Uniqueness of maximal subfields - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T23:26:09Z http://mathoverflow.net/feeds/question/35099 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/35099/uniqueness-of-maximal-subfields Uniqueness of maximal subfields unknown (yahoo) 2010-08-10T11:22:00Z 2010-08-11T20:16:48Z <p>Let D be a division ring with center Z. Let R and K be two maximal subfields of D, both purely inseparable of exponent one ( means the p power of each of them in Z). Why are R and K isomorphic?</p> <p>Or a better question, why there is only one maximal unique subfield which is purely inseparable of exponent one?</p> http://mathoverflow.net/questions/35099/uniqueness-of-maximal-subfields/35272#35272 Answer by Skip for Uniqueness of maximal subfields Skip 2010-08-11T20:16:48Z 2010-08-11T20:16:48Z <p>I think this question presupposes something that is incorrect. Here is an example. Fix a field $F$ of characteristic 2 and put $Z$ for the field $F(a,b)$ where $a$ and $b$ are indeterminates. Define $D$ to be the quaternion division algebra with center $Z$ generated by elements $i, j$ satisfying $$i^2 = a,\quad j^2 + j = b,\quad ij = (j+1)i.$$ Let us denote it by $\lt a,b]$, as PK Draxl does in his book "Skew Fields". This is a <em>division</em> algebra because $a$ is not a norm from the separable quadratic extension $E$ obtained from $Z$ by adjoining a root of $x^2 + x + b$.</p> <p>Obviously $D$ contains the purely inseparable quadratic extensions $Z(\sqrt{a})$. I claim it also contains the extension $Z(\sqrt{ab})$. To see this, we calculate in the Brauer group: $$\lt a,b] = \lt ab^2,b] = \lt ab,b] + \lt b,b] = \lt ab, b]$$ where the last equality is because $\lt b,b]$ is split, i.e. isomorphic to 2-by-2 matrices, which follows from the fact that $b$ itself is a norm from $E$ (in fact, the norm of the element $x$).</p> <p>Because $b$ is a nonsquare in $K$, we have found two non-isomorphic purely inseparable quadratic extensions in $D$ of exponent 1.</p>