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Let $R$ be a commutative ring with identity, such that $R$ has precisely three ideals, namely (0), a non zero proper ideal I, and $R$ itself. Is there any known result on the classification of such rings?

jimm
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    To find the duplicates, I put "exactly three ideals" into the search bar. Please also take this step yourself next time before posting. – rschwieb Oct 31 '24 at 13:08

1 Answers1

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Note that since $R$ has finitely many ideals, it must be Noetherian. Furthermore, $I$ is the unique maximal ideal (so $I = J(R)$), and $I \neq 0$.

Suppose $I^2 = I$, Nakayama's lemma implies that $I = (0)$ -- contradiction. Thus, $I^2 \subsetneq I$, i.e., $I^2 = (0)$.

Furthermore, there must be some non-zero non-unit $x \in I$ (since otherwise, $I = (0)$ or $I = R$), so in particular, $I = (x)$ for some (and indeed any) such $x$. Since $I^2 = 0$, we have $(x)^2 = (x^2) = (0)$, so $x^2 = 0$. In other words, $I$ is generated by some element $x$ such that $x \neq 0$, $x^2 = 0$.

Moreover, we have that for any $y \in R$, $(y) \in \{ 0, I, R \}$, so $y = 0$, $(y) = I = (x)$ (in which case, we have $y \neq 0$, $y^2=0$) or $y$ is a unit.

Thus, $R$ consists of zero, units and non-zero elements whose square is 0.

Johannes Kloos
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  • Some further consideration shows that for non-units $p, q$, we have $(p+q)^2 = 2pq$, so $p+q$ is a non-unit (and therefore $(p+q)^2 = 0$), and if the characteristic of $R$ is not 2, we even have that the product of non-units is 0. I suppose this could then be extended to prove that all non-zero elements with zero square are associated, giving a fairly strong structure result - but the details of the proof elude me at the moment. – Johannes Kloos Oct 31 '24 at 07:42
  • Or, you already showed that for any non-zero non-unit $x$, $(x) = I$. So if $p$ and $q$ are non-zero non-units, then $pq \in (p)(q) = I^2 = (0)$ and $pq=0$. Then, $p \in I = (q)$ and $q \in I = (p)$ means there exist $a,b$ so that $p=aq$ and $q=bp$. But $p=abp$ so $ab=1$, $a$ and $b$ are units, and $p$ and $q$ are associates. – aschepler Oct 31 '24 at 13:42
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    That won't work, since $p$ is a zero-divisor. – Johannes Kloos Oct 31 '24 at 14:51
  • Remember that $a | b$ and $b | a$ implies $a ~ b$ only iff at least one of $a$ and $b$ is cancellable. – Johannes Kloos Oct 31 '24 at 14:52