# $f \in K[X]$ of degree $n$ with Galois group $S_n$, why are there no non-trivial intermediate fields of $K \subset K(a)$ with $a$ a root of $f$?

Let $K$ be a field and $f \in K[X]$ of degree $n$ with Galois group $S_n$. Let $a$ be a root of $f$, $L = K(a)$, and let $E$ be a intermediate field of the extension $K \subset L$. Prove that $E = K$ or $E = L$.

Any suggestions on where to start? Thanks.

Let $M$ be the splitting field of $f$ and let $a_1,\ldots,a_n$ be the roots of $f$ where $a_1 = a$. Every automorphism of $M|K$ permutes the set $\{a_1,\ldots,a_n\}$ and the fact that the Galois group of $f$ is $S_n$ means that every permutation $\sigma \in S_n$ defines such an automorphism by sending $a_i$ to $a_{\sigma(i)}$. An automorphism $\sigma \in G(M|K)$ fixes $L$ if and only if $\sigma(a) = a$, i.e. $\sigma(1) = 1$ when considered as an element of $S_n$. That means that $G(M|L)$ as a subgroup of $S_n$ is the subgroup $S_{\{2,\ldots,n\}}$ of permutations fixing $1$.

The intermediate extensions $K \subseteq E \subseteq L$ now correspond to the subgroups $G$ of $S_n$ containing $S_{\{2,\ldots,n\}}$, so we have to show that every such subgroup is either $S_{\{2,\ldots,n\}}$ or $S_n$.

If $G$ contains an element $\sigma$ with $\sigma(1) \neq 1$, then it contains all transpositions of the form $$(1\; i) = \sigma^{-1} ( \sigma(1) \; \sigma(i)) \sigma,$$ (unless $\sigma(i) = 1$) and also all transpositions $(i\; j)$ with $i,j \neq 1$. Since $S_n$ is generated by the transpositions it follows that $G = S_n$.

Edit: To show that $(1\; i) \in G$ where $\sigma(i) = 1$ we may assume $n \geq 3$ (the case $n \leq 2$ is trivial). Take any $j$ with $j \neq 1$ and $j \neq i$. Then $(1\;i) = (i\;j)(1\;j)(i\;j) \in G$.

• Thank you. One question though: $G$ contains $(\sigma(1) \sigma(i))$ if $\sigma(i) \not = 1$, because $G$ contains $S_{{2,...,n}}$. But what if $\sigma(i) = 1$. Does $G$ contain $(1 i)$ then? – MrReese Apr 7 '13 at 20:54
• You're right, that case needs to be taken care of separately. See the edit in my answer. – marlu Apr 7 '13 at 22:30

Suggestion for where to start. $$M$$ stands for the splitting field:

1. Identify the subgroup $$H=\operatorname{Gal}(M/L)\le\operatorname{Gal}(M/K)\simeq S_n$$ as a group of permutations.
2. Prove that $$H$$ is a maximal subgroup of $$S_n$$, i.e. there are no subgroups properly between $$H$$ and $$S_n$$.
3. Apply Galois correspondence.

Adding details in response to a request.

Assume that $$f(x)\in K[X]$$ is a degree $$n$$ polynomial such that $$G=Gal(M/K)\simeq S_n$$, $$M$$ a splitting field over $$K$$. Let $$X=\{\alpha_1,\alpha_2,\ldots,\alpha_n\}$$ be the set of zeros of $$f(x)$$ in $$M$$. Let $$L=K(\alpha_1)$$ be an intermediate field. Because any automorphism $$\in G$$ is determined by the way it acts on $$X$$, we see that any permutation $$\pi\in Sym(X)$$ with the property $$\pi(\alpha_1)=\alpha_1$$ comes from an automorphism of $$M$$. We can continue to denote it $$\pi$$, and view it as an element $$\pi\in Gal(M/L)$$. So $$Gal(M/L)$$ is thus identified with the point stabilizer $$H=Stab_{Sym(X)}(\alpha_1).$$

Claim. $$H$$ is a maximal subgroup of $$G=Sym(X)$$.

Proof. Assume that there were a subgroup $$G_1\le G$$ properly containing $$H$$. The set $$X$$ is partitioned into two orbits of $$H$$, namely $$\{\alpha_1\}$$ and the rest of them. If $$\sigma\in G_1\setminus H$$, then $$\sigma(\alpha_1)\neq\alpha_1$$. Therefore the two $$H$$-orbits become a single $$G_1$$-orbit. In other words, $$G_1$$ acts transitively on $$X$$. By the orbit-stabilizer theorem $$|G_1|=|X|\cdot |Stab_{G_1}(\alpha_1)|.$$ But $$H\le Stab_{G_1}(\alpha_1)$$ has order $$(n-1)!$$. Therefore $$|G_1|\ge n(n-1)!=n!$$. OTOH $$G_1\le Sym(X)$$, so $$|G_1$$ can have order $$\ge n!$$ only when $$G_1=Sym(X)$$. QED.

Corollary. There are no intermediate fields between $$K$$ and $$L$$.

Proof. The Galois correspondence would associate such an intermediate field $$E$$ with a group $$Gal(M/E)$$ strictly between $$Gal(M/L)$$ and $$Gal(M/K)$$. But we just saw that no such group exists.

The above argument fails in the case $$K=\Bbb{Q}$$, $$f(x)=x^4-3$$ because the splitting field $$M=K(\root4\of3,i)$$ is a degree eight extension of $$K$$ only. Therefore $$Gal(M/K)$$ has order eight as well, and thus must be isomorphic to the dihedral group $$D_4\le S_4$$ of order eight. The point stabilizer of $$D_4$$ in $$S_4$$ has order two only (the orbit-stabilizer theorem again). More importantly, that point stabilizer is not a maximal subgroup of $$Gal(M/K)$$. This allows the existence of intermediate fields between $$L=K(\root4\of3)$$ and $$K$$.

But, all this depends heavily on basic results of Galois theory. I don't know of a way of communicating this argument without either referring to Galois theory (or possibly reproducing its relevant parts).