# Proving that the degree of transcendental extension is infinite

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!

• The definition of linear independence only involves finite linear combinations. You just have to deal with polynomials. – saulspatz Feb 13 at 18:54
• Yes but I am having difficulty showing how by assuming a finite basis for a transcendental extension, it somehow contradicts the transcendental property of $\alpha$ – Malcolm Feb 13 at 18:59
• This is the same as proving that a finite extension is algebraic – saulspatz Feb 13 at 19:03
• 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 :) – Malcolm Feb 13 at 19:11