Rings with exactly 3 ideals

Question

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?

In fact, they definitelty mean each alone is equivalent to three ideals. — Thomas Andrews, Oct 17 '21 at 22:09
So 1, 2 and 3 are all equivalent and so they’re are all equivalent individually to $R$ having exactly 3 ideals? — Anonmath101, Oct 17 '21 at 22:14
How does 1 imply 2 without knowing $R$ has exactly 3 ideals? — Anonmath101, Oct 17 '21 at 22:21

score 2 · Answer 1 · 2021-10-17T22:54:13.870

$(3)$ 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.

Condition (1) is equivalent to the ring being local, i.e., having a unique maximal ideal. Well I guess a local ring that is not a field, since it is assumed to be nonzero — metalspringpro, Oct 17 '21 at 22:53

Rushy · Accepted Answer · 2021-10-17T23:05:21.907

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 $r^2 = 0$.

@Ben Small note: This is not meant as an answer, since the necessity of 2) and 3) was already answered in the other answer. But I wasn't sure if your question was purely about the necessity of the conditions or also about how to prove the converse. In any case, I assume it could help other people looking for this question. — Rushy, Oct 17 '21 at 23:16

Rings with exactly 3 ideals

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