Consider the polynomial $f(t) = t^5 -4t+2$ over $\mathbb{Q}$. By Eisenstein's criterion, this is irreducible. By graphing this function, we see that it has 3 real roots, hence 2 (distinct, because they are conjugates) non-real roots. Therefore as 5 is prime, we can use a well known lemma to see that the Galois group of group of $f$ over $\mathbb{Q}$ is $S_5$. Now I want to look at the Galois group of $f$ over $\mathbb{Q}(i)$. Now I have managed to show that the group is either $A_5$ or $S_5$. Let the roots of $f$ be $\alpha_1,...,\alpha_5$ with $$\alpha_1 = x +iy$$ $$\alpha_2 = x-iy$$ $$y\ne0$$ Complex conjugates. Now let $\phi$ be the element of $Gal(f)$ over $\mathbb{Q}$ defined by complex conjugation. I.e. it fixes $\alpha_3,\alpha_4,\alpha_5$ and swaps $\alpha_1$ with $ \alpha_2$. This clearly does $not$ fix $i$, so $Gal(f)$ over $\mathbb{Q}(i)$ (call this group $G$) is not all of $S_5$, hence is $A_5$.

Although the line of reasoning here seems valid, it leads to some weird consequences. Given that $G$ = $A_5$, when we represent automorphisms by cycle types, any double transposition must be in the Galois group, i.e. must fix $i$. In particular $$(\alpha_1,\alpha_2)(\alpha_3,\alpha_4)$$ Must fix $i$. But when we apply this to $\alpha_1$, say, it would seem to involve conjugating it, and hence moving $i$. A contradiction. Yet if my line of reasoning is correct, this double transposition must leave $i$ fixed. How is this possible? Have I made an error somewhere, or just missing something obvious?

Many thanks.

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    $\begingroup$ I haven't worked out the discriminant of $f$, but I presume it is not the negative of a square, that is the quadratic subfield of the splitting field $L$ is not $\Bbb Q(i)$. Then $i\notin L$ so asking whether or not your $\phi$ fixes $i$ is pretty meaningless. $\endgroup$ – Lord Shark the Unknown Jan 13 '18 at 19:29
  • $\begingroup$ Are you saying that 'the' quadratic splitting field is unique? $\endgroup$ – Elie Bergman Jan 13 '18 at 19:37
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    $\begingroup$ The splitting field has a unique quadratic subfield. What is it? Is it $\Bbb Q(i)$ or not? $\endgroup$ – Lord Shark the Unknown Jan 13 '18 at 19:39
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    $\begingroup$ Indeed, you must have computed $\Delta$; is $\Bbb Q(\Delta)\ne\Bbb Q(i)$? $\endgroup$ – Lord Shark the Unknown Jan 13 '18 at 19:48
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    $\begingroup$ A nitpick :-). Because $S_5$ (or $S_4$ for that matter) has a unique subgroup of index two, the splitting field has a unique quadratic subfield. The Galois correspondence reverses order, and big subgroups correspond to small intermediate fields. +1 of course $\endgroup$ – Jyrki Lahtonen Jan 13 '18 at 20:09

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