate field when it has composite degree. $\endgroup$
    – Robin Chapman
    Commented Jun 21, 2010 at 6:10
  • $\begingroup$ Thank you. I don't know anything about the degree. I'm looking for a characterization of these division algebras, if there exists such a thing. For example, whould having a prime degree be a necessary condition? $\endgroup$
    – carlos
    Commented Jun 21, 2010 at 6:39
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    $\begingroup$ Having prime degree is a sufficient condition. I don't see any way to construct an example with composite degree, but I am no expert in the subject. $\endgroup$
    – Robin Chapman
    Commented Jun 21, 2010 at 6:44
  • $\begingroup$ Thanks. Well, I know, by Kaplasky's theorem, that a non-commtatuve division algebra cannot be a radical extension of its center. Thus there exists a non-central element x of D such that F[x]=F[x^n], for all positive integers n. This should give us something but I don't know how! $\endgroup$
    – carlos
    Commented Jun 21, 2010 at 7:14
  • $\begingroup$ Just to add something trivial to explain why I hope what I said in the previous post should give us something: the reason is that the degree of the minimal polynomial of x is obviously the same as the degree of D. So I thought maybe the condition F[x]=F[x^n], for all n, might force n to be a prime number. $\endgroup$
    – carlos
    Commented Jun 21, 2010 at 7:23

1 Answer 1

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The short answer is that not too much is known about this situation, beyond the easy observations that I will now list. I will call $D$ irreducible if it has the property you are interested in, i.e., every (commutative) subfield that properly contains the center is a maximal subfield.

  1. If $D$ has prime degree, then $D$ is irreducible. This is obvious because every subfield is contained in a maximal subfield, and the maximal subfields all have the same dimension over $F$.
  2. If the degree of $D$ has at least two prime factors, then $D$ is reducible. In this case you can factor $D$ as a tensor product of two division algebras of relatively prime degrees. Then you just take a maximal subfield in one of the two factors. This reduces us to considering algebras $D$ whose degree is a prime power.
  3. If $D$ has composite degree and $D$ is a crossed product, then $D$ is reducible. Recall that $D$ is a crossed product if it has a maximal subfield $L$ that is Galois over $F$. So suppose that the Galois group is $G$, necessarily of composite order. Then there is a nonzero proper subgroup of $G$, hence $D$ is reducible. (This deduction sounds foolish, because the theorem that something is a crossed product is much stronger than what you are asking about. But asking if something is a crossed product is a standard question, so in this way you can connect your question to standard results.)
  4. As a consequence of #3, every $D$ of degree 4 is reducible, and every $D$ of degree 8 and exponent 2 is reducible. That is because such algebras are cross products under $Z/2 \times Z/2$ (Albert) and $Z/2 \times Z/2 \times Z/2$ (Rowen) respectively.
  5. If $D$ has degree $p^r > p$ for some $p$ prime and it happens that every finite extension of $F$ has dimension a power of $p$, then $D$ is reducible. Indeed, by Galois theory every maximal subfield contains proper subfields.

So the first open case is where $D$ has degree 8 and exponent at least 4 in the Brauer group, and the base field has extensions of degree not a power of 2.

Translation in terms of algebraic groups

Your question is closely related to the question of whether the group $SL_1(D)$ has nonzero, proper connected subgroups. Well, $SL_1(D)$ always contains maximal tori. So the question is: Are there others? (If your field has nonzero characteristic, probably one should only consider reductive subgroups.) Subfields of $D$ correspond to tori in $SL_1(D)$, so your question is the same as asking: For what $D$ are maximal tori the only nonzero, proper reductive subgroups of $SL_1(D)$?

These sorts of questions are addressed in my joint paper with Philippe Gille Algebraic groups with few subgroups, J. London Math. Soc., vol 80 (2009), 405-430. http://dx.doi.org/10.1112/jlms/jdp030 See especially section 4.

Also, the paper Irreducible tori in semisimple groups by Gopal Prasad and Andrei Rapinchuk (IMRN 2001, #23, 1229-1242) http://ams.rice.edu/leavingmsn?url=http://dx.doi.org/10.1155/S1073792801000587 discusses a similar question for tori. Your maximal subfield has no proper intermediate fields if and only if the corresponding torus is irreducible in their sense. This is why I called your $D$ irreducible above.

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