Simple Module Over a Product Ring I am trying to prove the following:

If $A$ and $B$ are commutative, unital rings and $M$ is a simple module over $A \times B$. Then $M$ is a simple module over $A$ or $B$.

Suppose that $M$ is not a simple module over $B$. Then there exists a proper sub-$B$-module $N_0 \subseteq M$. Let $N_1 \subseteq M$ be a non-zero sub-$A$-module. I want to argue that $N_1 =M$, thereby proving $M$ is a simple $A$-module. I was thinking of showing that either $N_0 + N_1$ or $N_0 \cap N_1$ are non-zero sub-$A \times B$-modules, but I don't see how to do this. I don't even know if this is the right approach.
 A: As Eric Wofsey points out, there is no guarantee that we can transport an $A\times B$-module structure on $M$ to an $A$-module or $B$-module structure on $M$. This is related to that fact that, if we assume ring homomorphisms are unital, there are not canonical ring homomorphisms $A\rightarrow A\times B$ or $B\rightarrow A\times B$ in general. Indeed, if we have any ring morphism $f:R\rightarrow S$, there will in general be a natural way to turn an $S$-module into an $R$-module (let $r\cdot m=f(r)\cdot m$), but not the other way around.
Now, this is the statement in the chat that you linked to:

"if we have a simple module $M$ over a product ring $A\times B$, then one can show that either $AM = 0$ or $BM=0$ and in the first case $M$ is a simple $B$-module and in the second case a simple $A$-module"

So, the content of what Lukas is saying is that, if $M$ is a simple $A\times B$-module, then there will be a way of transporting the $A\times B$-module structure on $M$ to either an $A$-module structure or a $B$-module structures on $M$, and that the induced module structure in question will be simple. Let's prove it. We take $M$ to be a simple left $A\times B$-module.
Now, we cannot realize $A$ or $B$ as unital subrings of $A\times B$, but we can realize them as the respective ideals $A\times\{0\}$ and $\{0\}\times B$ of $A\times B$. We will use $\tilde{A}$ to refer to $A\times\{0\}\leqslant A\times B$ and $\tilde{B}$ to refer to $\{0\}\times B\leqslant A\times B$; you may have been confused because Lukas was referring to $\tilde{A}$ as "$A$" and $\tilde{B}$ as "$B$".
In general, if $R$ is a ring, $M$ is a (left) $R$-module, and $I$ is a (left) ideal of $R$, we let $I\cdot M$ be the submodule of $M$ generated by the set $\{i\cdot m:i\in I,m\in M\}$. You should check yourself that $I\cdot M$ is equivalently the abelian subgroup of $M$ generated by this set; this fact follows immediately from the fact that $I$ is closed under left-multiplication by elements of $R$. In any case, this construction behaves well with respect to addition of ideals: if $J$ is another left ideal of $R$, then we have $I\cdot M+J\cdot M=(I+J)\cdot M$. (Verify this yourself!) Thus, because $A\times B=\tilde{A}+\tilde{B}$, and because $M$ is non-zero, we cannot have $\tilde{A}\cdot M=\{0\}=\tilde{B}\cdot M$. Hence without loss of generality we may suppose $\tilde{B}\cdot M\neq 0$. Since $M$ is simple, this therefore means that $\tilde{B}\cdot M=M$.
We claim now that $\tilde{A}\cdot M=\{0\}$; this follows immediately from the fact that $\tilde{A}\tilde{B}=\{0\}$. Indeed, for any $m\in M$, because $M=\tilde{B}\cdot M$, there are some $(0,b_1),\dots,(0,b_k)\in\tilde{B}$ and $m_1,\dots,m_k\in M$ such that $m=\sum_{i=1}^k(0,b_i)\cdot m_i$. Then, for any $a\in A$, we have $$(a,0)\cdot m=(a,0)\cdot\sum_{i=1}^k(0,b_i)m_i=\sum_{i=1}^k(a,0)\cdot(0,b_i)m_i=\sum_{i=1}^k(0,0)\cdot m_i=0,$$ as desired.
Now we are ready to show that the action $b\cdot m:=(0,b)\cdot m$ makes $M$ into a $B$-module. As I mentioned in my comment, the issue you run into without the assumption that $M$ is a simple $A\times B$-module is that there is no guarantee that $(0,1)\cdot m$ equals $m$. In fact, if you look at the list of axioms for a module, it should be easy to verify all but that one, even in the case where $M$ is not simple. Go through all of these axioms and carefully check them yourself!
If $M$ is simple though, that final axiom does hold, and we can show it using the fact that $\tilde{A}\cdot M=\{0\}$. Indeed, let $m\in M$. Because $M$ is an $A\times B$-module, we have $(1,1)\cdot m=m$. On the other hand, we know that $(1,0)\cdot m=0$, so this means that $$(0,1)\cdot m=((1,1)-(1,0))\cdot m=(1,1)\cdot m-(1,0)\cdot m=m.$$ Thus $(0,1)$ acts as the identity on $M$ and so declaring $b\cdot m=(0,b)\cdot m$ indeed gives a well-defined $B$-module structure on $M$.
Finally, we just need to show that, equipped with this structure, $M$ is a simple $B$-module. This again follows immediately from the fact that $\tilde{A}\cdot M=\{0\}$. Indeed, let $m\in M$ be non-zero, and let $n\in M$ be arbitrary. We wish to find an element $b\in B$ such that $n=b\cdot m=(0,b)\cdot m$. Because $M$ is a simple $A\times B$-module, there is an element $(a,b)\in A\times B$ such that $(a,b)\cdot m=n$. On the other hand, we know that $(a,0)\cdot m=0$, and so we have $$(0,b)\cdot m=((a,b)-(a,0))\cdot m=(a,b)\cdot m-(a,0)\cdot m=n,$$ just as desired.
