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.
Best Answer
You need to define what you want the word lattice to mean in such an infinite-dimensional context. In any case, I want to point out that there is a way to remove all the use of these field embeddings
It might be cleaner (avoiding explicit mention of conjugates) if you use tensor products. Let $K$ have $r_1$ real embeddings and $2r_2$ complex (i.e., non-real) embeddings and set $N = [K:\mathbf Q] = r_1 + 2r_2$. Rather than combining all these embeddings into a single embedding of $K$ into $\mathbf R^{r_1} \times \mathbf C^{r_2}$ (which is non-canonically $\mathbf R^{N}$, even if $K$ is totally real, since it depends on the choice of the ordering of your embeddings), work instead with $\mathbf R \otimes_{\mathbf Q} K$, which is an $\mathbf R$-algebra of dimension $N$ as a real vector space. The image of the integers of $K$ inside here is a lattice, by which I mean it is a discrete subgroup with compact quotient (or, more concretely, it has a $\mathbf Z$-basis of size $N$ which is also a basis of the tensor product as a real vector space).
If the number field $K$ has some embedding into the number field $F$, you can base extend that field homomorphism $K \hookrightarrow F$ to an $\mathbf R$-algebra homomorphism $\mathbf R \otimes_{\mathbf Q} K \hookrightarrow \mathbf R \otimes_{\mathbf Q} F$ and the images of the integers of $K$ and $F$ inside these spaces will then appear as a lattice in one vector space that is a sublattice of a lattice in a larger vector space. This is a cleaner picture than the one you get using all those different real and complex embeddings of the number fields.
I am not directly addressing your question on the direct limit here, just suggesting a nicer way to think about the lattices in the number fields and how they can interact with one number field inside another.