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For a transcendental extension $K(\alpha):K$ for sub-field $K$ of $\mathbb{C}$, $[K(\alpha):K]=\infty$

Showing that the basis for $K(\alpha)$ that describes it as a vector space is infinite leads us nowhere since it would deal with infinite dimensional vector spaces.

So suppose that the degree of the trascendental extension is finite, say $N$. Then if $B$ is the basis for $K(\alpha)$ as a vector space, $|B|=N$ by definition. I do not know how to proceed from here and how to relate to the transcendence of $\alpha$ over $K$.

The book I am working on says that it is enough to show that $1,\alpha,\alpha^2,...$ is linearly independent. But I do not know how to use this without dealing with polynomials of infinite degree (which are not necessarily polynomials).

Any help would be much appreciated!

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  • $\begingroup$ The definition of linear independence only involves finite linear combinations. You just have to deal with polynomials. $\endgroup$ – saulspatz Feb 13 at 18:54
  • $\begingroup$ Yes but I am having difficulty showing how by assuming a finite basis for a transcendental extension, it somehow contradicts the transcendental property of $\alpha$ $\endgroup$ – Malcolm Feb 13 at 18:59
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    $\begingroup$ This is the same as proving that a finite extension is algebraic $\endgroup$ – saulspatz Feb 13 at 19:03
  • $\begingroup$ Thank you! I will be keeping this post up in case anyone wants to reference it in the future to find your very helpful comment :) $\endgroup$ – Malcolm Feb 13 at 19:11

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