Doubt: In this proof I know that we need to show that I is a subgroup of (R, +) and is I is closed under multiplication. How does the first part of the proof show the subgroup property? I did not understand the $$(r_1a_1-r'_1a_1) + (r_2a_2-r'_2a_2)+\cdots+(r_na_n-r'_na_n) = (r_1-r'_1)a_1+(r_2-r'_2)a_2+\cdots + (r_n-r'_n)a_n\in I$$ part, why are we performing a subtraction?
Let $R$ be a commutative ring with unity and let $a_{1}, a_{2}, \ldots, a_{n} \in R$. Then $$ I=\left\langle a_{1}, a_{2}, \ldots, a_{n}\right\rangle=\left\{r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n} \mid r_{i} \in R\right\} $$ is an ideal of $R$ called the ideal generated by $a_{1}, a_{2}, \ldots, a_{n}$. \PROOF. If $r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n}, r_{1}^{\prime} a_{1}+r_{2}^{\prime} a_{2}+\cdots+r_{n}^{\prime} a_{n} \in I$, $\left(r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n}\right)-\left(r_{1}^{\prime} a_{1}+r_{2}^{\prime} a_{2}+\cdots+r_{n}^{\prime} a_{n}\right)=$ $$ \left(r_{1} a_{1}-r_{1}^{\prime} a_{1}\right)+\left(r_{2} a_{2}-r_{2}^{\prime} a_{2}\right)+\cdots+\left(r_{n} a_{n}-r_{n}^{\prime} a_{n}\right)= $$ $$ \left(r_{1}-r_{1}^{\prime}\right) a_{1}+\left(r_{2}-r_{2}^{\prime}\right) a_{2}+\cdots+\left(r_{n}-r_{n}^{\prime}\right) a_{n} \in I . $$ If $r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n} \in I$ and $r \in R$, $$ \begin{aligned} r\left(r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n}\right)=\left(r_{1} a_{1}\right.&\left.+r_{2} a_{2}+\cdots+r_{n} a_{n}\right) r=\\ &\left(r r_{1}\right) a_{1}+\left(r r_{2}\right) a_{2}+1 \cdots+\left(r r_{n}\right) a_{n} \in I . \end{aligned} $$ Therefore $I$ is an ideal.
