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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. 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. 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).

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. 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. 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 $v_1$.

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. 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. 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: These lattices are sublattices, but not likely to be saturated in general.

One can even embed as a saturated sublattice without rescaling.

First consider the 1-dimensional case. One can embed $\mathcal{L}_m$ into $\mathbb{Z}^m$ as a saturated sublattice. Just send $v_1 \mapsto (1,\ldots,1) \in \mathbb{Z}^m$, which is clearly primitive and hence saturated.

In higher dimensions, we can embed in $\oplus_{i \leq j} \mathbb{Z}^{|\langle v_i,v_j \rangle |}$. Map $v_i$ to a vector which has $|\langle v_i,v_j \rangle |$ $1$s in the $ij$ subspace (or $-1$ if the inner product is negative if $i< j$).

One can check that this is a saturated subspace with the appropriate inner product, by considering the intersection with the diagonal subspaces to see that any vector in the subspace with integral coordinates has to be an integral linear combination of the image of the $v_i$s.

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. 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. 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).

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. 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. 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).

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. 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. 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: These lattices are sublattices, but not likely to be saturated in general.

One can even embed as a saturated sublattice without rescaling.

First consider the 1-dimensional case. One can embed $\mathcal{L}_m$ into $\mathbb{Z}^m$ as a saturated sublattice. Just send $v_1 \mapsto (1,\ldots,1) \in \mathbb{Z}^m$, which is clearly primitive and hence saturated.

In higher dimensions, we can embed in $\oplus_{i \leq j} \mathbb{Z}^{|\langle v_i,v_j \rangle |}$. Map $v_i$ to a vector which has $|\langle v_i,v_j \rangle |$ $1$s in the $ij$ subspace (or $-1$ if the inner product is negative if $i< j$).

One can check that this is a saturated subspace with the appropriate inner product, by considering the intersection with the diagonal subspaces to see that any vector in the subspace with integral coordinates has to be an integral linear combination of the image of the $v_i$s.