Rings with exactly 3 ideals So I am reading some notes which are a bit confusing so I’m trying to make sense of everything.
The statement is
A commutative ring $R$ has exactly 3 ideals if and only if:

*

*$I := R \setminus R^* $ is a non zero ideal.

*$I=\langle r \rangle $ for some non zero $r \in I $ .

*$r^2=0 $ for all $r\in I$.

Here $R^*$ is the unit group.
I can show that if $R$ has exactly 3 ideals then 1, 2 and 3 hold. But for the converse, do we need all 3 conditions to hold for $R$ to have 3 ideals or are the conditions s equivalent so we just need one of them?
 A: *

*$(3)$ doesn't exclude the possibility that $R\setminus R^*\subseteq\{0\}$. See for instance the zero ring and the fields. However, these are the only obstructions: condition $(3)$ in rings that are neither fields nor the zero ring implies that $R\setminus R^*$ is the nilradical of $R$ and that it isn't zero.


*$(2)\Rightarrow (1)$ tautologically.


*there are rings where $(1)\land (2)$ holds but $(3)$ doesn't. Namely consider $\Bbb Z_{2\Bbb Z}=\{\frac nm\in\Bbb Q\,:\, m\text{ odd}\}$. The set of non-invertible elements is the ideal generated by $2$. However, this is a domain and, in point of fact, there is a non-zero ideal for each power of $2$.


*if $R$ is a ring with exactly three ideals, then $R^2$ has $9$ ideals and it still satisfies $(3)\land (1)$. Therefore in $R^2$ condition $(2)$ doesn't hold.
Therefore, taking for granted that one ring with exactly three ideals exists (which is the case), you must require $(2)\land (3)$, while $(1)$ is redundant.
A: You only need conditions 2) and 3), since 2) implies 1). Let $R$ be a commutative ring.
The first/second condition tells us that $R$ contains at least three ideals $(0) \subset I \subset R$.
Now let $J$ be any non-zero, proper ideal of $R$. By definition of $I$ we have that $J \subset I$. Our goal is to show that $J = I$. Now, certainly $J$ contains some non-zero non-unit element $s \in J$, so we have the chain of inclusions $$
(s) \subset J \subset I
$$
Thus if we can show $(s) = I$ we are done.
Now, by condition 2) we can write $I = (r)$ for some non-zero $r \in I$. We claim that for any $a \in R$, if $a \notin R^*$ then $ar = 0$.
Assume that $a \notin R^*$. Then $a \in I$. Since $I = (r)$, you have $a = br$ for some $b \in R$. But then $ar = br^2 = 0$ by condition 3), as desired.
Now, since $s \in I$, we can write $s = ar$ for some $a \in R$. Since $s$ is nonzero, by the above discussion $a$ is a unit, thus $a^{-1}s = r$. So $I = (r) \subset (s)$.
Thus we have $I \subset J \subset I$, hence $J = I$ and $R$ has exactly three ideals.
Condition wise, it seems to be enough to say that $I = R\setminus R^* = (r)$ for some nonzero $r$ with $r^2 = 0$.
