eof
Mar28-10, 08:54 AM
This one step in a long problem (an example in algebraic geometry cast in another language) that I've condensed below (p is the characteristic):
Let \overline{k} be an algebraic closure of k and let L=\overline{k}(Y). Consider the monic polynomial
H(x)=x^s-(Y^p+Y^{-t})
in L[x]. Write s=p^ms' for some integer s' prime to p and some m\geq 0. Let x_0 be a root of H(x) in an algebraic closure \overline{L} of L. Show that the field L(x_0) is an inseparable degree p^m extension of L(x_1) where x_1=x_0^{p^m}.
Ok, so proving that the extension is inseparable is trivial. It then follows that the minimal polynomial of x_0 over x_1 has to be of the form (X-x_0)^{p^n}=X^{p^n}-x_0^{p^n}. This shows that x_0^{p^n}\in L(x_1). Unfortunately, this only gives that n\leq m, but it does not show equality. Does anyone have any ideas on how to approach this (I spent about 2 hours yesterday thinking about this...).
Let \overline{k} be an algebraic closure of k and let L=\overline{k}(Y). Consider the monic polynomial
H(x)=x^s-(Y^p+Y^{-t})
in L[x]. Write s=p^ms' for some integer s' prime to p and some m\geq 0. Let x_0 be a root of H(x) in an algebraic closure \overline{L} of L. Show that the field L(x_0) is an inseparable degree p^m extension of L(x_1) where x_1=x_0^{p^m}.
Ok, so proving that the extension is inseparable is trivial. It then follows that the minimal polynomial of x_0 over x_1 has to be of the form (X-x_0)^{p^n}=X^{p^n}-x_0^{p^n}. This shows that x_0^{p^n}\in L(x_1). Unfortunately, this only gives that n\leq m, but it does not show equality. Does anyone have any ideas on how to approach this (I spent about 2 hours yesterday thinking about this...).