l*****e
发帖数: 65 | 1
a=b^2+c^2
if charF=2, then Frobenius map x I-> x^2 is an isomorphism, nothing need to
prove.
If charF!=2, consider the set S={x^2: x\in F}, it contains (F-1)/2+1
elements.since half of non-zero elements is the square of some other element,
and 0 is in S.
Now the set {a-x^2: x\in F} also has (F-1)/2+1 elements, but there are only
(F-1)/2 elements are not in S, so there exists some y^2 st. a-x^2=y^2. done.
Just counting, hehe. |
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