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I am stuck on this exercise from David Cox's Galois Theory.

Let $F \subset L = F(\alpha_1, \dots, \alpha_n)$ be a finite extension, and suppose that $\alpha_1, \dots, \alpha_{n-1}$ are separable over $F$. Prove that $L$ has a primitive element.

By the primitive element theorem applied to $F(\alpha_1, \dots, \alpha_{n-1})$, there is a $\alpha$ in $L$ such that $F(\alpha_1, \dots, \alpha_{n-1}) = F(\alpha)$. So I just need to show that $F(\alpha, \alpha_n)$ has a primitive element, where $\alpha_n$ is algebraic.

But I don't see how to proceed from here, since I don't have that $\alpha_n$ is separable.

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  • $\begingroup$ You will never be able to show that $F(\alpha,\alpha_n)/F$ is separable, beacause there is no reason for $\alpha_n$ to be separable. The concept of "separable extension" and "simple extension (ie generated by one element) are not equivalent. It is easy to find inseparable simple extensions. $\endgroup$
    – GreginGre
    Commented Sep 24, 2019 at 9:40
  • $\begingroup$ I made a mistake, it should be "has a primitive element" Thanks for pointing that out. $\endgroup$
    – eatfood
    Commented Sep 24, 2019 at 10:29

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The thing is, when you prove the primitive element theorem, namely that $L:=F(\alpha,\beta)=F(\gamma)$ for some $\gamma\in L$, you need separability for only one of the roots, say $\beta$ (this is the only case that matters, for you can reduce inductively to it).

Take $\alpha_i, i=1,\dots, r$ and $\beta_j, j=1,\dots, s$ to be the distinct roots of minimal polynomials of $\alpha,\beta$ respectively in a splitting field. Since you can assume $F$ to be infinite (otherwise the proof is very easy), you can find $c\in F$ such that $\theta:=\alpha+c\beta$ differs from any $\alpha_i+c\beta_j$, the elements of form $\frac{\alpha-\alpha_i}{\beta-\beta_j}$ being finite.

If $\mu\in F[x]$ is the minimal polynomial of $\alpha$, you have that $\overline{\mu}(x):=\mu(\theta-cx)\in F(\theta)[x]$ verifies $\overline{\mu}(\beta)=0$ and $\overline{\mu}(\beta_j)\neq 0$ for all $\beta_j\neq\beta$. By the separability of $\beta$ on $F$ (and hence of $F(\theta)$), if $\nu(x)\in F[x]$ is the minimal polynomial of $\beta$, you obtain that $\text{gcd}(\nu(x),\overline{\mu}(x))= x-\beta$ in a splitting field $\overline{F}\supset F(\theta)$.

But $\text{gcd}$'s do not depend on the extension, and therefore you need $\beta\in F(\theta)$, which easily holds $\alpha\in F(\theta)$. Hence, $F(\theta)=F(\alpha,\beta)$.

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  • $\begingroup$ OK, so it is a careful modification of the proof for primitive element. Thanks! $\endgroup$
    – eatfood
    Commented Sep 24, 2019 at 12:05

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