So I have a question from an exercise but don't know how to proceed. The question is,

Let L be a finite Galois extension of K such that G := Gal(L/K) ≃ A5.

a) Prove that for every α ∈ L \ K, L is a splitting field of $m_{α,K(x)}$ over K.

b) Use Galois theory to prove that G does not have any subgroups of order 15 or 20.

c) Prove that there exists a subfield M of L containing K such that [M : K] = 12. Describe Aut(M/K). (Hint: Use the previous question and part b).)

d) Let f(x) ∈ K[x] be monic, irreducible and separable of degree 10. Assume that f(x) has no roots in L and f(x) is reducible in L[x]. Determine the number of irreducible factors and each of their degrees in a factorization of f(x) in L[x].

any help on any part would be appreciated.


1 Answer 1


We use the Galois correspondence to turn information about subgroups of $A_5$ into informations of subextensions $L/M/K$ and vice-versa.

Part (a): For any $\alpha \in L \backslash K$, there is a splitting field of the minimal polynomial $m_{\alpha, K}(x)$ contained in $L/K$ (this is true in general for any Galois extension $L/K$, since $L/K$ is normal). Denote this subextension by $L/M_{sp. \alpha}/K$. We have that $M_{sp.\alpha} / K$ is a Galois extension, hence the corresponding subgroup $\text{Gal}(L/M_{sp.\alpha}) \le \text{Gal}(L/K)$ is normal. It is well-known that $A_5$ is a simple group, so it follows $\text{Gal}(L/M_{sp.\alpha}) = 1$ or $=A_5$. In the second case, $M_{sp.\alpha} = K$, contradicting the choice of $\alpha$. Hence, we conclude $M_{sp.\alpha} = L$.

Part (b): For any $\alpha \in L\backslash K$, a standard result in field theory tells us that $[M_{sp. \alpha} : K] \le n!$ where $n = [K(\alpha) : K]$. By part $(a)$, we know $M_{sp. \alpha} = L$, so $60 = |A_5| = [L: K] \le n!$. It follows that $n\ge 5$. This implies that every nontrivial subextension $L/M/K$ has $[M:K] \ge 5$ (since $M/K$ certainly contains an extension of the form $K(\alpha)/K$). Writing this in the language of groups (using the Galois correspondence), we see that every subgroup of $\text{Gal}(L/K)$ has index at least $5$, i.e. every subgroup has size at most $|A_5|/5 = 60/5 = 12$. In particular, it follows that $A_5$ has no subgroups of size $15$ or $20$.

Part (c): There is certainly a subgroup $H\le A_5$ of index $12$: simply let $H$ be a subgroup generated by a 5-cycle. By the Galois correspondence, this yields a subextension $L/M/K$ with $[M:K] = 12$ and $\text{Gal}(L/M) = H$. Let $M_{fix}$ denote the fixed field of $\text{Aut}(M/K)$, so we have the subextension $M/M_{fix}/K$. Since $M/K$ is not Galois, the extension $M_{fix}/K$ is nontrivial, hence $[M_{fix} : K] \ge 6$. It follows that $[M:M_{fix}] \le 2$. But a result of Artin says $[M:M_{fix}] = |\text{Aut}(M/K)|$. Thus, $\text{Aut}(M/K)$ is either trivial or $\mathbb{Z}/2\mathbb{Z}$. We can do better and pinpoint the automorphism group precisely. The subgroup of $\text{Gal}(L/K)$ consisting of automorphisms sending $M$ to itself is the normalizer of $H$, which we denote $N(H)$. There is a surjective homomorphism $N(H) \to \text{Aut}(L/K)$ with kernel precisely $H$, so $\text{Aut}(L/K) = N(H)/H$. If $|N(H)| > |H|$, then the automorphism group is nontrivial, hence is $\mathbb{Z}/2\mathbb{Z}$. But certainly any $5$-cycle in $A_5$ is normalized by something that isn't its power; for example $(12345)$ is normalized by $(25)(34)$.

Part (d): Let $J$ denote a splitting field of $f$ over $K$. Then the composite $JL$ is a Galois extension of $K$ with $L/K$ a Galois subextension. Hence $\text{Gal}(JL/L) \le \text{Gal}(JL/K)$ is a normal subgroup acting on the set of roots of $f$. Its orbits correspond to the sets of roots of irreducible factors of $f$ over $L$. Since $f$ is irreducible over $K$, its Galois group $\text{Gal}(JL/K)$ acts transitively on the set of roots; a well-known result of group theory says that if $G$ acts transitively on a set and $N$ is a normal subgroup of $G$, then all the orbits of $N$ have equal cardinality. Thus, every irreducible factor of $f$ over $L$ has the same degree. A computation with the orbit-stabilizer theorem gives that the number of irreducible factors of $f$ over $L$ is $[K(\alpha) \cap L : K]$, where $\alpha$ is a root of $f$. Since $[K(\alpha) \cap L : K ] \ge 5$, we conclude that $f$ splits into five quadratic factors over $L$.


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