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Let $K$ be a number field. $K$ is a normal extension of $\mathbb{Q}$ iff $\exists f(x)\in\mathbb{Q}[x]: K$ is the splitting field for $f(x)$. A field extension is Galois iff it is normal and separable, but since all number fields are separable extensions of $\mathbb{Q}$, normal and Galois coincide in this case. I also recall a theorem saying Galois number fields must have either all totally real or all complex infinite places (embeddings).

A common example is $K=\mathbb{Q}(\sqrt[3]{2})$. I understand why this is not normal: it lacks the complex cube roots of 2. But what if we instead extend $\mathbb{Q}$ by one of the complex roots?

Consider the case where $K=\mathbb{Q}(\zeta_3\sqrt[3]{2})$, where $\zeta_3$ is the primitive cube root of unity. This is a degree 3 extension of $\mathbb{Q}$, with 2 of those degrees corresponding to the identity/conjugation complex place. Since complex places contribute multiples of 2 to the degree, the remaining 1 must correspond to a real place (in particular, having as image the field in the previous paragraph), meaning $K$ cannot be normal. But $K$ is the splitting field for $x^3-2$, [Note: $K$ is not the splitting field - that was the nature of the confusion here. -JQ]. which is the minimal polynomial of $\zeta_3\sqrt[3]{2}$, is it not?

In general I'd like to understand the case of number fields with exactly 1 complex place, and at least 1 real place. I'm able to produce lots of examples, but I don't get why they are not considered normal.

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    $\begingroup$ I don't understand your argument, but anyway, $K$ is not the splitting field, as it does not contain the real cubic root of $2$. $\endgroup$ Commented Oct 13, 2014 at 20:48
  • $\begingroup$ I think this answers my question in that case. I overlooked that $K$ doesn't include the real cube root. So the splitting field is actually $\mathbb{Q}(\sqrt[3]{2}, \zeta_3\sqrt[3]{2})$? The problem is I don't know enough about non-Galois field extensions. What is the primitive element of $\mathbb{Q}(\sqrt[3]{2}, \zeta_3\sqrt[3]{2})$? $\endgroup$
    – j0equ1nn
    Commented Oct 13, 2014 at 21:17
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    $\begingroup$ Some linear combination over $\mathbb{Q}$ of $\sqrt[3]{2}$ and $\zeta_3$, by the primitive element theorem. $\endgroup$ Commented Oct 13, 2014 at 21:53
  • $\begingroup$ Okay, fair enough. $\endgroup$
    – j0equ1nn
    Commented Oct 13, 2014 at 21:55
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    $\begingroup$ Oh, and Galois theory tells you that $z=\root3\of2+\zeta_3$ is a primitive element. For all non-identity automorphisms $\sigma\in Gal(\Bbb{Q}(\root3\of2,\zeta_3)/\Bbb{Q})$ we have $\sigma(z)\neq z$. Thus $Gal(\Bbb{Q}(\root3\of2,\zeta_3)/\Bbb{Q}(z))$ is trivial, so by Galois correspondence $\Bbb{Q}(z)=\Bbb{Q}(\root3\of2,\zeta_3)$. $\endgroup$ Commented Oct 14, 2014 at 12:12

1 Answer 1

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In general I'd like to understand the case of number fields with exactly 1 complex place, and at least 1 real place. I'm able to produce lots of examples, but I don't get why they are not considered normal.

One of the several possible equivalent definitions of "Galois," for a number field $K$, is that every embedding $K \to \overline{\mathbb{Q}}$ has the same image. (If $K$ is the splitting field of some polynomial $f$, then this image is generated by the roots of $f$.) Composed with the natural map $\overline{\mathbb{Q}} \to \mathbb{C}$, yet another equivalent is that every embedding $K \to \mathbb{C}$ has the same image. In particular, either they're all contained in $\mathbb{R}$ or none of them are.

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  • $\begingroup$ Yes I'm aware of that; you phrase it nice and concisely though. Since posting the question, I've remembered a lot of what I'd forgotten about Galois theory - it had been a while. $\endgroup$
    – j0equ1nn
    Commented Oct 23, 2014 at 19:28

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