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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.

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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 $x^{(2)}y$.

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Very nice answer! I think that my above statement is true for an algebraically closed field, but somehow I'd tricked myself into thinking it worked over any field. –  Chuck Hague Nov 13 '12 at 14:16

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