Embedding algebraic integers as a lattice

Question

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.

Answer 1

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.

Answer 2

I assume that you have chosen an algebraic closure $\overline{\mathbb{Q}}$ of $\mathbb{Q}$, and are considering only the finitely generated subfields, since your declaration that number fields form a directed set ordered by inclusion skips over some murky waters concerning large isomorphism classes.

I'm not sure what definition of lattice you're using, so I'll conveniently interpret your question as the following:

Given the integral closure $\overline{\mathbb{Z}}$ of $\mathbb{Z}$ in our chosen $\overline{\mathbb{Q}}$, is the image of $\overline{\mathbb{Z}}$ in $\overline{\mathbb{Z}} \otimes_\mathbb{Z} \mathbb{R}$ discrete, and is the canonical inclusion $\overline{\mathbb{Z}} \otimes_\mathbb{Z} \mathbb{R} \to \overline{\mathbb{Q}} \otimes_\mathbb{Z} \mathbb{R}$ an isomorphism?

The answer is "yes". The embedding is discrete if we choose the canonical metric on the system of finite extensions (see Neukirch, Algebraic Number Theory, Chapter 1, section 5), since distances are bounded below by 1. Now, we consider an arbitrary element of $\overline{\mathbb{Q}} \otimes_\mathbb{Z} \mathbb{R}$, written as a finite sum $\sum q_i \otimes r_i$, with $q_i \in \overline{\mathbb{Q}}$. There is a nonzero integer $N$ such that $Nq_i \in \overline{\mathbb{Z}}$ for all $i$. You can obtain $N$ using suitable variable substitutions in the minimal polynomials of the $q_i$. This implies $\sum q_i \otimes r_i = \sum Nq_i \otimes \frac{r_i}{N} \in \overline{\mathbb{Z}} \otimes_\mathbb{Z} \mathbb{R}$.

Why this may not be a complete answer: As we saw in the comments, lattices are often defined as discretely embedded subgroups of a Lie group with compact (or finite covolume) quotients. I don't know of a canonical measure on $\mathbb{R} \otimes_\mathbb{Q} \overline{\mathbb{Q}}$, and translation-invariant analogues of Lebesgue measure in infinite dimensions seem to have trouble existing. The quotient under the colimit topology is homeomorphic to $\underset{n}{\operatorname{colim}}(S^1)^n$, but this is not a product of circles, since it has countable dimension.