= 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 \b[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).
Back to article page
