Lattices over number field correspond to ideal classes Let $K$ be a number field with ring of integers $A$. Let. A lattice in $K^n$ is a sub-$A$-module of $K^n$, that generates $K^n$ as sub-$K$-module, that is of finite type over $A$.
In a paper by Borel, he states on page 11 that, to give a lattice in $K^n$ is to give the ideal class of a fractional ideal in $K$. For this result, he refers to a book by Eichler in German. I have tried to read the pertinent paragraphs, but my knowledge of German is not optimal and I am having difficulties understanding his notation.
Is this result well-known? If so, do you have another reference where I can find a proof this?
 A: Yes, there are certainly more modern references. These results are part of the theory of modules over a Dedekind domain, which is covered in Dummit and Foote $\S16.3$, for instance.
The first part of his claim is known as the Fundamental Theorem of Finitely Generated Modules over a Dedekind Domain, Theorem 22 of $\S16.3$ in Dummit and Foote.
Theorem. Suppose $M$ is a finitely generated module over a Dedekind domain $R$. Let $n \geq 0$ denote the rank of $M$ and let $\newcommand{\Mtors}{M_\text{tors}} \Mtors$ be the torsion submodule of $M$. Then
$$
M \cong \underbrace{R \oplus \cdots \oplus R \oplus I}_{n \text{ factors}} \oplus M_\text{tors}
$$
for some ideal $I$ of $R$. The ideal $I$ is unique up to multiplication by a principal ideal.
(The full theorem also characterizes the structure of $\Mtors$.)
The fact that the class of the ideal $I$ is determine by $M$ (or $X$ in Borel's notation) is actually proved earlier, in Proposition 21. One can reduce to the case of just two fractional ideals: $I \cong J$ iff $I = (a) J$ for some nonzero $a \in K$. The reverse direction is straightforward (the isomorphism is just multiplication by $a$), and if $I \cong J$, then $I J^{-1} \cong R$. Letting $a$ be the image of $1 \in R$ under this isomorphism, then $I J^{-1} = aR = (a)$, so $I = (a) J$.
