The following is claimed (without much proof) during the the proof of Prop 9.2 in Atiyah & MacDonald. Saurabh commented below giving the proof that was probably intended by A&M (thank you!). I have now come up with a different proof and would be very grateful if someone would tell me whether it works.
Many thanks!
Lemma. Suppose $(A,\mathfrak{m})$ is a local Noetherian domain of dimension one. Any proper, non-zero ideal $\mathfrak{a}$ of $A$ is $\mathfrak{m}$-primary.
Proof. Since $A$ is local and every non-zero prime ideal of $A$ is maximal, it follows that $\mathfrak{m}$ is the only non-zero prime ideal of $A$. Hence the quotient ring $A/\mathfrak{a}$ has only one prime ideal and, as such, it is a local Artinian ring (since the quotient $A/\mathfrak{a}$ is also Noetherian and "Artinian = Noetherian of dimension zero"). Since the maximal ideal of a local Artinian ring is nilpotent, it follows that $\mathfrak{m}^n\subseteq\mathfrak{a}$ for some $n \in \mathbb{N}$. We also have $\mathfrak{a}\subseteq\mathfrak{m}$ and so, by taking radicals, we see that $\sqrt{\mathfrak{a}}=\mathfrak{m}.$ Hence $\mathfrak{a}$ is indeed $\mathfrak{m}$-primary. //
The unique maximal ideal is prime, and the zero ideal is prime. Every other proper ideal of the ring lies somewhere between. Since it is one dimensional, there are no prime ideals between these two.
Thus the radical of any proper nonzero ideal (the intersection of primes containing said ideal) is the maximal ideal.