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>> No.9915316 [View]
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9915316

Can anyone answer the second part of the question? If [math]\alpha[/math] is an infinite extension, the clearly it is an isomorphism [math]k(\alpha)\cong k(t)[/math] sending [math]\alpha[/math] to [math]t[/math]. So I have to assume that the extension is then finite to find some counterexample, but any element of [math]k(\alpha)[/math] is a rational expression with the top and bottom of the fraction being linear combinations of powers of [math]\alpha[/math]... how is it not going to be an onto function?

Maybe there is a typo and the question meant to have the arrow the other way around? Or they meant to say not injective?

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