# One dimensional noetherian domain with every maximal ideal being principal: any radical ideal is principal

Let $$R$$ be a one-dimensional Noetherian domain such that every maximal ideal is principal. If $$I$$ is a radical ideal ($$\sqrt I = I),$$ show that $$I$$ is principal.

Since $$R$$ is a domain, $$\{0\}$$ is prime; so given $$P$$ prime ideal, since $$\{0\}\subseteq P$$, then $$P$$ must be maximal by the fact that $$R$$ is one-dimensional. Now, $$R$$ being Noetherian gives us

$$I = \bigcap Q_i, \quad Q_i \text{ a }P_i\text{ - primary ideal.}$$

Since $$\sqrt I = I$$ and each $$P_i$$ is principal (by being maximal), then

$$I = \bigcap (a_i), \quad a_i\in R.$$

And now? How do I prove that $$I$$ is principal? If I manage to prove that $$I$$ is prime, then I would be done as it would be equal to some of the $$(a_i)$$. Or if I could prove that $$I = (a_1...a_n)...$$

Hints? Thank you.

EDIT: After a while, I've found out that any ring whose prime ideals are principal is a PIR. That's the case in the exercise above. But I would like to know if there's a more direct proof without using the result I have just mentioned.

Hint: Let $$I,J$$ be comaximal ideals, i.e. $$I+ J = R$$. Then $$I \cap J = IJ$$ because $$IJ \subseteq I \cap J = (I + J)( I \cap J) \subseteq IJ$$.