I can't solve this problem. Please help me .

Let $F$ be a field of characteristic $0$ , contained in its algebraic closure $A$. Let $a∈A$ and suppose $a∉F$, but every finite extension $E$ of $F$ with $E\neq F$ contains $a$. In other words, $F$ is a maximal subfield of $A$ not containing $a$. Prove that every finite extension of $F$ is cyclic.

I tried to solve it like below. Let $f(x)$ be a minimal polynomial of $a$ over $F$. Let $deg(f)=n$. Let $E/F$ be finite $m$ dimension such that $(m,n)=1$. Because $E/F$ is finite dimension, $a∈E$. So $F⊂F(a)⊂E$. Because $(m,n)=1$, $F=F(a)$ so $a∈F$. Contradict to $a∉F$.

    I had a contradiction. What is wrong? Does it mean no finite extension of $F$ such that $(m,n)=1$? if so

I suppose any finite extension of $F$ is not $(m,n)=1$. Let $σ$ be Automorphism of $F(a)/F$. then $σ(a)$ be a root of $f(x)$. Then $F(σ(a))$ is finite extension of $F$ and $F(σ(a))⊂A$.So,$a∈F(σ(a))$.So,$F(a)⊂F(σ(a))$. Because $[F(a):F]=deg(f)=n$. So ,$F(a)=F(σ(a))$. So $F(a)$ is minimal field of $f(x)$. Because $F$ is a field of characteristic $0$, $F(a)$ is separable over $F$. So $F(a)/F$ is a Galois extension. If n is not prime , there exist nutural number $r$ such that $1<r<n, r|n$. Let G be Galois group of $F(a)/F$. Let H be a subgroup of $G$ with order $r$. Let K be fix field of $H$ over $F$. Then $a∉K$and $K/F$ is finite extention. So contradiction .So n need to be a prime number. Because of n is prime number, Galois group of $F(a)/F$ is cyclic.

After that what should I do.

I tried to solve it again.

Let $N$ be a Galois group of $E/F(a)$. Since order of $G/H$ is $p$, there exist $g∈G$ such that $g∉H$. $g$ generates a cyclic group $<g>$. Let $K$ be the fix field of $<g>$, then $E/K$ is finite dimention, so $E/F$ is finite dimention . If not $K=F$, then $F(a)⊂K$. So, $g∈H$. Contradiction. Hence, $K=F$. So $G=<g>$. Hence $E/F$ is cyclic. Q.E.D.

I could solve it. Is it correct?


1 Answer 1


Let $L/F$ be finite and Galois, $G$ be its Galois group.

We first show that for any two proper subgroups $H_1,H_2$ of $G$, the join $\langle H_1,H_2 \rangle$ is still proper. Let $K_i = L^{H_i}$, the fixed fields of $H_i$. Then $K_1 \cap K_2 = L^{\langle H_1,H_2 \rangle}$, by assumption, $K_1 \cap K_2 \neq F$, hence $\langle H_1,H_2 \rangle$ is proper.

Now, let $|G|=p_1^{r_1}\cdots p_m^{r_m}$. If $m>1$, then we can select subgroups $H_i$ with order $p_i^{r_i}$. The subgroups generated by proper subgroups $H_i$ will be $G$, contradiction. Thus $|G|$ is a prime power.

Now let $H$ be the subgroup generated by all the proper subgroups of $G$, then by above, $H$ is still proper, and contains all proper subgroups. Choose $g\notin H$, then $\langle g \rangle$ must be the whole group $G$, thus $G$ is cyclic.

Let $K/F$ be finite extension, $N$ be its Galois closure, then by above, $N/F$ is cyclic, thus $K/F$ is also Galois and cyclic.

  • $\begingroup$ Thank you for answer. About lemma 1 I couldn't understand some part. "Since a is contained in all the nontrivial extension of F, the intersection of all nontrivial subgroups of G is nontrivial." Why non trivial? $\endgroup$
    – tarou
    Mar 12, 2018 at 16:11
  • $\begingroup$ I thought it could be happens that the intersection of all nontrivial subgroups of G is trivial. $\endgroup$
    – tarou
    Mar 12, 2018 at 17:01
  • $\begingroup$ @tarou I revamped my answer, it turns out to be quite simple. $\endgroup$
    – pisco
    Mar 12, 2018 at 18:04

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