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^m $. 


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 divided-power ring $k\langle x\_1, \ldots, x\_n \rangle := A \otimes\_{\mathbb Z} k$ is spanned as a $k$-vector space by the divided powers of degree-1 polynomials). EDIT: Following Scott Carnahan's answer below, I need to take $k$ algebraically closed here.



On the divided power ring over the integers
If we let $k = \mathbb{F}\_2$, then there are only three polynomials of total degree 1 in the variables $x$ and $y$, namely $x,y,x+y$. The space of degree 3 elements in the divided power ring over $k$ is 4 dimensional, spanned by $x^{(3)}, x^{(2)}y, xy^{(2)}, y^{(3)}$. Unless I have made a mistake, I am forced to conclude that even over a field, the divided power ring is not spanned by divided powers of polynomials of total degree 1.


By choosing small examples of $a$ and $b$, we see that the span of $\{ (ax + by)^{(3)} \}\_{a,b \in \mathbb{Z}}$ in $\mathbb{Z}\langle x, y \rangle$ contains $x^{(3)}$, $y^{(3)}$, $x^{(2)}y + x y^{(2)}$, and $x^{(2)}y - x y^{(2)}$. However, reduction mod 2 shows that it doesn't contain $x^{(2)}y$.


