Let $R$ be a PID and $A$ an ideal of $R$ such that for any $r^2 \in A$ where $r \in R$, then $r \in A$.
I'm trying to show that $R / A$ is isomorphic to a finite direct product of fields. But I'm a bit stuck.
I was trying to employ the fact that if $A_1, A_2, \ldots, A_n$ are ideals of $R$ and $A = \bigcap_{ i = 1 }^n A_i$, then $R / A$ is isomorphic to a subring of $R / A_1 \times R / A_2 \times \cdots \times R / A_n$. In order to do so, I set $A = \langle a \rangle$ for some $a \in A$ and, using the fact that $R$ is a UFD, decompose $a$ into $$a \sim p_1^{ k_1 } p_2^{ k_2 } \cdots p_n^{ k_n }$$ I had originally thought that I would be able to use $\langle p_1^{ k_1 } \rangle, \langle p_2^{ k_2 } \rangle, \ldots, \langle p_n^{ k_n } \rangle$ as my ideals as $A = \bigcap_{ i = 1 }^n \langle p_i^{ k_i } \rangle$ however, these $p_i^{ k_i }$ are not prime so I can't bet on the fact that $R / A_i$ is a field.
I was wondering if I could get a hint.
Your argument would work just fine if $k_i = 1$ for each $i$, right? In other words, you want to show that $p_1\cdots p_n\in A$. On the other hand, you know that $(p_1\cdots p_n)^k\in A$ for large enough $k$. Thus it suffices to do the following:
To do this, I suggest letting $k$ be minimal such that if $r^k\in A$. If $k =1$, you're done. Otherwise, if $k>1$, consider $(r^{k-1})^2$.