 Solution

No. A domain with a nonzero idempotent maximal ideal is a counterexample. ConsiderThis can be realized concretely as a ring of functions on $t>0$: its elements are finite sums of rational multiples of nonnegative dyadic rational powers of $t$. Each stage $\mathbb{Q}[u_n]$ is a polynomial ring, because a nonzero polynomial cannot vanish for all $u_n>0$. Any two elements lie in one stage, which is an integral domain, so their product is nonzero if both are nonzero. Thus $R$ is an integral domain.

The constant term defines a surjective ring homomorphism $ϵ:R\to \mathbb{Q}$. It is well defined because replacing $u_n$ by $u_{n+1}^2$ preserves the constant term. Its kernel of a ring homomorphismis a maximal ideal, since $R/\mathfrak{m}\cong \mathbb{Q}$ is a field, by the maximal ideal quotient criterion. Also $u_0=t\in \mathfrak{m}$ is nonzero.

If $f\in \mathfrak{m}$, choose a stage in which $f=u_{n}g(u_n)$ for a polynomial $g\in \mathbb{Q}[z]$. ThenBoth factors belong to $\mathfrak{m}$, so $f\in {\mathfrak{m}}^2$. The reverse inclusion holds for every ideal, proving that $\mathfrak{m}$ is an idempotent ideal. HenceThe Noetherian ring hypothesis in c is therefore essential even for a maximal ideal.