$R/I$ where if $I\ne 0$ contains $r^2$ for $r\in R$ then I contains $r$ Let $R$ be a PID and let $A \ne 0$ be an ideal of $R$ that satisfies the condition that $r^2\in A$, $r\in R$ implies that $r\in A$. Show that $R/A$ is isomorphic to a finite direct product of fields.
My initial thought is take some $a$ such that $A=<a>$ by the PID property and then decompose $a$ into a product of its prime factors, but I'm not sure whether this is a good approach.
 A: As @ReneSchoof notes in the comments (I have edited the question to reflect this), you must assume $A \ne 0$, so that $a\ne 0$. I figure you left that out accidentally.
Write $a = p_1^{e_1}\cdots p_n^{e_n}$ with $p_1,\dots,p_n$ prime. Can you say anything about the $e_i$? Notice that $R/\langle p_1\cdots p_n \rangle \cong R/p_1 \times \dots \times R/p_n$, a product of fields, by the Chinese remainder theorem. Moreover, if any $e_i > 1$, then $R/\langle p_1^{e_1}\cdots p_n^{e_n} \rangle \cong R/p_1^{e_1} \times \dots \times R/p_n^{e_n}$ is not a product of fields. So you know what you need to prove.
A: You might want to use primary decomposition. Since $R$ is a PID, it's noetherian, so there are finitely many minimal primes above $A$. This just means you can write $A$ as an intersection of finitely many prime ideals:
$$A=P_1\cap\cdots \cap P_n$$
(Actually, you should be using $\sqrt{A}$ instead of $A$ here, but these are the same because of your condition on $A$.) Notice that prime ideals are maximal in a PID, so all these $P_i$'s are maximal (assuming $A\neq 0$).
Now use the Chinese Remainder Theorem.
EDIT: You need $A\neq 0$. The statement isn't even true if $A=0$.
