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Let $K$ be a Galois extension of $\mathbb{Q}$ whose Galois group is isomorphic to $S_5$. Prove that $K$ is the splitting field of some polynomial of degree $5$ over $\mathbb{Q}$.

Since $K$ is a finite Galios extension over $\mathbb{Q}$ we know that $K$ is the splitting field of a separable polynomial $f$ over $\mathbb{Q}$. Let $n$ be the degree of this separable polynomial. Since the Galois group acts on the roots $f$ via permutation we know that the Galois group is isomorphic to a subgroup of $S_n$ and hence $n \geq 5$. Let $\alpha$ be a root of $f$. Since since $|K : \mathbb{Q}| = |K :\mathbb{Q}(\alpha)| |\mathbb{Q}(\alpha):\mathbb{Q}| = n|K :\mathbb{Q}(\alpha)|$, we have that $n|5! = 120$. Therefore $n \in \{5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120\}$

Any help would be appreciated.

GuySa
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Samantha Wyler
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    The subject line of your post is rather bland. – KCd Jul 19 '20 at 19:29
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    Using Galois theory, think about what subgroup of ${\rm Gal}(K/\mathbf Q)$ should correspond to a subfield generated by a root of the desired $5$th-degree polynomial. If $F$ is an intermediate field corresponding to the subgroup $H$ of ${\rm Gal}(K/\mathbf Q)$, a Galois extension of $\mathbf Q$ containing $F$ corresponds to a normal subgroup of ${\rm Gal}(K/\mathbf Q)$ contained in $H$. – KCd Jul 19 '20 at 19:33
  • I don't think I see what you are getting at. Since $S_4$ is normal subgroup of $S_5$, we have that there exists an element $b \in K$ such that $S_4 =$ Gal$(K/\mathbb{Q}(b))$, and then deg(min$_b(\mathbb{Q})) = 5$ and $\mathbb{Q}(b)$ is Galois over $\mathbb{Q}$. – Samantha Wyler Jul 19 '20 at 20:24
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    $S_4$ is not a normal subgroup of $S_5$. –  Jul 19 '20 at 21:47
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    First, be careful about saying $S_4$ is a subgroup of $S_5$, normal or not. There are multiple subgroups of $S_5$ that are isomorphic to $S_4$: for each $i \in {1,2,3,4,5}$, the permutations in $S_5$ fixing $i$ are a subgroup isomorphic to $S_4$. And they are all conjugate to each other, so none is normal in $S_5$. Second, when $n \geq 5$ the only normal subgroups of $S_n$ are ${(1)}$, $A_n$, and $S_n$. (In $S_4$ there are two nontrivial proper normal subgroups.) – KCd Jul 19 '20 at 22:47
  • @TokenToucan you right, can't believe I just said $S_4$ is normal. – Samantha Wyler Jul 20 '20 at 14:51

1 Answers1

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So I think this is it.

Since there are only finitely may subgroups of $S_5$, there are only finitely many intermediate fields of $K|\mathbb{Q}$, and so each intermediate field is a simple extension of $\mathbb{Q}$. Since $S_4$ is a subgroup of $S_5$, there exists an intermediate field $\mathbb{Q}(b)$ such that $S_4 = \mathrm{Gal}(K/\mathbb{Q}(b))$. Since $24 = |K : \mathbb{Q}(b)|$ we have $5 = |\mathbb{Q}(b): \mathbb{Q}|$ and hence $\deg \min(b,\mathbb{Q}) = 5$.

Let $g =\min (b,\mathbb{Q})$. Now since $K$ is Galois over $\mathbb{Q}$, we know that $g$ splits in $K$, and furthermore the splitting field $F$ of $g$ over $\mathbb{Q}$ contained in $K$, is a Galois extension of $g$ and so Gal$(K/F)$ must be a normal subgroup of $S_5$ that is contained in $S_4$. Since the only such subgroup is the trivial subgroup is the trial subgroup, it follows that $K = F$, and hence $K$ is the splitting field of $g$.

Jyrki Lahtonen
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Samantha Wyler
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