Must integer lattice have vector of norm equal to determinant?

Let $$\mathcal{L}\subseteq\mathbb{Z}^n$$ be an integer lattice (not necessarily full-rank). Is it guaranteed that there is a vector $$\mathbf{v}\in\mathcal{L}$$ such that $$\|\mathbf{v}\|=\det(\mathcal{L})$$?

If $$\mathcal{L}$$ is full-rank, then the answer is certainly yes. Let $$\mathbf{B}$$ be a matrix whose columns form a basis for $$\mathcal{L}$$. Then $$\|\det(\mathbf{B})\|=\det(\mathcal{L})$$. Then the adjugate of $$\mathbf{B}$$, namely $$\mathbf{B}^{adj}=\det(\mathbf{B})\mathbf{B}^{-1}$$, is a matrix of integer entries. Therefore, the columns of $$\mathbf{B}\cdot\mathbf{B}^{adj}=\det(\mathbf{B})\mathbf{I}_n$$ are in $$\mathcal{L}$$. But the columns are just $$\det(\mathbf{B})$$ times the standard basis vectors. Each of these scaled standard basis vectors has norm $$\|\det(\mathbf{B})\|=\det(\mathcal{L})$$.

What about integer lattices that are not full rank?

(Note that the requirement that $$\mathcal{L}\subseteq \mathbb{Z}^n$$ is an integer lattice is important here. For arbitrary real lattices $$\mathcal{L}\subseteq\mathbb{R}^n$$, the statement is false. For example, the lattice $$\frac{1}{2}\cdot \mathbb{Z}^n$$ has determinant $$2^{-n}<1/2$$ for $$n\geq 2$$. However all vectors have the norm at least $$1/2$$.)

• How is $\det(\mathcal{L})$ defined if $\mathcal{L}$ does not have full rank? Commented Jul 10 at 0:06
• $\det(\mathbf{B}^T\cdot\mathbf{B})^{1/2}$ where the columns of $\mathbf{B}$ form a basis for $\mathcal{L}$. Geometrically, this is the volume of any fundamental parallelpiped of the lattice.
– AAA
Commented Jul 10 at 1:18

Here is a counterexample: Let $$\mathcal{L} \subseteq \mathbb{Z}^3$$ have basis $$\{(1,1,0),(0,1,1)\}$$. Then $$\det(\mathcal{L})^2 = 3$$, but $$\|\mathbb{v}\|^2$$ is even for any $$\mathbb{v} \in \mathcal{L}$$, so $$\|\mathbb{v}\| = \det(\mathcal{L})$$ is impossible.