Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 10 / 5 / b / Solution

The assertion requires a proper ideal. If $I=A$, no proper ideal can contain it, so the literal statement without that restriction is false.

Let $I$ be proper, and consider the partially ordered collection of proper ideals containing $I$. It is nonempty. The union of a chain is an ideal, because any two of its elements belong together to some member of the chain. It is still proper: if it contained $1$, some member of the chain would contain $1$ and equal $A$. Thus every chain has an upper bound in the collection. The Zorn lemma gives a maximal member, and any proper ideal containing it still belongs to the collection. Hence this is a maximal ideal of $A$.

Therefore every proper ideal is contained in a maximal proper ideal. This algebraic argument only needs a nonzero unital commutative ring; the Banach algebra assumptions become important for closedness of maximal ideals in part (c).