# Simple algebraic extension of fields [closed]

Let $$L/K$$ be a finite extension of fields such that for every $$a\in L\setminus K$$, we have $$[L:K]=[K(a):K]$$. Can we say that $$[L:K]$$ is a prime number?

The answer is no: if $$n>1$$ is an integer, and the field $$K$$ is such that it admits a Galois extension with group $$S_n$$, then there exists a degree $$n$$ extension $$L$$ of $$\mathbb{Q}$$ such that there are no subextensions $$K \subsetneq K' \subsetneq L$$. It follows that, in that case, for any $$a \in L$$ with $$a \notin K$$ we have $$K(a)=L$$.

To show this, let $$F/K$$ be a Galois extension with group $$S_n$$, and let $$L \subset F$$ be the subfield which corresponds to the subgroup $$S$$ of permutations $$\pi \in S_n$$ which satisfy $$\pi(1)=1$$ (or which have any other fixed point, it doesn't matter which one). It is easy to convince yourself that there are no subgroups $$S \subsetneq S' \subsetneq S_n$$, hence by the main theorem of Galois theory there are no subextensions $$K \subsetneq K' \subsetneq L$$.

As an example, the field $$\mathbb{Q}$$ admits $$S_n$$ extensions for every $$n$$, and hence there are field extensions $$\mathbb{Q} \subset L$$ of every degree $$n>1$$ that have no proper subextensions.

For an explicit example: the polynomial $$f = x^4+x+1$$ has Galois group $$S_4$$ over $$\mathbb{Q}$$. If $$F=\mathbb{Q}(\alpha_1,\alpha_2,\alpha_3,\alpha_4)$$ is the splitting field of $$f$$, where the $$\alpha_i$$ are the zeros of $$f$$, then the subextension corresponding to the permutations which keep the $$i$$-th zero fixed is just $$\mathbb{Q}(\alpha_i)$$. So for any zero $$\alpha$$ of $$x^4+x+1$$, the field extension $$\mathbb{Q}(\alpha)/\mathbb{Q}$$ is an example.

• In that case I think the answer is positive, because a group of non-prime order will have non-trivial subgroups (just take the subgroup generated by any element of prime order), hence by the Galois correspondence there must be non-trivial subextensions. Of course, the field extensions constructed by the method described in my answer will not be Galois (unless $n=2$).
– R.P.
Commented Apr 11, 2023 at 20:12
• The field extensions confronting me are Galois. According to your answer, I conclude the answer is yes in the Galois extensions! Thank you so much. You helped me a lot.
– Mary
Commented Apr 11, 2023 at 20:21
• To try to convince myself that there is no subgroup $S \subsetneq S' \subsetneq S_n$: I see $S$ has index $n$ in $S_n$ (as it's essentially $S_{n-1}$). Adding in at least one more element means $S'$ will have index less than $n$. But there is a nontrivial fact that there is no subgroup of $S_n$ with index strictly between $2$ and $n$ for $n \neq 4$ (and one can treat $4$ separately). Alternately, this claim is equivalent to the fact that $S_{n-1}$ is a maximal subgroup of $S_n$, which I also think of as lightly nontrivial. Do you have an easier verification in mind? Commented Apr 11, 2023 at 20:57
• @davidlowryduda What I had in mind was this: suppose $S'$ is a subgroup which strictly contains $S_{n-1}$ (which here means the subgroup of permutations keeping $1$ fixed). Let $s\in S'$ be an element outside $S_{n-1}$, say $s(1)=i \neq 1$. Then $sS_{n-1}$ consists exactly of all the permutations which send $1$ to $i$. But by post-composing $s$ by elements of $S_n$ I can get permutations sending $1$ to any $j$ that I want. This would seem to show that $S_{n-1}sS_{n-1}$ equals the whole $S_n$.
– R.P.
Commented Apr 11, 2023 at 21:37
• Correction: not the whole $S_n$, but the whole complement of $S_{n-1}$ in $S_n$, which is what needed to be shown.
– R.P.
Commented Apr 11, 2023 at 22:33