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