# Commutative local ring with $10$ ideals

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.

• 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. – Pratyush Sarkar Dec 30 '18 at 6:15
• @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 ... – user521337 Dec 30 '18 at 6:38
• 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. – Pratyush Sarkar Dec 30 '18 at 7:39

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.
• thanks ... and yes next would be $19$ ... of course any such ring is local ... do you have any non PIR example in mind .. ? – user521337 Dec 31 '18 at 1:09
• @user521337 I think $\mathbb{F}_7[x,y]/(x^3,xy,y^2)$ works. – Jeremy Rickard Dec 31 '18 at 9:43
• @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 $s(t+1)$. – Jeremy Rickard Dec 31 '18 at 13:50