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Why are the splitting fields of a polynomial over $\mathbb{Q}$ always Galois? Is that a theorem? Some knowledge too common to write down? I'm just curious because I don't think I've come across that statement in my textbook.

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    $\begingroup$ Often it's by definition. There are many definitions of "Galois." Which one(s) are you familiar with? I'm guessing "normal + separable." Understanding the equivalence of definitions of Galois is (or should be) a rite of passage in Galois theory. The most efficient way to prove it is to first equip the primitive element theorem $-$ do you happen to have that handy? $\endgroup$ – anon Dec 6 '13 at 4:31
  • $\begingroup$ "Let $K$ be a field, $f(x)\in K[x]$ & $F$ be splitting field for $f(x)$ over $K$. Then $Gal(F/K)$ is the \textbf{Galois Group} of $f(x)$ over $K$, or the \textbf{Galois group of the equation} $f(x)=0$ over $K$." I'm still working through what it means exactly. $\endgroup$ – Desperate Fluffy Dec 6 '13 at 4:34
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    $\begingroup$ That's the definition of the Galois group of a splitting field or equivalently of a polynomial. I'm asking you what the definition (your text has) of a field extension being Galois - you have already insinuated that your text does not use the definition "is a splitting field." Surely you wouldn't ask why splitting fields are "Galois" if you don't have a definition for the adjective "Galois"? $\endgroup$ – anon Dec 6 '13 at 4:36
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    $\begingroup$ Okay. Then what do you mean by "Galois" when you ask the question "why are splitting fields Galois"? Should we just go with "normal + separable" as I suggested? I don't understand why you would ask without knowing what the word Galois means. It might also help to actually state what book you use. $\endgroup$ – anon Dec 6 '13 at 4:39
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    $\begingroup$ It should be clear to you that for your question to make sense (to you!) you need to have a definition of what a Galois extensión is! $\endgroup$ – Mariano Suárez-Álvarez Dec 6 '13 at 5:07

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