# Let $L = F(\alpha_1, \dots, \alpha_n)$ be a finite extension, with all $\alpha_i$ except $\alpha_n$ separable over $F$, then $L$ has primitive element

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.

• 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. Commented Sep 24, 2019 at 9:40
• I made a mistake, it should be "has a primitive element" Thanks for pointing that out. Commented Sep 24, 2019 at 10:29

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, $$F(\theta)=F(\alpha,\beta)$$.