I have established that the Galois group for $x^4+7x+1$ is $S_4$ and I have shown that it has real roots by showing that it goes from negative to positive twice and that it can't have 3 real roots because its derivative does not have 3 real roots.
My idea was to use the Galois correspondence, but I can not see why we could not have $\mathbb{Q}(\alpha)$ such that it is a degree 4 or degree 2 extension over $\mathbb{Q}$.
Like if it is of degree 2, then it would just correspond to the group $A_4$ and i cant see the problem with that.
And if it had degree 4 it would correspond to $D_8$ which i also see no issue with either?
I have an upcoming Galois Exam and these questions are in some exams but the solution only says
"this is equivalent to showing that a subgroup of $S_4$ of order 6 is not contained in a subgroup of order 12." which I find unclear. I don't get why that is enough.
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$S_4$ is solvable...Thought that meant that its roots were all constructable. – lulu Jan 02 '25 at 13:39
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2@lulu You are not correct. Constructible requires a sequence of adjunctions of square roots. – ancient mathematician Jan 02 '25 at 13:41
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3@ancientmathematician Oh, of course. $X^3-2$ and all. Thanks. – lulu Jan 02 '25 at 13:43
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I think that $x^3-2$ is a counter example for that, it has Gal group $S_3$ which is solvable but the real root, third sqrt of 2, is not – Not_Bernhard_Riemann Jan 02 '25 at 13:45
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1Linking this related question. May be also this? – Jyrki Lahtonen Jan 03 '25 at 07:57
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There is well known criterion for constructibility. If $\alpha$ is algebraic with minimal polynomial $p(x) \in\mathbb{Q} [x] $ then $\alpha$ is constructible if and only if the degree of splitting field of $p(x) $ over $\mathbb{Q} $ is a power of $2$ (or in other words the order of the Galois group is a power of $2$). – Paramanand Singh Jan 04 '25 at 13:56
1 Answers
$X^4+7X+1$ is irreducible: If it were reducible then it would be reducible in $\mathbb{F}_2[X]$ and so would have either a linear or irreducible quadratic factor there; but none of $X+0,X+1, X^2+X+1$ divide $X^4+X+1$ in $\mathbb{F}_2[X]$.
Hence $\mathbb{Q}(\alpha):\mathbb{Q}=4$.
You say that you have proved the Galois Group is $S_4$ so I assume that. So we have that the subgroup of $S_4$ corresponding to $\mathbb{Q}(\alpha)$ is of index $4$, and so order $6$, and hence is an $S_3$ by the structure of $S_4$.
If the $\mathbb{Q}(\alpha)$ were constructible then we would have to have an intermediate extension of degree $2$ and this would correspond to a subgroup of $S_4$ of order $12$ containing $S_3$. But the only subgroup of $S_4$ of index $2$ is $A_4$. So constructibility is impossible.
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Ok thanks but my confusion arises because of this "If the Q(α) were constructible then we would have to have an intermediate extension of degree 2", i thought degree 4 was fine since it is a power of 2? – Not_Bernhard_Riemann Jan 02 '25 at 14:00
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6I think that if you look at the definition of constructible you'll see that it needs a sequence of adjunctions of square roots. So to be constructible forces the degree to be a power of $2$ but that isn't sufficient. – ancient mathematician Jan 02 '25 at 14:12