# Closure of a number field with respect to roots of a cubic

Consider the following property of subfields ${\mathbb K}$ of ${\mathbb C}$ :

$$\text{Any polynomial of degree 3 with coefficients in} \ {\mathbb K} \ \text{has a root in } {\mathbb K} \ \ \ \ (*)$$

Thus, $\mathbb R$ itself satisfies (*), but it is easy to see that there are many other subfields with this property.

If ${\mathbb K} \subseteq {\mathbb R}$ satisfies $(*)$, does ${\mathbb K}(i)$ satisfy $(*)$ also ?

This question was inspired to me by that other one.

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Many other subfields? The only other subfield that comes to mind is the real algebraic numbers. What did you have in mind? –  JSchlather Jan 21 '13 at 6:55
@JacobSchlather Given any coutable field $\mathbb T$, we can construct a tower of degree 3 extensions of $\mathbb T$, whose union $F({\mathbb T})$ is again countable and satisfies (*). So we see that there are at least countably many such subfields. –  Ewan Delanoy Jan 21 '13 at 7:40
Since my answer doesn't work and this question hasn't gotten any more attention. You might consider posting this problem on mathoverflow. –  JSchlather Jan 23 '13 at 4:58
@JacobSchlather I believe the problem in your answer can be fixed, but I'm working on other problems right now. I'll post more about this when I have the time. –  Ewan Delanoy Jan 23 '13 at 5:17

## 2 Answers

I have a counter-example.

Consider the integer polynomial

$$f(x) = (x^3 + x + 1)^2 + 1$$

$f$ is irreduible over $\mathbb{Q}$ and factors into two cubic polynomials over $\mathbb{Q}(i)$. Let $g(x)$ be one of those polynomials$. Using sage, I compute that the splitting field$E$of$f$is degree 72: exactly what you would expect due to the factorization$f = g \bar{g}$. Of particular note is that the Galois group has no subgroups of order 24, and thus the splitting field contains no cubic subfields. Now suppose$K$is any field constructed by starting with$\mathbb{Q}$and iteratively making cubic extensions (transfinite iteration, unless you carefully order the extensions). Since$K$has no degree 3 extensions, every cubic polynomial over$K$has a root in$K$. Also, every finitely generated subfield$L \subseteq K$satisfies$[L:\mathbb{Q}] = 3^n$, and contains a subfield of degree 3 over$\mathbb{Q}$In particular, let$L = E \cap K$. Because$E$has no subfields of degree$3$, we must have$E \cap K = \mathbb{Q}$. However, this would imply$E \cap K(i) = \mathbb{Q}(i)$, and thus$g$is irreducible over$K(i)$. - Thanks for your answer. Can you say more about the Galois group ? Its structure is S3 x S3 x Z/2Z, right ? It would be nice also if you explained in a little bit more detail why it has no subgroup of index 3 (a sage command is fine). – Ewan Delanoy Jan 26 '13 at 9:19$Z/2$comes in as a semidirect product. It's the subgroup of$S_6$generated by the symmetric group on 123, the symmetric group on 456, and by the permutation$(14)(25)(36)$. After constructing$f$, the command to see there were no index 3 subgroups was G = f.galois_group() and then listing for A in G.subgroups(): print len(A). – Hurkyl Jan 26 '13 at 13:16 By the way, you might be interested in a related question I asked here recently : math.stackexchange.com/questions/286655/… – Ewan Delanoy Jan 26 '13 at 16:37 @Ewan: I am; the very same question had went through my mind. – Hurkyl Jan 26 '13 at 21:34 So the exact place where I got stuck is why it's not true. Interesting. I couldn't decide whether or not my approach was too abstract and it actually did work in the specific case of$K(i)$. – JSchlather Jan 27 '13 at 16:11 Let$K$be a perfect field such that every cubic polynomial over$K$has a root in$K$. Let$i \in \overline{K}$be a solution to$x^2+1 \in K$, if$i \in K$there is nothing to prove so we assume not. We claim every cubic polynomial over$K(i)$also takes a root. Suppose not then we can find cubic polynomial$p(x) \in K(i)[x]$such that$p(x)$is irreducible over$K(i)[x]$. Let$L$be the splitting field of p(x), then$L$is also an extension field of$K$. In particular$[L:K]=6,12$. In both cases the Sylow 2-subgroup of$\mathrm{Gal}(L/K)$has index$3$, so its fixed field is a degree$3$extension of$K$. This is impossible, so every cubic polynomial over$K(i)$has a root. Notice that this proof works for any degree$2$extension of$K$. It seems likely to me that such a field does not have any extensions of degree divisible by$3$. But I don't think it's true that every group whose order is divisible by$3$has a subgroup of index$3$, so this technique will likely not generalize. Edit: It was pointed out in the comments that this approach is incorrect because$L/K$need not be Galois. So you have to look at normal closure$E$of$L$over$K$. In order to salvage this approach in this case one would need to show that every transitive subgroup of$S_6$whose order is divisible$3$and$2$has a subgroup of index$3$. I'll leave this up in case its useful to anyone else attempting to solve the problem. - Right. You can kill two birds with one stone : I believe this answer will also fit in the question linked in my OP (a bountied question by the way). – Ewan Delanoy Jan 21 '13 at 7:23 @EwanDelanoy Hmm. The question linked is at a part of Stewart's book that hasn't talked about field extensions. So the OP there probably wouldn't appreciate a proof using Galois theory too much. – JSchlather Jan 21 '13 at 7:33 My mistake, the book is about Galois theory though. – Ewan Delanoy Jan 21 '13 at 7:37 @JacobSchlather : the extension$L/K$may not be Galois. You have to take the splitting field of$x$over$K$, but this field is of degree dividing$6!=720\$. –  user10676 Jan 21 '13 at 10:35
@user10676, Good point. This method probably isn't salvageable then. –  JSchlather Jan 21 '13 at 17:01