I need to find an example of a ring that is not a PID but every ideal is principal. I know that $\mathbb Z\times\mathbb Z$ is not an integral domain, so certainly is not a PID, but here every ideal is principal. I already proved that if $R$ and $S$ are ring every ideal in $R \times S$ is $I \times J$ with ideals in the original ring. But I cant follow from that that $\mathbb Z\times\mathbb Z$ has only principal ideals.
Explicitly, if $I$ is an ideal $(a)\times(b)$ which would be its generator $(c,d)$ in $\mathbb Z\times\mathbb Z$?
Thanks.
Best Answer
Suppose $\def\ZZ{\mathbb Z}I\subseteq\ZZ\times\ZZ$ is an ideal.
If $(x,y)\in I$, then $(x,0)=(1,0)(x,y)$ and $(0,y)=(0,1)(x,y)$ are also in $I$. If we write $I_1=\{x\in\ZZ:(x,0)\in I\}$ and $I_2=\{y\in\ZZ:(0,y)\in I\}$. then it follows easily from this that $I=I_1\times I_2$. Now $I_1$ and $I_2$ are ideals of $\ZZ$, so that there are $a$, $b\in\ZZ$ such that $I_1=(a)$ and $I_2=(b)$, and then $I$ is generated by $(a,b)$.