**Some initial clarifications**


By [lattice](https://en.wikipedia.org/wiki/Lattice_(group)) 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.




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**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 *sublattice* of $\mathbb Z^n$.
> 
> 
> 


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


$$\langle \phi(v),\phi(w)\rangle=\alpha \langle v,w\rangle,\quad\text{for all $v,w\in\mathcal L\_1$}.$$



Does every positive-definite integral lattice admit an angle-preserving homomorphism into $\Bbb Z^n$ for some $n$?
Yes, this is true. There's some fancy number theory that one can apply (the Hasse-Minkowski invariant and embedding of quadratic forms), but one can see this directly without number-theoretic machinery. 


First, notice that one can choose a basis for the lattice which is orthogonal. Just start with any basis and apply Gram-Schmidt [orthogonalization](https://en.wikipedia.org/wiki/Orthogonalization). The new orthogonal basis for the lattice may no longer be integral, but we may multiply the inner product by an integer to clear denominators make it integral (and so that the original lattice is a sublattice up to scaling). 


Now one is left with the problem of embedding the 1-dimensional quadratic form/lattice $\mathcal{L}\_m$ given by $\mathbb{Z}v\_1, \langle v\_1,v\_1\rangle = m$ into an integral lattice. This may be done using [Lagrange's 4-square theorem](https://en.wikipedia.org/wiki/Lagrange%27s_four-square_theorem). Let $m=w^2+x^2+y^2+z^2$. Then we may embed $\mathcal{L}\_m$ into the lattice $\mathbb{Z}^4$ via $$v\_1 \mapsto (w,x,y,z).$$ Do this for each basis vector individually, and take the orthogonal direct sum of these $n$ 4-dimensional lattices (so this embeds into the lattice $\mathbb{Z}^{4n}$ up to scaling, which is likely far from optimal in general). 


**Addendum:** The answer above shows how to embed as a sublattice up to scaling, whereas the question asks for a *saturated* sublattice (intersection of $\mathbb{Z}^n$ with a subspace). I don't know how to obtain this in general, but in dimensions $1$ and $2$ one can make a construction by hand *without scaling*.


In dimension $1$, send the basis vector $v\_1$ in the lattice $\mathcal{L}\_m$ with $\langle v\_1, v\_1 \rangle = m$ to the vector $(1,\ldots,1)\in\mathbb{Z}^m$. This vector is primitive, and hence the image is a saturated sublattice. 


In dimension $2$, one may do a variation on this embedding. 
Suppose that one has a 2-dimensional lattice with $\langle v\_1,v\_1\rangle =a >0, \langle v\_1,v\_2\rangle =b\geq 0, \langle v\_2,v\_2\rangle = d >0$, with $ad-b^2>0$. By reduction theory, we may in fact assume that $a \leq d$ and $ ad \geq 2b^2$ (by taking $v\_1$ to have minimal norm, and the acute angle between $v\_1$ and $v\_2$ to be $\geq \pi/3$). 


Let $b=(a-1)q+r$, $0\leq r < a-1$. Then send $v\_1\mapsto (1,\ldots,1,0,\dots,0)$ and $v\_2\mapsto (q,\ldots, q, r, 1, \ldots, 1) \in \mathbb{Z}^n$, where $n=a+d- (a-1)q^2 -r^2$ (so the second vector has $(a-1)$ $q$s)
. For this to work, one has to check that $d > (a-1)q^2+r^2$ which can be deduced from the inequalities $ad \geq 2b^2$ and $d\geq a$. Moreover, one may see that this is a saturated sublattice of $\mathbb{Z}^n$: if any of the last $d-(a-1)q^2-r^2$ coordinates is non-zero, then we may subtract off the corresponding multiple of $v\_2$ to make these coordinates $=0$. What's left must be a multiple of $v\_1$. 


