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→Proof of Lang's theorem: Punctuation of maths: do not think '.' is required at all |
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Define:
:<math>f_a: G \to G, \quad f_a(x) = x^{-1}a\sigma(x)</math>
Then
:<math>(d f_a)_e = d(h \circ (x \mapsto (x^{-1}, a, \sigma(x))))_e = dh_{(e, a, e)} \circ (-1, 0, d\sigma_e) = -1 + d \sigma_e</math>
where <math>h(x, y, z) = xyz</math>. It follows <math>(d f_a)_e</math> is bijective since the differential of the Frobenius <math>\sigma</math> vanishes. Since <math>f_a(bx) = f_{f_a(b)}(x)</math>, we also see that <math>(df_a)_b</math> is bijective for any ''b''.<ref>This implies that <math>f_a</math> is [[étale morphism|étale]].</ref> Let ''X'' be the closure of the image of <math>f_1</math>. The [[smooth point]]s of ''X'' form an open dense subset; thus, there is some ''b'' in ''G'' such that <math>f_1(b)</math> is a smooth point of ''X''. Since the tangent space to ''X'' at <math>f_1(b)</math> and the tangent space to ''G'' at ''b'' have the same dimension, it follows that ''X'' and ''G'' have the same dimension, since ''G'' is smooth. Since ''G'' is connected, the image of <math>f_1</math> then contains an open dense subset ''U'' of ''G''. Now, given an arbitrary element ''a'' in ''G'', by the same reasoning, the image of <math>f_a</math> contains an open dense subset ''V'' of ''G''. The intersection <math>U \cap V</math> is then nonempty but then this implies ''a'' is in the image of <math>f_1</math>.
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