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 $(1\;i) = (i\;j)(1\;j)(i\;j) \in G$.

  • 1
    $\begingroup$ 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? $\endgroup$ – MrReese Apr 7 '13 at 20:54
  • $\begingroup$ You're right, that case needs to be taken care of separately. See the edit in my answer. $\endgroup$ – 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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.