Theorem. Let $L/K$ be a finite extension. If $L$ is separable over $K$, then a primitive element $c\in L$ exists.

The Proof I have access to:

By induction, the general case is reduced to $L=K(\alpha,\beta)$ where $\alpha,\beta\in L$ are separable over $K$. Let $f_{1},\ldots,f_{n}$ be distinct embeddings of $K$ into the normal closure of $L$. Since $L/K$ is separable, $[L:K]=n$. Set $$P(x)=\prod_{i\neq j}((f_{i}(\alpha)+xf_{i}(\beta))-(f_{j}(\alpha)+xf_{j}(\beta))).$$ Clearly $P(x)$ is a non-zero polynomial and so $\exists\ c\in K:P(c)\neq0$. Then $f_{i}(\alpha+c_{i}\beta)$ are distinct, and so $K(\alpha+c\beta)$ has degree at least $n$ over $K$. But $[K(\alpha,\beta):K]=n$ and so $K(\alpha,\beta)=K(c)$.

My Questions:

  1. What is the induction to reduce the problem to $L=K(\alpha,\beta)$?
  2. Why are $f_{i}(\alpha+c_{i}\beta)$ distinct?
  3. Why is $[K(\alpha+c\beta):K]\geqslant n$?
  • 1
    $\begingroup$ This proof has all the details $\endgroup$ – reuns Apr 23 '17 at 13:00
  • $\begingroup$ I will have a read. Thank you. $\endgroup$ – PercyF2519 Apr 23 '17 at 13:02
  1. If for instance $L=K(\alpha,\beta,\gamma)$ then applying the two-variable version to $L/K(\gamma)$ we get there is $\delta$ with $L=K(\delta,\gamma)$. Now apply two-variable version again.
  2. Because $P(c)\ne0$.
  3. Because all the $f_i(\alpha+c\beta)$ are distinct (so the minimum polynomial of $\alpha+c\beta$ has at least $n$ zeros).
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