good night!
Make sure that $m \mathbb{Z}\times n\mathbb{Z}$ in the ring $\mathbb{Z} \times \mathbb{Z} $ is an ideal.
I couldn’t come up with a convincing answer. Could you help me? Thanks in advance!
I know that ideal is a product $R\times S$ of rings with identity has the form $I\times J$, where $I, J$ are ideals in $R, S$ respectively. Moreover, since $\mathbb{Z}$ is a principal ideal domain, every ideal of $ \mathbb{Z} $ has the form $m\mathbb{Z}$ for some $m$. Hence every ideal of $ \mathbb{Z}\times \mathbb{Z} $ has the form $m\mathbb{Z} \times n\mathbb{Z}$ for soem $m, n \in \mathbb{Z}.$
But, Is it true?
First, make sure that $m\mathbb Z$ and $n\mathbb Z$ are ideals of $\mathbb Z$. How would you do that? You just take generic elements $a\in m\mathbb Z$ and $b\in\mathbb Z$ and check that $ab\in m\mathbb Z$, right?
Now, assuming that operations on products are defined coordinate-wisely, you can use the same method to prove that $I=m\mathbb Z\times n\mathbb Z$ is an ideal of $R=\mathbb Z\times\mathbb Z$. In fact, a generic element of $I$ has form $(a,a')$ with $a\in m\mathbb Z$ and $a'\in n\mathbb Z$, and for any $(b,b')\in\mathbb Z\times\mathbb Z$, their product is defined as $$ (a,a')(b,b')=(ab,a'b'). $$ I believe that you're able to continue by yourself from here :)