# Prime submodules of $\mathbb{Z}×\mathbb{Z}$ as $\mathbb{Z}$-module

Let $$R$$ be a ring and $$M$$ a left $$R$$-module, a proper submodule $$P$$ of $$M$$ is said to be prime submodule if for any ideal $$I$$ of $$R$$ and any submodule $$N$$ of $$M$$, if $$IN\subseteq P$$, either $$N\subseteq P$$ or $$IM\subseteq P$$.

The prime submodule of $$R$$ as a left $$R$$-module are same as prime ideal of $$R$$.

My question: If $$R=\mathbb{Z}$$ and $$M=\mathbb{Z}×\mathbb{Z}$$, then the prime submodules of $$M$$ are of the form $$p\mathbb{Z}×q\mathbb{Z}$$, where $$p,q$$ are prime numbers.

How we can prove it by using the definition?

Perhaps, if you use that $$(N:M)\subseteq{R}$$ is a prime ideal whenever $$N\subseteq{M}$$ is a prime submodule, you get more info on the prime submodules of $$\mathbb{Z}\times\mathbb{Z}$$. Afterthat, distinguish between zero and nonzero prime ideals of $$\mathbb{Z}$$.
In the case of $$p\mathbb{Z}\subseteq{\mathbb{Z}}$$, since $$p\mathbb{Z}\times{p\mathbb{Z}}\subseteq{N}\subset{M}$$, then $$N/(p\mathbb{Z}\times{p\mathbb{Z}})\subset\mathbb{Z}_p\times\mathbb{Z}_p$$, study over the field $$\mathbb{Z}_p$$, and go up to $$\mathbb{Z}$$.
In the case of $$0\subseteq\mathbb{Z}$$, localizes at $$\Sigma=\mathbb{Z}\setminus\{0\}$$, then $$\Sigma^{-1}N\subset\mathbb{Q}\times\mathbb{Q}$$, study over $$\mathbb{Q}$$, and go back to $$\mathbb{Z}$$.
Since $$\mathbb{Z}$$-modules correspond to abelian groups, you have to look at proper subgroups of $$\mathbb{Z} \times \mathbb{Z}$$. The subgroups of $$\mathbb{Z} \times \mathbb{Z}$$ are well known, see for example here. As also all the ideals of $$\mathbb{Z}$$ are given by $$n\mathbb{Z}$$ for some $$n \geq 0$$, you can explicitly check the conditions you want for prime submodules.
• what about if i want to prove the prime submodules of $R×R$ are the Cartesian product of prime submodules of $R$? – Naheel Ghaith Jul 3 at 18:16