Consider the divided-power ring $A := \mathbb Z \langle x_1, \ldots, x_n \rangle$ consisting of $\mathbb Z$-linear combinations of divided-power monomials of the form $x_1^{(a_1)} \cdots x_n^{(a_n)}$; this can be defined as the subring of the polynomial ring $\mathbb Q[ x_1, \ldots, x_n ]$ which is generated as a $\mathbb Z$-algebra by the elements $ x_i^{(m)} := \displaystyle \frac 1 {m!} x_i $. 

Next consider the lattice $V \subseteq A$ generated by $x_1, \ldots, x_n$; that is, $V$ is the subspace of $A$ consisting of degree-1 polynomials. Then divided powers of elements of $V$ are elements of $\mathbb Q[ x_1, \ldots, x_n ]$, and one can check that they are in fact in $A$. Now my question is the following: Do the divided powers of elements of $V$ generate $A$ as an abelian group? (This is motivated by the fact that for a field $k$, the $k$-vector space of degree-$m$ polynomials in the polynomial ring $k[x_1, \ldots, x_n]$ is spanned by the $m^{th}$ powers of degree-1 polynomials).