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
i2=a,quadj2+j=b,quadij=(j+1)i.
Let us denote it by lta,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 x2+x+b.
Obviously D contains the purely inseparable quadratic extensions Z(sqrta). I claim it also contains the extension Z(sqrtab). To see this, we calculate in the Brauer group:
lta,b]=ltab2,b]=ltab,b]+ltb,b]=ltab,b]
where the last equality is because ltb,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.
No comments:
Post a Comment