Existence of a maximal element in a set of Ideals

Question

Let R be a commutative Ring with $1$ and $S\subset R$ multiplicatively completed ($s_1,s_2\in S\implies s_1 s_2 \in S$) and $I\subset R$ an ideal, such that $I\cap S=\emptyset$. Consider now $M:=\{J\subset R\ : J \ \text{is an ideal with} \ J\cap S = \emptyset \ \text{and} \ I\subseteq J \} $. Prove now, that $M$ has maximal Elements (with the partial order $\subseteq$).
The lemma of Zorn states, if every chain of elements is upper bounded, then there is a maximal element, but I'm struggeling to prove, that an upper bound must exist.

Answer 1

Note that $M$ is non-empty since $\{0\} \in M$.

Let $I_1 \subseteq I_2 \subseteq \ldots$ be a chain in $M$. Then define $J = \cup_{n=1}^\infty I_n$.

Let $x, y \in J$ and $r \in R$. Then there are integers $n_x,n_y$ such that $x \in I_{n_x}$ and $y \in I_{n_y}$, so taking $n = \max (n_x, n_y)$ we have $x, y \in I_n$, so $x - ry \in I_n\subseteq J$, so $J$ is an ideal of $R$. Clearly $J$ is an upper bound on the chain, so we may apply Zorn's Lemma.

Edit: My original argument was unclear. The upper bound does actually have to be in $M$. To see that this is the case, note that $I\subseteq J$, and also that for each $x \in J$, we have $x \in I_n$ for some $n$, and hence $ x\not\in S$ since $I_n \in M$ so $I_n \cap S = \varnothing$. So $J$ does not contain any elements of $S$, and hence $J \in M$.