Division algebras in which every proper subfield is maximal - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T19:27:53Z http://mathoverflow.net/feeds/question/28917 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/28917/division-algebras-in-which-every-proper-subfield-is-maximal Division algebras in which every proper subfield is maximal carlos 2010-06-21T05:23:33Z 2010-06-24T00:18:28Z <p>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? </p> <p>Is there any characterization of such division algebras? </p> <p>Does anybody know any book or paper that discusses this? </p> <p>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.</p> <p>(The degree of D is the square root of the dimension of D over F.)</p> <p>I hope my question is not too trivial! Thanks. </p> http://mathoverflow.net/questions/28917/division-algebras-in-which-every-proper-subfield-is-maximal/29276#29276 Answer by Skip for Division algebras in which every proper subfield is maximal Skip 2010-06-23T20:31:21Z 2010-06-24T00:18:28Z <p>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 $D$ <em>irreducible</em> if it has the property you are interested in, i.e., every (commutative) subfield that properly contains the center is a maximal subfield.</p> <ol> <li>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 $F$.</li> <li>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.</li> <li>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.)</li> <li>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.</li> <li>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.</li> </ol> <p>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.</p> <h2>Translation in terms of algebraic groups</h2> <p>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 $SL_1(D)$?</p> <p>These sorts of questions are addressed in my joint paper with Philippe Gille <em>Algebraic groups with few subgroups</em>, J. London Math. Soc., vol 80 (2009), 405-430. <a href="http://dx.doi.org/10.1112/jlms/jdp030" rel="nofollow">http://dx.doi.org/10.1112/jlms/jdp030</a> See especially section 4.</p> <p>Also, the paper <em>Irreducible tori in semisimple groups</em> by Gopal Prasad and Andrei Rapinchuk (IMRN 2001, #23, 1229-1242) <a href="http://ams.rice.edu/leavingmsn?url=http://dx.doi.org/10.1155/S1073792801000587" rel="nofollow">http://ams.rice.edu/leavingmsn?url=http://dx.doi.org/10.1155/S1073792801000587</a> 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.</p>