Proof that on infinite fields, finite separable extensions are all simple.

Question:

Let $$k$$ be a infinite field. If $$F = k(\alpha_1,...,\alpha_r)$$, with each $$\alpha_i$$ separable over $$k$$, prove that there exist $$c_1,...,c_r \in k$$ such that $$F = k(c_1\alpha_1+...+c_r\alpha_r)$$ .

I want to prove it by embedding $$F$$ into $$\mathcal{M}_n(k)$$. I think the proof would be neater than the ones using Galois Theory or explicitly constructing irreducible polynomials.

Let $$[F:k] = n$$. $$c_1\alpha_1+...+c_r\alpha_r$$ can be embedded into $$\mathcal{M}_n(k)$$ and all I have to prove is that the corresponding matrix has rank $$n$$, but do not know how to proceed my proof. Please help.

The standard proof for this uses Galois theory, I hope you are familiar with this. The theorem you're trying to prove is more or less the Primitive Element Theorem, and it seems a difficult problem coming up with the proof on your own. Here is a sketch. I simplify to $$F=k(\alpha,\beta)$$, the general case follows by induction.

Lemma. $$F/k$$ has only finitely many intermediate fields.

Proof. As $$F/k$$ is finite and separable, since $$\alpha$$ and $$\beta$$ are separable, we may take the Galois closure to get a finite Galois extension $$L/k$$ with $$F\subset L$$. This has a finite Galois group, and from the fundamental theorem of Galois theory, we know that intermediate fields correspond to subgroups of the Galois group, of which there must be finitely many. As $$L/k$$ has finitely many intermediate fields, then certainly $$F/k$$ must as well. //

Now, form the subfields $$k(\alpha+c\beta)$$ for $$c\in k$$. These are clearly intermediate fields. As $$k$$ is infinite, but there are only finitely many intermediate fields by the lemma, we see that $$k(\alpha+c\beta) = k(\alpha+c'\beta)$$ for some $$c\neq c'$$. We then see $$(c-c')\beta$$ must lie in this field, and as $$c-c'$$ is non-zero, thus invertible, so must $$\beta$$, and then so must $$\alpha$$. Thus, we get $$k(\alpha+c\beta)=k(\alpha,\beta)=F$$, and we are done.

• Thank you! This proof is neat and helpful. However, the exercise belongs to a chapter before Galois Theory. There should be a proof independent from Galois Theory and I am seeking to find one. I already found a proof by explicitly constructing a minimal polynomial for $(\alpha+c\beta)$ but I thought a proof by the embeddings of field extensions into matrix rings would be nicer and neater. Commented Jul 5, 2023 at 10:29
• I see. One solution could be that, since the above lemma does not at all use the full strength of Galois theory, there should be a more elementary proof of the same result, and then the proof would avoid Galois theory altogether. If what you are really after is a proof by matrix theory, I suggest you update your question to reflect that. Commented Jul 5, 2023 at 11:13
• In addition to SomeCallMeTim’s comment, you should perhaps be more specific about what tools you’re okay with using. We can’t guess what you already know. Commented Jul 5, 2023 at 11:17
• Nice answer. It will be nice to argue that Galois closure of $F/k$ is also a finite extension of $F$ or $k$ in your answer for completeness. This is the only missing piece i see. Is it just a matter going to splitting field of $\alpha,\beta$ on top of $F$ ? Commented Jul 6, 2023 at 1:40
• @Balajisb Yes, taking the splitting field of $\alpha$ and $\beta$ (in any order) will yield the Galois closure, and it is easy to show that splitting fields are finite extensions. Commented Jul 6, 2023 at 13:14