Lemma: Let $L/K$ be a finite extension with $\text{char}(K)=0$. Then there is an $\alpha\in L$ such that $K(\alpha)=L$.

My Proof. Any extension of a field of characteristic zero is separable, so $\alpha$ exists by Existence of Primitive Element in Separable Extension.

My Problem: Isn't it true that my first sentence is only true when $L$ is either algebraic of finite? I haven't assumed either here.


Separability, as a concept, only applies to algebraic extensions. You may define some arbitrary extension to be separable if its algebraic subextension is separable, but this is still simply using the same characterization of separability without introducing anything new.

Infinite separable (and thus, algebraic) extensions need not admit primitive elements, indeed, if an algebraic extension $ L/K $ admits a primitive element, then it must actually be of finite degree (the degree would be equal to the degree of the minimal polynomial of the primitive element over $ K $). The existence of primitive elements in separable extensions holds only for finite extensions.

  • $\begingroup$ Oh ok, so by saying that the extension is separable, I am already implying that it is algebraic? Thanks. $\endgroup$ – PercyF2519 Apr 24 '17 at 20:52
  • $\begingroup$ @PercyF2519 Yes, that's correct. $\endgroup$ – Starfall Apr 24 '17 at 20:53

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