Let \(\Lambda\subset\mathbb R^n\) be a lattice. Choose \(a_1,\ldots,a_n\in\Lambda\) successively so that \(a_1\) is a shortest nonzero vector and, for \(i>1\), \(a_i\) is a shortest vector linearly independent from \(a_1,\ldots,a_{i-1}\).
Must one always have \[ \Lambda=\mathbb Za_1\oplus\cdots\oplus\mathbb Za_n? \]
Solution
No. A counterexample is provided by the \(D_4\) lattice \[ \Lambda =D_4 :=\left\{(x_1,x_2,x_3,x_4)\in\mathbb Z^4: x_1+x_2+x_3+x_4\equiv0\pmod2\right\}. \]
Choose \[ \begin{aligned} a_1&=e_1+e_2,\\ a_2&=e_1-e_2,\\ a_3&=e_3+e_4,\\ a_4&=e_3-e_4. \end{aligned} \]
1. The shortest nonzero vectors of \(D_4\) have length \(\sqrt2\).
Every vector of \(D_4\) has integral coordinates. A vector of Euclidean length \(1\) would therefore have to be one of \[ \pm e_1,\ldots,\pm e_4. \] Each of these has odd coordinate sum, so none lies in \(D_4\). Hence every nonzero vector of \(D_4\) has squared norm at least \(2\).
On the other hand, \[ e_i\pm e_j\in D_4 \qquad(i\ne j) \] and has squared norm \(2\). Thus the lattice minimum is exactly \[ \sqrt2. \]
2. The displayed \(a_1,a_2,a_3,a_4\) are valid successive choices in the problem.
Each \(a_i\) has norm \(\sqrt2\), so by 1 every \(a_i\) is a shortest nonzero lattice vector. They are linearly independent: relative to the standard basis their column matrix is block diagonal, \[ \begin{pmatrix} 1&1&0&0\\ 1&-1&0&0\\ 0&0&1&1\\ 0&0&1&-1 \end{pmatrix}, \] whose determinant is \(4\ne0\).
Consequently, after \(a_1,\ldots,a_{i-1}\) have been chosen, the vector \(a_i\) is linearly independent from them and has the smallest possible nonzero lattice length \(\sqrt2\). Hence it is certainly a shortest lattice vector with the required independence property.
3. These four vectors do not generate \(D_4\) as a lattice.
Their integer span is \[ L=\mathbb Za_1+\mathbb Za_2+\mathbb Za_3+\mathbb Za_4. \] From \[ m(e_1+e_2)+n(e_1-e_2) =(m+n,m-n,0,0), \] the first two coordinates of every vector of \(L\) have the same parity. Similarly the last two coordinates have the same parity. Thus \[ L =\{x\in\mathbb Z^4: x_1\equiv x_2\pmod2, \ x_3\equiv x_4\pmod2\}. \] This is a proper sublattice of \(D_4\): for example, \[ e_1+e_3=(1,0,1,0) \] has even coordinate sum and hence lies in \(D_4\), but it does not lie in \(L\).
Equivalently, the determinant of the four \(a_i\) is \(4\), whereas \(D_4\) has index \(2\) in \(\mathbb Z^4\) and therefore covolume \(2\). Hence \[ [D_4:L]=\frac4{2}=2. \] Thus valid successive shortest independent vectors need not form a lattice basis.