Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In a broad sense, what relationships are there between the torsion subgroup $G$ of the multiplicative group of non-zero elements of a field $K$ and whether or not certain polynomials in $K[x]$ have roots in $K$ (the failure of $K$ to be algebraically closed)? How much can you use $G$ to determine exactly which polynomials in $K[x]$ will have roots in $K$?

For example, the multiplicative group $\mathbb{R}\setminus\{0\}$ has the torsion subgroup $\{1, -1\}$, which is isomorphic to $C_2$, and the only irreducible polynomials in $\mathbb{R}[x]$ have degrees $1$ and $2$. By comparison, the multiplicative group $\mathbb{C}\setminus\{0\}$ has a torsion subgroup consisting of the $n$th roots of unity for all $n$, which is isomorphic to $\mathbb{Q}/\mathbb{Z}$, and $\mathbb{C}$ is algebraically closed. These two properties would seem to be closely related. However, $\mathbb{Q}\setminus\{0\}$ has the same torsion subgroup as $\mathbb{R}\setminus\{0\}$, but $\mathbb{Q}$ has irreducible polynomials of every degree. So what relation is there?

This is an open-ended question, so I'm just looking for some useful information, not necessarily comprehensive treatments. Thanks!

share|cite|improve this question
Perhaps I miss something, but I don't think there is any relation at all...and I can't see any straightforward reason to think there should be one. – DonAntonio Nov 11 '12 at 3:58
up vote 2 down vote accepted

You’re stumbling towards something very interesting. If you had asked about the Galois group of the maximal abelian extension of a field, then that would have been better. Of course the max ab of $\mathbb R$ is $\mathbb C$.

But just look at $\mathbb C\supset \mathbb R$ as a finite abelian extension of complete fields. Now the right way to think of the Galois group is not that it’s the group of roots of unity in $\mathbb R$, but rather ${\mathbb R}^*/\mathbf N(\mathbb C^*)$, where $\mathbf N$ is just the field-theoretic norm, in this case sending $z=a+bi$ to $z\overline z=a^2+b^2$. The interesting thing is that if $K\supset F$ is a finite abelian extension of $p$-adic fields (so, finite extensions of ${\mathbb Q}_p$) you’d have the very same relation, that the Galois group is isomorphic to $F^*/\mathbf N(K^*)$, where of course now the norm $\mathbf N$ is given by a much different formula. This relation can be extended to a description of the Galois group of the maximal abelian extension of a $p$-adic field; all this is Local Class-Field Theory, you can look it up here and there.

Beyond all that, I’d like to say that you need to have more different fields in your bag of tricks. Get to know intimately a few finite extensions of $\mathbb Q$, especially some cyclotomic ones, all the finite fields, and a good number of $p$-adic fields, not just ${\mathbb Q}_p$. Then you’ll start to make some interesting guesses.

share|cite|improve this answer

Let $K$ be the subfield of $\mathbb{C}$ generated by all the $n$-th roots of unity for all $n$ over $\mathbb{Q}$. Then $Gal(K/\mathbb{Q})$ is abelian. Since there are infinitely many irreducible polynomials in $\mathbb{Q}[X]$ whose Galois groups are not abelian, there are infinitely many irreducible polynomials of degree greater than one in $K[X]$. In particular $K$ is not algebraically closed.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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