Double Density Theorem?

Question

A colleague asks me the following: "I wonder if you can give me a reference - or a guidance where to look – from a fact I recall from graduate school. I’m sure it can be generalized quite a bit but in the simplest form is it is the 'double density theorem':

Let $F$ be a number field and $j_1$ and $j_2$ two distinct embeddings of $F$ into the complex plane $\mathbb C$ which are NOT related by complex conjugation: i.e. $j_2$ is not equal $\bar{j_1}$. Then then map $$ j_1 \times j_2 : \Delta\subset F \times F \to \mathbb C \times\mathbb C, $$ (where $\Delta$ is the diagonal) has a dense image ( in the usual metric topology)."

Answer 1

Here is a more elementary explanation than others that have been given in comments or answers. When $j_1$ and $j_2$ are two complex (i.e., non-real) embeddings of $F$ that are neither equal nor complex-conjugate, you could use them as part of the components in the Euclidean embedding of $F$ into ${\mathbf R}^{r_1} \times {\mathbf C}^{r_2}$. We know the image of ${\mathcal O}_F$ under that embedding is a lattice (full-rank discrete subgroup), so $F$ is the ${\mathbf Q}$-span of a lattice in a Euclidean space, and therefore $F$ has dense image in ${\mathbf R}^{r_1} \times {\mathbf C}^{r_2}$. Now project to the two components corresponding to $j_1$ and $j_2$.

Answer 2

I assume that you mean complex embeddings as oppposed to real emebedding, then it follows from approximation theorems for the adèles.