3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Dec 30 '18 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 Dec 30 '18 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 Dec 30 '18 at 7:39
1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Next time. If somebody asks for twelve ideals, I will know what to do :-) $\endgroup$ – Jyrki Lahtonen Dec 30 '18 at 9:47
  • $\begingroup$ @JyrkiLahtonen But I don’t want twelve ideals. I want nineteen. $\endgroup$ – Jeremy Rickard Dec 30 '18 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 Dec 31 '18 at 1:09
  • $\begingroup$ @user521337 I think $\mathbb{F}_7[x,y]/(x^3,xy,y^2)$ works. $\endgroup$ – Jeremy Rickard Dec 31 '18 at 9:43
  • 1
    $\begingroup$ @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 Dec 31 '18 at 13:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.