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It is known that every element in a ring with finitely many ideals is either a unit or a zero divisor.

Now consider a ring $R$ with exactly three ideals. Take an element $x$ such that $(x)\neq (1)$. For $y\in R$ and $y\notin (x)$, there are two possible cases:

  1. $(y)=(1)$

  2. $(y)\neq(1)$. Then $(y)=(x)$, that is $x=ys=xts$ for some $s,t\in R$. It follows that $$x(1-ts)=0$$ which implies that $x$ is a zero divisor.

However, if there is no such a $y$ of the second case, how to prove $x$ is a zero divisor?

sysMirror
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    Actually, this question can be interpreted as "is there a ring $R$ with three exactly ideals ${0},~{0,x}$ and $R$". – sysMirror Jul 14 '17 at 21:29

3 Answers3

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Suppose $R$ is a commutative ring with only one nonzero proper ideal, $I$ say.

Let $x \in I$, where $x \ne 0$. Since $I$ is a proper ideal, $x$ is not a unit.

Then $(x)$ is a nonzero proper ideal of $R$, hence we must have $I=(x)$.

Suppose $x$ is not a zero divisor.

Then $x^2 \ne 0$, and $(x^2) \subseteq (x)$, so $(x^2)$ is a nonzero proper ideal of $R$.

Hence we must have $I = (x^2)$, and thus, $(x^2) = (x)$.

Then $x = rx^2$ for some $r \in R$, so $x(rx-1)=0$.

But $x$ is not a zero divisor, hence we must have $rx = 1$, contradiction, since $x$ is not a unit.

Thus, $x$ must be a zero divisor.

Hence, noting that $0$ is also a zero divisor, all elements of $I$ are zero divisors.

Next suppose $y \in R,\;y \notin I$.

Then $y \ne 0$, so $(y)$ is a nonzero ideal with $(y) \ne I$.

It follows that $(y)=(1)$, hence $y$ is a unit.

Thus we've shown that all elements of $I$ are zero divisors, and all elements of $R\setminus I$ are units.

quasi
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  • Could I just say let $x\ne 0$; then if $(x)=(1)$, we are done. If not, $(x)=I$. Suppose $x$ isn't a zero divisor, we must prove it's a unit. Then apply your argument: $(x^2)\ne(0)$. If $(x^2)=(1)$, we are done. Otherwise $(x^2)=(x)$ and $x=rx^2\iff x(rx-1)=0\implies rx=1$, so $x$ is a unit. – user557 Jul 19 '18 at 19:58
  • @user437309: Yes, that's the essence of my argument. – quasi Jul 19 '18 at 20:30
  • So there is no need to consider the cases $x\in I$ and $y\notin I$ separately, right? – user557 Jul 19 '18 at 20:34
  • @user437309: Effectively, you're doing that. – quasi Jul 19 '18 at 20:38
  • Yes, my original question was whether I can consider only one case, and I gave a modification of your proof in which I consider only one case. Now I see I can. Thanks! – user557 Jul 19 '18 at 20:41
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Either $(x^2)=(x)$ or $(x^2)=\{0\}$. The former case is impossible, because, by Nakayama's lemma, $(x)$ is generated by an idempotent $e$, but then in that case there would be at least four ideals, $\{0\}$, $eR$ $(1-e)R$ and $R$ among them.

(Alternatively you can just work out the consequences of $x^2r=x$ assuming $x$ isn't a zero divisor or unit.)

So you only have the first case, where $x^2=0$.

is there a ring $R$ with three exactly ideals $\{0\}$, $\{0,x\}$, and $R$

Yes, of course: $F_2[x]/(x^2)$.

Actually, for any commutative ring and nonidempotent maximal ideal $M\lhd R$, $R/M^2$ has exactly three ideals, but you went further and specified that the maximal ideal only has two elements...

rschwieb
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If there is no such a $y$ of the second case, we have to look at $(x)$.

If there exists $z\in (x)$ and $z\neq x$, then we have $(z)=(x)$. Thus, $x=zs=xts$ for some $s,t\in R$. It follows that $x(1−ts)=0$. So $x$ is a zero divisor.

If there is no other element besides $x$, that is, the ring has three ideals： $$\{0\},~\{0,x\},~ R$$ If there exists $y\neq x$ and $y\in R$, then $y=1$. Otherwise $\{0,x\}$ is not a ideal. And $x^2=0$, which implies that $x$ is a zero divisor.

sysMirror
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