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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?

Thanks in advance.

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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 $\mathbb{Z}$.

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This is probably not the easiest idea since it is sort of bruteforce, but here:

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.

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  • $\begingroup$ what about if i want to prove the prime submodules of $R×R$ are the Cartesian product of prime submodules of $R$? $\endgroup$ – Naheel Ghaith Jul 3 at 18:16
  • $\begingroup$ I do not think that you will be able to do that for arbitrary rings as one does not even necessarily has a good grasp on ideals and prime ideals for arbitrary rings. $\endgroup$ – ThorWittich Jul 3 at 19:44

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