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$\newcommand\Q{\mathbb Q}$Is it possible to find an irreducible polynomial $f\in \Q[x]$ of degree $4$ such that the following holds:

  1. All roots of $f$ are non-real
  2. The splitting field of $f$, $K_f$, contains "the" imaginary unit $\mathrm i$
  3. The maximum real subfield of $K_f$, i.e. $K_f\cap \mathbb R$, is not Galois over $\Q$
  4. $\mathrm{Gal}(K_f/\Q)\cong A_4$

I have the feeling that this cannot be done but I'm not sure how to prove this. Item 3. suggests that $K_f$ should not be some sort of cyclotomic extension but that is all I have figured. Can it be some radical extension? there is not much room to work with because I have only degree $4$.

In general, I would also like to have an intuition what the implication of having the imaginary unit is in a Galois number field. Will the Galois group have some specific property or will there be some characterizations for such fields?

Edit: I managed to find an example that satisfies 1,3 and 4. For instance $f=x^4 - x^3 - 3x + 4$.

quantum
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    Do you know the Fundamental Theorem of Galois Theory? You may want to consider the fixed subfield of the complex conjugation. – daruma Aug 09 '18 at 05:55
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    We have $i \in K_f$ if and only if $\mathbf Q(i) \subset K_f$. For number fields $E$ and $F$ that are Galois over $\mathbf Q$, Bauer's Theorem implies that $F \subset E$ if and only if the prime numbers splitting completely in $E$ also split completely in $F$. Taking $F = \mathbf Q(i)$ we get that $\mathbf Q(i) \subset E$ if and only if the primes splitting completely in $E$ all satisfy $p \equiv 1 \bmod 4$. So that is one characterization of Galois number fields containing $i$. – KCd Aug 09 '18 at 06:05
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    If you satisfy (4) and your field is not totally real, then you automatically satisfy (3). This is because the fixed field of complex conjugation is precisely $K_f\cap \mathbb R$ and for this field to be galois over $\mathbb Q$, the subgroup (of order 2) generated by complex conjugation should be normal in $A_4$. But no subgroup of order $2$ is normal in $A_4$. – Asvin Aug 09 '18 at 06:07
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    @KCd I think your answer deserves to be promoted to an answer post (as opposed to just a comment). It answer my second question. – quantum Aug 09 '18 at 07:09

1 Answers1

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Items 2 and 4 are incompatible. Assume that $$ \Bbb{Q}\subset \Bbb{Q}(i)\subset K_f, $$ where $K_f$ is Galois over $\Bbb{Q}$, $G=Gal(K_f/\Bbb{Q})\simeq A_4$. By Galois correspondence $\Bbb{Q}(i)$ is the fixed field of a subgroup $H\le G$. Because $[\Bbb{Q}(i):\Bbb{Q}]=2$ it follows that $|H|=|A_4|/2=6$.

But $A_4$ has no subgroups of order six.

Jyrki Lahtonen
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    More generally, if the Galois group is $A_4$, there cannot be any quadratic subfields at all. So we can also conclude that for example $\sqrt2,\sqrt3,\sqrt{-2}\notin K_f$. – Jyrki Lahtonen Aug 09 '18 at 06:38
  • Wonderful argument. Thank you. – quantum Aug 09 '18 at 07:06
  • What if I replace $A_4$ with $S_4$? I'm thinking that an example is possible. – quantum Aug 10 '18 at 09:47
  • @quantum I cannot immediately think of a reason why one would not exist. I don't have an off-the-shelf example though. You need the discriminant of the polynomial to be the negative of the square. That may be worth another question. – Jyrki Lahtonen Aug 10 '18 at 10:14
  • I know that for $A_n$ the discriminant is the square of an integer. But how did you concluded that for $S_4$ with the given conditions? At the moment I don't have an example from databases, but I'll update once I see one (if it exists). Thank you for the intuition though! – quantum Aug 10 '18 at 10:46