I have problems understanding the equivalence of two definitions of generated ideals. If $A\subset R$ is a subset of a Ring, the generated ideal $(A)\subset R$ is defined as the intersection of all ideals $I$ with $A\subset I$: $$ (A)=\bigcap_{A\subset I \subset R} I $$ Now the book says that if $R$ is abelian, then $$ (A)=\{x_1a_1+...+x_na_n:x_i \in R, a_i\in A \} $$ which is the set of all finite linear combinations of elements of $A$.
Now I have problems showing this equality. I know that I need to show that
i) $M:= \{x_1a_1+...+x_na_n:x_i \in R, a_i\in A \}$ is an ideal and $A\subset M$, because then $(A)\subset M$.
ii) $M \subset (A)$
For i) I have shown that $a,b\in M$ also $a-b\in M$ and $a\cdot b\in M$. It is also easy to show that $ra\in M,ar\in M$ for $r \in R, a\in M$. But I have problems showing that $A\subset M$. If $a\in A$ I could only show that $a\in M$ if $1\in R$ but this is nowhere to be found in the book. Is there another way of showing this, maybe using that $R$ is commutative?
For ii) If $R$ is abelian $\{x_1i_1+...+x_na_n:i_i \in R, a_i\in I \}\in I$ for every ideal $I$. So since $a_i\in \bigcap_{A\subset I \subset R} I$ and $\bigcap_{A\subset I \subset R} I$ is an ideal itself because the intersection of ideals is an ideal, then for every $a \in \{x_1a_1+...+x_na_n:x_i \in R, a_i\in A \}$als $a \in \bigcap_{A\subset I \subset R} I = A$.