**Some initial clarifications**

By lattice I mean an additive subgroup of $\mathbb R^n$ which is isomorphic to $\mathbb Z^n$ and has full rank (i.e. spans $\Bbb R^n$ when considered as set of vectors). A lattice $\mathcal L$ is *integral* if $\langle v,w\rangle\in\mathbb Z$ for all $v,w\in\mathcal L$, where $\langle\cdot,\cdot\rangle$ denotes the standard inner product.

Above terms are standard, but some of the terms I will use in the following, like "sublattice" and "lattice isomorphism", are probably not standard. Feel free to correct my terminology, as I simply do not know enough of that subject.

**The actual question**

I wonder whether every integral lattice $\mathcal L$ is isomorphic to a sublattice of $\mathbb Z^n$ in the following sense:

Consider some vectors $v_1,..., v_k\in\mathbb Z^n$. Then $$\mathcal L(v_1,...,v_k):=\mathrm{span}\{v_1,...,v_k\}\cap \mathbb Z^n$$ is an integral lattice in $\mathrm{span}\{v_1,...,v_k\}\subseteq\mathbb R^n$ (with the inner product inherited from $\mathbb Z^n$), and will be called a

sublatticeof $\mathbb Z^n$.

I care about whether I can find $\mathcal L$ in $\Bbb Z^n$ with the right angles, not necessarily the right scale. So I need the following kind of "isomorphism": two lattices $\mathcal L_1$ and $\mathcal L_2$ are isomorphic, if there is a group isomorphism $\phi:\mathcal L_1\to\mathcal L_2$ and a constant $\alpha\in\mathbb R$ with

$$\langle \phi(v),\phi(w)\rangle=\alpha \langle v,w\rangle,\quad\text{for all $v,w\in\mathcal L_1$}.$$

Euclideanlattice". This has been ameliorated somewhat by adding routine definitions, at the expense of readability. However, the notions ofisomorphismandsublatticein the current version of the question are non-standard: it is more common to define isomorphism to be anisometry, so that, in particular, a bijective isomorphism satisfies the definition of an embedding and the image of an isomorphism ${\mathcal L}\to {\Bbb Z}^n$ is a sublattice according to the chosen definitions. $\endgroup$ – Victor Protsak Aug 5 '19 at 21:41