Let $f(x) \in \mathbb{Q}[x]$ an irreducible polynomial with the splitting field $E$ and let the group $Gal(E/\mathbb{Q})$ be abelian. If $a$ is a root of $f(x)$ then $E=\mathbb{Q}(a)$.
Could you give me some hints how to show this??
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This is in many ways a duplicate of this question/answer. Not casting my powervote in this case, but... – Jyrki Lahtonen Dec 15 '14 at 14:01
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Note that $Gal(E|\mathbb{Q}(a))$ is a subgroup of $Gal(E|\mathbb{Q})$. As $Gal(E|\mathbb{Q})$ is Abelian, $Gal(E|\mathbb{Q}(a))$ is a normal subgroup. So, by the fundamental theorem of Galois theory, $\mathbb{Q}(a)/\mathbb{Q}$ is a normal extension. Thus, all roots of the minimal polynomial of $a$, which happens to be $f(x)$ lie in $\mathbb{Q}(a)$. Thus, the splitting field of $f(x)$ is in fact $\mathbb{Q}(a)$ i.e $E = \mathbb{Q}(a)$. Hope this helps.
SMG
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Could you explain me why $Gal(E/\mathbb{Q}(a))$ is a subgroup of $Gal(E/\mathbb{Q})$ ?? – Mary Star Dec 15 '14 at 01:50
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1Because any automorphism which fixes elements of $\mathbb{Q}(a)$ also fixes elements of $\mathbb{Q}$ as $\mathbb{Q}\subseteq \mathbb{Q}(a)$. – SMG Dec 15 '14 at 01:52
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Do you mean that we have then the following??
$$Gal(E/\mathbb{Q}(a))\triangleleft Gal(E/\mathbb{Q}) \Rightarrow \mathbb{Q}(a) \leq \mathbb{Q} \text{ Galois }$$
– Mary Star Dec 15 '14 at 01:57 -
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1No. I mean that as $Gal(E|\mathbb{Q}(a))$ is normal, $\mathbb{Q}(a)/\mathbb{Q}$ is a normal extension. That is the fundamental theorem of Galois theory. $\mathbb{Q}(a)$ is never a subfield of $\mathbb{Q}$. That happens only when $a \in \mathbb{Q}$. – SMG Dec 15 '14 at 02:00
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1$\mathbb{Q}$ is certainly a subfield of $\mathbb{Q}(a)$. But that is not really an implication of normality of $Gal(E|\mathbb{Q}(a))$. – SMG Dec 15 '14 at 02:02
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1A very nice argument, but this has appeared earlier on the site. – Jyrki Lahtonen Dec 15 '14 at 14:03