3
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ I don't understand your argument, but anyway, $K$ is not the splitting field, as it does not contain the real cubic root of $2$. $\endgroup$ – Amitai Yuval Oct 13 '14 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 Oct 13 '14 at 21:17
  • 2
    $\begingroup$ Some linear combination over $\mathbb{Q}$ of $\sqrt[3]{2}$ and $\zeta_3$, by the primitive element theorem. $\endgroup$ – Amitai Yuval Oct 13 '14 at 21:53
  • $\begingroup$ Okay, fair enough. $\endgroup$ – j0equ1nn Oct 13 '14 at 21:55
  • $\begingroup$ Hopefully you know about this, but just in case: $K$ itself is not closed under complex conjugation. It contains only one of the roots of $p(x)=x^3-2$. Also, if $\alpha_1$, $\alpha_2$ and $\alpha_3$ are the three zeros of $p(x)$, then all the fields $\Bbb{Q}(\alpha_i)$, $i=1,2,3$, are isomorphic to each other (and to $\Bbb{Q}[x]/\langle p(x)\rangle$). Thus it would be surprising if they had a different set of embeddings to $\Bbb{C}$. $\endgroup$ – Jyrki Lahtonen Oct 14 '14 at 12:04
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Oct 23 '14 at 19:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.