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What kind of rings(commutative, w/ unity) have exactly three ideals? I know that those with exactly two ideals are "the fields", but what about three? Is there a fancy name for them?

sacch
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    Because any non-zero non-unit generates a proper ideal, all those need to generate the same ideal. IOW that third ideal consist of precisely the non-units. A ring with the property that non-units form an ideal is called a local ring (motivation for the name comes from algebraic geometry). However, more often than not local rings have more ideals (such as the ideal generated by the squares of non-units), so they don't usually (nearly always) have this property. – Jyrki Lahtonen Jul 03 '14 at 05:52
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    Why «"the fields"»? I pictured you doing finger quotes... – Mariano Suárez-Álvarez Jul 03 '14 at 05:53
  • Examples of rings with three ideals I can think of would be $R=\Bbb{Z}_{p^2}[x]/\langle f(x)\rangle$, where $f(x)$ is a monic polynomial with the property that when it's coefficients are reduced modulo $p$, it is irreducible in $\Bbb{Z}_p[x]$. The only ideals of $R$ are $0,pR$ and $R$. – Jyrki Lahtonen Jul 03 '14 at 05:55
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    Not much more generally, let $R$ be a local ring with (unique) maximal ideal $\mathfrak m$. Then $R/\mathfrak m^2$ has exactly three ideals. And each ring with exactly three ideals can be obtained this way. – Hagen von Eitzen Jul 03 '14 at 06:50
  • @HagenvonEitzen what makes your comment not an answer? – Mark Bennet Jul 03 '14 at 07:12

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Let $R$ be a commutative with exactly three ideals. These must be $0 \subset \mathfrak{m} \subset R$ for some ideal $\mathfrak{m}$. It follows $\mathfrak{m}$ is a maximal ideal, in fact the unique one, so that $R$ is local.

Choose $x \in \mathfrak{m} \setminus \{0\}$, then $0 \neq \langle x \rangle \neq R$, hence $\mathfrak{m} = \langle x \rangle$. We conclude that $R$ is a special principal ideal ring. Now look at $\mathfrak{m}^2$. If $\mathfrak{m}^2=\mathfrak{m}$, then the Nakayama Lemma implies $\mathfrak{m}=0$, a contradiction. Otherwise we have $\mathfrak{m}^2=0$.

Conversely, let $R$ be a special principal ideal ring with maximal ideal $\mathfrak{m}$ satisfying $\mathfrak{m}^2=0$ and $\mathfrak{m} \neq 0$. Then $R$ has exactly three ideals.

Martin Brandenburg
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  • Could you explain why $\mathfrak{m} ^2 = \mathfrak{m}$ implies $\mathfrak{m} =0$ without using Nakayama Lemma? I haven't studied modules yet. – Sahiba Arora May 29 '17 at 13:18