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For the following polynomials I need to find out if the splitting field over $\mathbb {Q}$ has a subfield M such that M:$\mathbb {Q}$ is not normal.

1) $ x^6-7$

2)$ x^3 + 3x +3 $

3)$x^{100} - 1$

I know that by Eisenstein's criterion the first 2 are irreducible but the third is reducible. I can't think of any simple way of finding if these polynomials contain such subfields.

I was thinking of possibly using the fundamental theorem of galois theory which states that M:$\mathbb {Q}$ is normal iff Gal(L:M) is a normal subgroup of G but I'm not sure how.

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For the first two, you can find a real, non-rational root $\alpha \in \mathbb{R} \setminus \mathbb{Q}$. (E.g. $\alpha = \sqrt[6]{7}$ for the first one; and $\alpha$ exists for the second one by the intermediate value theorem applied to $x = -1$ and $x = 0$, and isn't rational by the rational roots theorem.) Then $Q[\alpha]:Q$ is an intermediate extension that isn't normal, for the given polynomials have $\alpha$ as roots but don't split completely over $\mathbb{Q}[\alpha]$.

For the last one, you can use the fact that the Galois group of $\mathbb{Q}[\zeta]:\mathbb{Q}$, where $\zeta$ is a primitive $100$th root of unity, is isomorphic to the group of units of $\mathbb{Z}/100$, hence abelian. Then use the fundamental theorem relating the normal extensions with the normal subgroups of the Galois group.

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  • $\begingroup$ So is $Q[\alpha]:Q$ only normal if you can't form a polynomial in $Q$ for which $\alpha$ is a root but the other roots lie outside of the extension? $\endgroup$ – user2973447 Jan 3 '16 at 1:08
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    $\begingroup$ @user2973447 Yes, an irreducible polynomial. This is essentially the definition of a normal extension: $L:K$ is normal if every irreducible polynomial in $K[x]$ that admits a root in $L$ splits completely over $L$. $\endgroup$ – Alex Provost Jan 3 '16 at 1:26
  • $\begingroup$ for the last part do you mean the integers modulo 100 or something else? Also is there a way of identifying quickly if a given group is isomorphic to a well known one? I see people saying "this is because G is isomorphic to H therefore..." quite a lot but that relation isn't one which is immediately obvious to me and usually takes me some time to check $\endgroup$ – user2973447 Jan 3 '16 at 15:05
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    $\begingroup$ @user2973447 Yes, I mean the integers mod 100. While in general there is no easy way of telling if two groups are isomorphic, if people use that phrasing then usually there is an "obvious" isomorphism, or a classical theorem that you can use. In this case, there is the well-known result that the Galois group of the $m$th cyclotomic polynomial embeds into the group of units of $\mathbb{Z}/m$ by considering what power of itself a primitive root of unity is mapped to by an automorphism; and since the cyclotomic polynomial is irreducible over $\mathbb{Q}$, that embedding is surjective, thus iso. $\endgroup$ – Alex Provost Jan 3 '16 at 17:04
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Hint: An idea here is to find if you have a complex root and a root in $R-Q$ for example $x^6-7$, you have the root $7^{1/6}$ and $e^{2i\pi/6}$ so $Q(7^{1/6})$ is a subfield which is not normal.

For the second, there is a real root $\alpha$ since the degree of the polynomial is 3, its derivative is $3x^2+3$ thus it is a strictly increasing real function who has only one real roots, the two others roots are complex, since $x^3+3x+3$ is the minimal polynomial of $\alpha$, $Q(\alpha)$ is not normal.

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  • $\begingroup$ So does that mean that if we have complex roots then none of the real roots will be normal? If I'm understanding that right is there a reason for this? $\endgroup$ – user2973447 Jan 3 '16 at 0:33
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    $\begingroup$ If the minimal polynomial of the real root is the given polynomial, then the subfield it generates over $Q$ can't be normal. $\endgroup$ – Tsemo Aristide Jan 3 '16 at 0:34
  • $\begingroup$ So if you extend over a root, the extension is only normal if it also includes all other roots which are solutions to a polynomial including the original root? $\endgroup$ – user2973447 Jan 3 '16 at 0:57

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