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Let $E$ be a field extension of $F$. Suppose $\alpha\in E$ is transcendental over $F$. Is it true that $\{\alpha^n:n\in\Bbb Z \}$ is a basis for the simple extension $F(\alpha)$ over $F$? In fact, I want to write a basis explicitly for $F(\alpha)$ over $F$. Thanks!

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    $\begingroup$ You are basically asking for an $F$-basis of the field $F(x)$ where $F$ is an indeterminate. The powers $1, x, x^2, \ldots$ span $F[x]$ and are linearly independent. To get rational functions with denominators you should think about how the partial fraction decomposition works in $F(x)$: you'll want to use certain rational functions whose denominators run over powers of monic irreducibles in $F[x]$. $\endgroup$ – KCd May 30 '18 at 17:44
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No, it is not true. How would you express, say $\frac1{1+\alpha}$ as a linear combination of elements of the form $\alpha^n$ with $n\in\mathbb Z$? If you could do it, then $\alpha$ would be algebraic.

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