A ring with a unique nonzero proper ideal?

Question

Is there any commutative unital ring $R$ with a unique nonzero proper ideal? In particular, there must be a nonzero proper ideal, so fields don't work.

Clearly, such a ring must be local, and the unique (maximal) ideal $I$ contains every non-unit. If the ring is not reduced (i.e., there exists a nonzero nilpotent), then the nilradical is a nonzero proper ideal, so it must be equal to $I$.

I know $R$ can't be an integral domain, since if $a\in R$ is a nonzero nonunit, then $(a^2)\neq (a)$ are distinct nonzero proper ideals.

I'm not quite sure how to proceed from here. Can such an $R$ exist?

I believe $\Bbb Z/4\Bbb Z$ is a ring with one non-trivial ideal. — Gregory Grant, Nov 30 '15 at 21:06
Oh, right...are there any others besides $\mathbb Z/p^2\mathbb Z$ for $p$ prime? — Nishant, Nov 30 '15 at 21:06
@Nishant if there is exactly one nontrivial ideal, that must be the unique prime ideal. If The zero ideal were prime then it is an integral domain. There would be a nontrivial principal ideal that squares to a different nontrivial ideal, and that is a contradiction. So the zero ideal is not prime. Then the unique nonzero prime ideal is the nil radical, so such a ring cannot be reduced. — rschwieb, Dec 01 '15 at 01:07
Alternativel,y the ring is artinian, so the unique maximal ideal is the only prime ideal, is the nil radical etc. — rschwieb, Dec 01 '15 at 01:11

score 2 · Answer 1 · answered Nov 30 '15 at 21:32

2

$K[X]/(X^2)$, $K$ a field. It has only one prime ideal, generated by the class of $X$.

answered Nov 30 '15 at 21:32

Bernard

175,478

score 2 · Accepted Answer · answered Nov 30 '15 at 21:55

In the given setup , let $k=R/I$.

Following your argument $(a^2)\ne (a)$ for non-zero non-units $a$ implies $a^2=0$ for all non-units. (In particular, $R$ is reduced).

Pick $0\ne a\in I$. Then all elements of $I$ can be written as $ax$ with $x\in R^\times\cup\{0\}$ for if $x$ is a non-unit, the $x=ay\in I$ and $ax=a^2y=0$. Define a multiplication on $I$ by $ax\odot ay=axy$ for $x,y\in R^\times \cup\{0\}$. This makes $(I,+,\odot,0,a)$ a field. In fact, $(R,+,\cdot)\to (I,+,\odot)$, $x\mapsto ax$ is a ring homomorphism with kernel $I$, so the field $I$ is isomorphic to $R/I$. Thus as rng, $R$ is an extension of $k$ by $k$: $$0\to k\to R\to k\to 0$$

The simplest case (i.e., where $R=k\oplus k$ holds for the additive groups) is $R=k[\epsilon]=k[X]/(X^2)$. (Bernard's example)
For $k=\Bbb Z/p\Bbb Z$, we can easily write down $R=\Bbb Z/p^2\Bbb Z$. (For $p=2$, Nishant's example)
For $k=\Bbb F_{q}=\Bbb F_p[X]/(f)$, we can wite down $R=(\Bbb Z/p^2\Bbb Z)[X]/(pf)$

I suppose more can be said if one has a closer look at the Ext functor ...

So this classification works for rings where $(a^2)\neq (a)$ for some $a\in I$? Also, if $a^2=0$, then $R$ isn't reduced, right?
I don't quite understand how your third example is an extension of $k$ by $k$...what is $I$ in that case? — Nishant, Nov 30 '15 at 22:10

A ring with a unique nonzero proper ideal?

2 Answers2