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.
Best Answer
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.