Let us verify that the definition of $\circ$ is well-defined. Let $\left[\!\left[a\right]\!\right]_m=\left[\!\left[b\right]\!\right]_m$. Then we need to show that: :$\forall x \in R : \left[\!\left[a\right]\!\right]_m \circ x = \left[\!\left[b\right]\!\right]_m \circ x$ By the definition of congruence: :$\left[\!\left[a\right]\!\right]_m = \left[\!\left[b\right]\!\right]_m \iff \exists k \in \Z : a = b + k m$ Then: {{begin-eqn}} {{eqn | l = \left[\!\left[a\right]\!\right]_m \circ x       | r = a \cdot x       | c = Definition of $\circ$ }} {{eqn | r = \left(b+km\right) \cdot x }} {{eqn | r = b \cdot x + km \cdot x       | c = Powers of Group Elements: Sum of Indices }} {{eqn | r = b \cdot x + k \cdot \left( m \cdot x \right)       | c = Powers of Group Elements: Product of Indices }} {{eqn | r = b \cdot x + k \cdot 0_R       | c = Characteristic times Ring Element is Ring Zero }} {{eqn | r = b \cdot x + 0_R       | c = Power of Identity is Identity }} {{eqn | r = b \cdot x }} {{eqn | r = \left[\!\left[b\right]\!\right]_m \circ x       | c = Definition of $\circ$ }} {{end-eqn}} Thus, the definition of $\circ$ is well-defined. {{qed|lemma}} Let us verify that $\left({R, +, \circ}\right)_{\Z_m}$ is a unitary $\Z_m$-module by verifying the axioms in turn. === Axiom $(1)$ === We need to show that: :$\left[\!\left[{a}\right]\!\right]_m \circ \left({x + y}\right) = {\left[\!\left[{a}\right]\!\right]_m \circ x} + {\left[\!\left[{a}\right]\!\right]_m \circ y}$ {{begin-eqn}} {{eqn | l = \left[\!\left[a\right]\!\right]_m \circ \left({x + y}\right)       | r = a \cdot \left(x +y\right)       | c = Definition of $\circ$ }} {{eqn | r = a \cdot x + a \cdot y       | c = Power of Product in Abelian Group }} {{eqn | r = \left[\!\left[a\right]\!\right]_m \circ x + \left[\!\left[a\right]\!\right]_m \circ y       | c = Definition of $\circ$ }} {{end-eqn}} {{qed|lemma}} === Axiom $(2)$ === We need to show that: :$\left( {\left[\!\left[a\right]\!\right]_m +_m \left[\!\left[b\right]\!\right]_m}\right) \circ x = \left[\!\left[a\right]\!\right]_m \circ x + \left[\!\left[b\right]\!\right]_m \circ x$ {{begin-eqn}} {{eqn | l = \left( {\left[\!\left[a\right]\!\right]_m +_m \left[\!\left[b\right]\!\right]_m}\right) \circ x       | r = \left[\!\left[ a+b \right]\!\right]_m \circ x       | c = {{Defof|Modulo Addition}} }} {{eqn | r = \left(a+b\right) \cdot x       | c = Definition of $\circ$ }} {{eqn | r = a \cdot x + b \cdot x       | c = Powers of Group Elements: Sum of Indices }} {{eqn | r = \left[\!\left[a\right]\!\right]_m \circ x + \left[\!\left[b\right]\!\right]_m \circ x       | c = Definition of $\circ$ }} {{end-eqn}} {{qed|lemma}} === Axiom $(3)$ === We need to show that: :$\left({\left[\!\left[a\right]\!\right]_m \times_m \left[\!\left[b\right]\!\right]_m}\right) \circ x = \left[\!\left[a\right]\!\right]_m \circ \left({\left[\!\left[b\right]\!\right]_m \circ x}\right)$ {{begin-eqn}} {{eqn | l = \left({\left[\!\left[a\right]\!\right]_m \times_m \left[\!\left[b\right]\!\right]_m}\right) \circ x        | r = \left[\!\left[ a \times b \right]\!\right]_m \circ x       | c = {{Defof|Modulo Multiplication}} }} {{eqn | r = \left(a \times b\right) \cdot x       | c = Definition of $\circ$ }} {{eqn | r = a \cdot \left({b \cdot x}\right)       | c = Powers of Group Elements: Product of Indices }} {{eqn | r = \left[\!\left[a\right]\!\right]_m \circ \left({\left[\!\left[b\right]\!\right]_m \circ x}\right)       | c = Definition of $\circ$ }} {{end-eqn}} {{qed|lemma}} === Axiom $(4)$ === We need to show that: :$\left[\!\left[1\right]\!\right]_m \circ x = x$ since [[Modulo Multiplication has Identity|$\left[\!\left[1\right]\!\right]_m$ is the unity of $\Z_m$]]. That is, that $1 \cdot x = x$. This follows from the definition of power of group element. {{qed|lemma}} Having verified all four axioms, we have shown that $\left({R, +, \circ}\right)_{\Z_m}$ is a unitary $\Z_m$-module. {{qed}} Category:Unitary Modules Category:Group Theory ipltlbrpyn2938lcspwwaozmqnj4eza