One question regarding a subring of $Q$ which will contain 1. Let $R$ be a subring of $Q$ containing 1. Then which of the following is/are true?

$R$ is  a PID.

$R$ contains infinitely many prime ideals.

$R$ contains a prime ideal which is not a maximal ideal.

For every maximal ideal $m$ , $R/m$ will be finite.


My Try : 2. This is not true. $Q$ itself is the counter example. $Q$ has no ideal.

This is correct. Because {0} is in every $R$.

If we take the subring [$a/3^k$: where k is non negative integer]. Then $Z$ is a maximal ideal of this subring but $R/Z$ is not finite. so False.
I have no idea about option 1.


Have I gone wrong anywhere? Please correct me if I have and tell me what will happen for the option 1.
2, 3, and 4 are obviously false, since $\Bbb{Q}$ is an obvious counterexample in each case.

1 is true since the subrings are various localizations of $\Bbb{Z}$, and localizations of PIDs are PIDs.
2 is false, let $p$ be a prime in $\mathbb Z$ and let $R$ be the ring generated by $\mathbb Z$ and $\{n^{-1}:n\in\mathbb Z, p\nmid n$}.  In, other words every member of $\mathbb Z$ is a unit unless it is in the prime ideal $(p)$.  It's not hard to show that the only non-trivial ideal is $(p)$. So, $R$ has only one prime ideal.