# What do lattices correspond to in Algebraic Number Theory?

The following setup is standard studied in Algebraic Number Theory:

Let $$K$$ be a number field and $$\mathcal{O}_K$$ be its ring of integers. Suppose $$[K : \mathbb{Q}] = n$$ so we have $$n$$ embeddings of $$K$$ into $$\mathbb{C}$$. Suppose there are $$r$$ real embeddings and $$2s$$ complex embeddings (nonreal), then we have

$$K \hookrightarrow \mathbb{R}^r \times \mathbb{C}^{2s} \cong \mathbb{R}^n$$ by mapping pairs of complex embeddings to the real and imaginary parts. Then we know that $$\mathcal{O}_K$$ maps into a (full-rank) lattice in $$\mathbb{R}^n$$ (that is, $$\mathbb{Z}$$ combinations of an $$\mathbb{R}$$ basis for $$\mathbb{R}^n$$). Similarly, any fractional ideal also maps into a lattice in $$\mathbb{R}^n$$.

I want to understand the following question:

What are the algebraic structures within $$K$$ that map to lattices in $$\mathbb{R}^n$$?

I understand that any free abelian group (free $$\mathbb{Z}$$-module) would be necessary but insufficient (because lattices in $$\mathbb{R}^n$$ has this discrete property, where accumulation points are disallowed). A finitely generated nonzero $$\mathcal{O}_K$$-module in $$K$$ (aka fractional ideals) would be sufficient but probably not necessary.

If there are any references that would be greatly appreciated!

You're not going to get every lattice of $$\mathbf R^n$$ this way, since for instance when $$K = \mathbf Q$$ you're not going to obtain $$\mathbf Z\pi$$ in this way.
The lattices in $$\mathbf R^n$$ are precisely the $$\mathbf Z$$-spans of bases of $$\mathbf R^n$$. You're asking what the $$\mathbf Z$$-span of a $$\mathbf Q$$-basis is in $$K$$. These could be described as the fractional ideals of orders in $$K$$.
An order in $$K$$ is a finite-index subring of $$\mathcal O_K$$. Let $$M \subset K$$ be the $$\mathbf Z$$-span of a $$\mathbf Q$$-basis of $$K$$. Then $$\mathcal O(M) := \{\alpha \in K : \alpha M \subset M\}$$ turns out to be an order in $$K$$, and $$M$$ is an $$\mathcal O(M)$$-module. Let $$d \in \mathbf Z^+$$ be a common denominator of the $$\mathbf Q$$-basis spanning $$M$$ over $$\mathbf Z$$ (each member of the basis is an algebraic integer divided by $$d$$), so $$dM \subset \mathcal O_K$$. Note $$\mathcal O(dM) = \mathcal O(M)$$. Let $$r = [\mathcal O_K:\mathcal O(M)] = [\mathcal O_K:\mathcal O(dM)]$$, so $$rdM \subset r\mathcal O_K \subset \mathcal O(dM)$$. Thus $$rdM$$ is a nonzero $$\mathcal O(dM)$$-module inside $$\mathcal O(dM)$$, so $$\mathfrak a := rdM$$ is an ideal in $$\mathcal O(dM)$$. Hence $$M = (1/(dr))\mathfrak a$$ is a fractional $$\mathcal O(dM)$$-module.