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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?

Or a better question, why there is only one maximal unique subfield which is purely inseparable of exponent one?

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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 division algebra because $a$ is not a norm from the separable quadratic extension $E$ obtained from $Z$ by adjoining a root of $x^2 + x + b$.

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 $x$).

Because $b$ is a nonsquare in $K$, we have found two non-isomorphic purely inseparable quadratic extensions in $D$ of exponent 1.

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Why is the field $Z(\sqrt{a})$ not isomorphic to $Z(\sqrt{ab})$? In fact, I believe those fields are isomorphic as $F$-algebras (though not as $Z$-algebras), since $Z = F(a,b) = F(ab,b)$. –  George McNinch Aug 12 '10 at 14:56
    
I meant "not isomorphic as $Z$-algebras" (which is obvious because $b$ is not a square in $Z$); I agree they are isomorphic as $F$-algebras. I think $Z$-algebra isomorphism is what was intended in the original question. –  Skip Aug 12 '10 at 18:16
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