= Solution
The matrix with rows $u,v,e_3$ has determinant
$$
\det\begin{pmatrix}1&2&3\\2&3&1\\0&0&1\end{pmatrix}=3-4=-1.
$$
Its inverse has integer entries by the adjugate formula, so these rows form an integer <basis> of $L$. The <unimodular basis test for an integer direct summand> now gives
$$
\boxed{L=N\oplus\mathbb Z(0,0,1).}
$$
Thus \b[$N$ is a <direct summand>], with the displayed complement. This proves integral splitting, not just real or rational linear independence.
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