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Let $R$ be a commutative ring with unity with exactly $10$ ideals (including $\{0\}$ and $R$ ) . Then is it true that $R$ is a Principal Ideal Ring ?

My Work: I know that any commutative ring with $5$ or less ideals is a PIR. Indeed, suppose $R$ has $5$ or less ideals. Then $R$ is Noetherian so every ideal is finitely generated. If $R$ is not PIR, then there exists a $2$-generated ideal which is not principal, call it $J=(a,b)$. Then $(0);(a); (b); ( a+b); (a,b) $ and $R$ are all distinct ideals of $R$ giving at least $6$ ideals, contradiction ! Hence any ring with $5$ or less ideals is PIR. Now using this it follows that if $R$ is not local and has $10$ ideals, then by structure thm of Artinian rings, $R\cong R_1 \times R_2$ , where $R_1, R_2$ are non-zero Artinian rings having say $n_1$ and $n_2$ many ideals respectively. Then $n_1n_2=10$. Since $n_1,n_2 \ge 2$, we get w.l.o.g. $n_1=2, n_2=5$, thus both $R_1,R_2$ are PIR, hence $R\cong R_1 \times R_2$ is a PIR.

Thus, we only need to check the case for local rings.

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  • $\begingroup$ I looked up some properties of Artinian rings and I think may be you can adjust your proof like this: 1. I think you need local ring for the case with 5 ideals or less just so that we know $(a + b)$ is distinct from $R$. 2. Then for any ring with 10 ideals, it is Artinian and then proceed as you did breaking it up into finite product of local Artinian rings. I think that should do it? Let me know if I made a mistake. $\endgroup$
    – Pratyush Sarkar
    Commented Dec 30, 2018 at 6:15
  • $\begingroup$ @PratyushSarkar: We don't need local to ensure $(a+b)\ne R$ ... if $(a+b)=R$, then $(a,b)=R=(1)$ is principal, which we assumed it was not ... and your suggestion doesn't help ... if $R$ is local with $10$ ideals, then we can't break it as a product of further non-zero rings ... $\endgroup$
    – user521337
    Commented Dec 30, 2018 at 6:38
  • $\begingroup$ Right, that makes sense. Ok then may be suppose not PIR and look at the ideal $(a, b)$? If this is not the maximal ideal then there is another nonunit $c \in R \setminus (a, b)$ and then you have the distinct ideals $(0), (a), (b), (c), (a + b), (b + c), (c + a), (a, b), (b, c), (c, a), (a, b, c), R$ which is a contradiction. Then look at the case that $(a, b)$ is the maximal ideal. $\endgroup$
    – Pratyush Sarkar
    Commented Dec 30, 2018 at 7:39

1 Answer 1

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I count exactly ten ideals in $R=\mathbb{F}_5[x,y]/(x^2,y^2)$. Namely $$0,(xy),(x),(x+y),(x+2y),(x+3y),(x+4y),(y),(x,y),R.$$ And $(x,y)$ is not principal, so $R$ is not a PIR.

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    $\begingroup$ Next time. If somebody asks for twelve ideals, I will know what to do :-) $\endgroup$
    – Jyrki Lahtonen
    Commented Dec 30, 2018 at 9:47
  • $\begingroup$ @JyrkiLahtonen But I don’t want twelve ideals. I want nineteen. $\endgroup$
    – Jeremy Rickard
    Commented Dec 30, 2018 at 10:05
  • $\begingroup$ thanks ... and yes next would be $19$ ... of course any such ring is local ... do you have any non PIR example in mind .. ? $\endgroup$
    – user521337
    Commented Dec 31, 2018 at 1:09
  • $\begingroup$ @user521337 I think $\mathbb{F}_7[x,y]/(x^3,xy,y^2)$ works. $\endgroup$
    – Jeremy Rickard
    Commented Dec 31, 2018 at 9:43
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    $\begingroup$ @user521337 $\mathbb{F}_q[x,y]/(x^2,y^2)$ gives $q+5$, and $\mathbb{F}_q[x,y]/(x^{n+1},xy,y^2)$ gives $n(q+1)+3$. Using this, I can get from $6$ to $24$. And if you can do $25$ then the minimal counterexample must be a prime, as if $R$ does $s$ then $R\times k[x]/(x^t)$ does $s(t+1)$. $\endgroup$
    – Jeremy Rickard
    Commented Dec 31, 2018 at 13:50

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