1
$\begingroup$

Let $K/\Bbb Q$ be a finite Galois extension.

Denote by $U_K$ the units group of $\mathcal O_K$.

I am studying the impact of the finiteness of $U_K/(U_K\cap \Bbb R)$ on the existence of non real embeddings of $K$ in $\Bbb C$.

I have the following guess:

$U_K/(U_K\cap \Bbb R)$ finite $\Leftrightarrow$ there are no non real embeddings of $K$ in $\Bbb C$

If my guess is correct, I don't see how to prove it from the following manipulation:

By the Dirichlet Unit theorem

$U_K\cong T\times \Bbb Z^{r+s-1}$ with $r,s$ respectively the number of real, complex pair embeddings and $T$ the subgroup of roots of unity in $K$.

Then:

$U_K/(U_K\cap \Bbb R) \cong (T\times \Bbb Z^{r+s-1})/((T\cap \Bbb R)\times (\Bbb Z^{r+s-1}\cap \Bbb R)$ (step 1)

$\cong (T\times \Bbb Z^{r+s-1})/((T\cap \Bbb R)\times \Bbb Z^{r+s-1})$ (step 2)

$\cong T/(T\cap \Bbb R)$ (step 3).

I am not sure if something is incorrect in the steps above or otherwise, how to go further.

Thank you for any help to see this more clearly.

$\endgroup$
9
  • 3
    $\begingroup$ Did you forget to write something on the left hand side of that $\Leftrightarrow$ sign? Because as it's written, the left hand side does not have a truth value. $\endgroup$
    – Lee Mosher
    Nov 27, 2019 at 4:15
  • 1
    $\begingroup$ Another strange detail in a possibly cool question is the tag finite-fields. The field $\Bbb{Q}$ already has infinitely many elements, so the same holds for $K$. I don't see any finite fields here. $\endgroup$ Nov 27, 2019 at 4:26
  • $\begingroup$ But as far as I can tell the real problem in your thinking is the following. Let $M=K\cap\Bbb{R}$. You have not considered the parameters $r$ and $s$ of the subfield $M$? If $K=\Bbb{Q}(i)$ with $r=0,s=1$, then $M=\Bbb{Q}$ with $r=1, s=0$. If $K=\Bbb{Q}(\root3\of2)=M$ then $r=s=1$. If $K=\Bbb{Q}(\root3\of2,e^{2\pi i/3})$ then $r=0,s=3$ but with $M$ you get $r=s=1$ as in the previous example. $\endgroup$ Nov 27, 2019 at 4:33
  • $\begingroup$ In other words, it is anything but clear what can be said about $\Bbb{Z}^{r+s-1}\cap\Bbb{R}$. Remember that this particular copy of $\Bbb{Z}^{r+s-1}$ is a subgroup of $K^*$. $\endgroup$ Nov 27, 2019 at 4:35
  • 1
    $\begingroup$ Actually my original problem has to do with proving that that quotient is finite iff the complex conjugation is in the center of Gal(K/\Bbb Q). This last condition I know it's equivalent to the fact that there are non non real embeddings from $K\cap \Bbb R$ to $\Bbb C$ $\endgroup$
    – Conjecture
    Nov 27, 2019 at 20:29

1 Answer 1

3
$\begingroup$

Let $K/K^{+}$ be any non-trivial extension of number fields. Then the natural statement is that $U_K/U_{K^+}$ is infinite unless $K^{+}$ is totally real and $K$ is a totally complex quadratic extension, i.e. a CM field. In particular, if $K$ is Galois over $\mathbf{Q}$, then every conjugate of complex conjugation fixes $K^{+}$ and so must lie in $\mathrm{Gal}(K/K^{+})$, which is thus normal and so central. Let's prove this.

Suppose that $K^{+}$ has signature $(r,s)$, and that $K/K^{+}$ has degree $d \ge 2$. The complex places $s$ of $K^{+}$ split into $ds$ complex places of $K$. $m \le r$ of the real places of $K^{+}$ remain real, and so give $dm$ real places of $K$, and the other $(r-m)$ real places split, and thus give $d/2(r-m)$ complex places of $K$ (the real places can only split when $d$ is even). It follows that $K$ has signature $(R,S) = (dm,ds + d/2(r-m))$. For the unit quotient to be finite, the ranks of the groups must be the same, and thus by Dirichlet's theorem $R+S=r+s$. But

$$R+S-r-s = dm + ds + d/2(r-m) - r- s = \frac{1}{2} \left(d(m+s) + (d-2)(r+s)\right).$$

This can only be zero if $d = 2$ and $m+s=0$, or if $(r,s) = (r,0)$, and $(R,S) = (0,2r)$, or if $K^{+}$ is totally real and $K$ is totally complex.


For your actual question, you can deduce the following:

Let $\sigma$ be an embedding of $K$ into the complex numbers, let $K^{+}$ denote the intersection of $\sigma(K)$ with $\mathbf{R}$, so $U_{K^{+}} = U_K \cap \mathbf{R}$. Then either $K = K^{+}$, or $K$ is a CM extension.

Note that $K \ne K^{+}$ exactly when $\sigma$ is not a real embedding, so one can rephrase this as:

If $\sigma$ is a complex (non-real) embedding of $K$ and $U_K/U_K \cap \mathbf{R}$ is finite, then $K$ is CM.

Of course you can't say much more than that, because if $\sigma$ is a real embedding then $K^{+}=K$ and the condition you give is trivial. Perhaps you have the condition for all $\sigma$, which would imply that either $K$ is totally real (and so complex conjugation is trivial) or $K$ is CM.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .