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!


1 Answer 1


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.

A reference: Sections 2 and 6 of Chapter 2 of Borevich and Shafarevich’s Number Theory.


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