# On the non existence of non real embeddings

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.

• 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. Nov 27, 2019 at 4:15
• 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. Nov 27, 2019 at 4:26
• 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. Nov 27, 2019 at 4:33
• 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 $K^*$. Nov 27, 2019 at 4:35
• 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 $\Bbb C$ Nov 27, 2019 at 20:29

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.