Let $I(X)= \cap \lbrace I \subset R | X \subsetneq I \rbrace $ where $I \subset R$ is an ideal. I want to prove that $I(X)$ is equal to:
$$A= \lbrace a \in R |a=\sum_{i=1}^{n}r_{i}x_{i} \quad r_{i} \in R, x_{i} \in X \rbrace$$
As intesection of ideals is an ideal, I got no problem showing that $A \subseteq I(X)$ just by taking an element $a \in A$ and to check is contained in $I(X)$ follow straightforward from the fact ideals are closed under sums and scalar products where the scalar is in $R$.
Im troubled checking the other contention, lets take $a \in I(X)$, then $x \in I$ for every $I \subseteq R$ such $X \subsetneq I$, and from here im stuck as i cannot how can $a$ is a finite linear combination of elements of $X$ and $R$? Thanks
To show $I(X) \subset A$, it suffices to show that $A$ is an ideal containing $X$, then by definition of $I(X)$, you will obtain the inclusion.