I am sorry for this rather dumb-sounding question. But I am thinking of it for the last two days and am unable to find an answer.

Let $K, L$ be an algebraic number fields, ie a finite extensions of $\mathbb Q$. Let $F = KL$ be their compositum. Let $\mathcal O_K, \mathcal O_L, \mathcal O_F$ be the rings of integers of these fields. Let $n(K), n(L), n(F)$ be the degrees of these number fields.

Now, we have a triplet $(K, \mathcal O_K, i_{K} : \mathcal O_K \hookrightarrow \mathbb{R}^{n(K)})$, where $i_K$ is the embedding $x \mapsto (x^{(1)}, x^{(2)}, \ldots , x^{(n(K))})$, where $x^{(1)}, x^{(2)}, \ldots , x^{(n(K)}$ are the conjugates of $x$.

What bothers me in this situation is that the conjugates depend on the number field. To alleviate this situation, I go ahead as follows.

We consider the similar triplets $(L, \mathcal O_K, i_{L} : \mathcal O_L \hookrightarrow \mathbb{R}^{n(L)})$ and $(F, \mathcal O_F, i_{F} : \mathcal O_F \hookrightarrow \mathbb{R}^{n(F)})$. There is a natural embedding $\mathcal O_K \hookrightarrow \mathcal O_F$ and similarly $\mathcal O_L \hookrightarrow \mathcal O_F$ and these respect the inclusions of the number rings into their respective Euclidean spaces. So we can "include" the triplets for $K$ and $L$ into $F$.

Now, we order all number fields via inclusion. This is a directed set. And the set of all triplets considered above of the form $(K, \mathcal O_K, i_{K} : \mathcal O_K \hookrightarrow \mathbb{R}^{n(K)})$, is a directed system of such triplets. So we take the direct limit. The result should be some embedding of the ring of all algebraic integers into a countable dimension Euclidean space.

Question:

Does this embedding give a lattice?

I would be grateful for answers. Again I am sorry if this is a stupid question.