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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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Ewan Delanoy
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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 06:55
  • @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 07:40
  • 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 04:58
  • @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 05:17

2 Answers2

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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 09:19
  • $Z/2$ 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). –  Jan 26 '13 at 13:16
  • By the way, you might be interested in a related question I asked here recently : http://math.stackexchange.com/questions/286655/number-fields-with-all-degrees-equal-to-a-power-of-three – Ewan Delanoy Jan 26 '13 at 16:37
  • @Ewan: I am; the very same question had went through my mind. –  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
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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.

JSchlather
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  • 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 07:23
  • @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 07:33
  • My mistake, the book is about Galois theory though. – Ewan Delanoy Jan 21 '13 at 07:37
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    @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