Exercise 1, § 11

Let $\Delta$ be a commutative monoid, written additively, with its identity element denoted by 0. Let $\mathbf{Z}^{(\Delta)}$ be the algebra of $\Delta$ over $\mathbf{Z}$ (III, § 2, no. 6), that is a commutative ring, which is a free $\mathbf{Z}$-module with a basis $(X^\sigma)_{\sigma \in \Delta}$ and the multiplication $(aX^\sigma)(bX^\tau) = abX^{\sigma + \tau}$. For every $\mathbf{Z}$-module, we write $E^{(\Delta)} = E \otimes_\mathbf{Z} \mathbf{Z}^{(\Delta)}$, every element of $E^{(\Delta)}$ hence being written uniquely in the form $\sum_{\sigma \in \Delta} z_\sigma \otimes X^\sigma$, with $z_\sigma \in E$, the family $(z_\sigma)$ having finite support. For every $\mathbf{Z}$-module homomorphism $f : E \to E'$, let $f^{(\Delta)}$ denote the $\mathbf{Z}^{(\Delta)}$-module homomorphism $f \otimes 1 : E^{(\Delta)} \to {E'}^{(\Delta)}$. Similarly, if $\Delta'$ is another additive monoid with its identity element denoted by 0 and if $\alpha : \Delta \to \Delta'$ is a monoid homomorphism, $\alpha(E)$ denotes the homomorphism $E^{(\Delta)} \to E^{(\Delta')}$ such that $\alpha(E)(z_\sigma \otimes X^\sigma) = z_\sigma \otimes X^{\alpha(\sigma)}$. $(E^{(\Delta)})^{(\Delta')}$ is identified with $E^{(\Delta \times \Delta')}$ by associating with $\sum_{\sigma' \in \Delta'} \left( \sum_{\sigma \in \Delta} z_{\sigma, \sigma'} \otimes X^\sigma \right) \otimes X^{\sigma'}$ the element $\sum_{\sigma, \sigma'} z_{\sigma, \sigma'} \otimes X^{(\sigma, \sigma')}$. Finally, let $\varepsilon(E)$ denote the homomorphism $E^{(\Delta)} \to E$ mapping $\sum_\sigma z_\sigma \otimes X^\sigma$ to $\sum_\sigma z_\sigma$.

(a) Let $E$ be a $\mathbf{Z}$-module, $(E_\sigma)_{\sigma \in \Delta}$ a graduation on $E$ of type $\Delta$ and, for all $\sigma \in \Delta$, let $p_\sigma$ be the projector $E \to E_\sigma$ corresponding to the decomposition of $E$ as the direct sum of the $E_\sigma$. For all $x \in E$, let $\phi_E(x) = \sum_{\sigma \in \Delta} p_\sigma(x) \otimes X^\sigma$. Show that the homomorphism $\phi_E : E \to E^{(\Delta)}$ thus defined has the following properties:

(1)

(2) where $\delta : \Delta \to \Delta \times \Delta$ is the diagonal mapping.

Show that a bijection of the set of graduations on $E$ of type $\Delta$ is thus defined on to the set of homomorphisms of $E$ into $E^{(\Delta)}$ satisfying conditions (1) and (2).

(b) Let $E, F$ be two graded $\mathbf{Z}$-modules of type $\Delta$. For a homomorphism $f : E \to F$ to be graded of degree 0, it is necessary and sufficient that $\phi_F \circ f = f^{(\Delta)} \circ \phi_E$. For a submodule $E'$ of $E$ to be graded, it is necessary and sufficient that $\phi_E(E') \subset {E'}^{(\Delta)}$.

(c) Let $A$ be a ring; $A^{(\Delta)} = A \otimes_\mathbf{Z} \mathbf{Z}^{(\Delta)}$ is given a ring structure such that $(a \otimes X^\sigma)(b \otimes X^\tau) = (ab) \otimes X^{\sigma + \tau}$. Let $(A_\sigma)_{\sigma \in \Delta}$ be a graduation of the additive group $A$; for this graduation to be compatible with the ring structure on $A$ and $1 \in A_0$, it is necessary and sufficient that $\phi_A : A \to A^{(\Delta)}$ be a ring homomorphism. State and prove an analogous result for graded $A$-modules.