Exercise 12.3.3

Proof. Write $I=⟨M_1,\dots ,M_s⟩$. $I$ is a subgroup of $R$, since $0\in I$, and if $M,N\in I$, then $M={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^s{A}_i{M}_i,A_i\in R,N={\sum }_{i=1}^s{B}_i{M}_i,B_i\in R$, so $M-N={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^s{C}_i{M}_i$, where $C_i=A_i-B_i\in R$, so $M-N\in I$.

Moreover, if $M\in I$ and $A\in R$ , then $M={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^s{A}_i{M}_i,A_i\in R$, therefore $\mathit{AM}={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^s{D}_i{M}_i$, where $D_i=A{A}_i\in R$.

$I=⟨M_1,\dots ,M_s⟩$ is an ideal of $R$. □