how to prove topological module with discrete topology is a topological module

Question

I have read about the connection between injectivity in topological modules and topological discrete in the paper Goldman and Sah, and deadlock with this property: if R is a topological ring and M is untopologized R-module, then if M is assigned the discrete topology, M will be a topological module if, and only if, the annihilator of each element of M is an open ideal in R. Proof: $(\Leftarrow)$ Let he annihilator of each element of M is an open ideal in R. If $\tau$ is discrete topology on $M$, then $(M,\tau)$ is topological group. We have to prove that $f\colon R\times M\to M$ is continuous. Let $U$ be an open set in $M$ and $(r,m)$ be any point in $f^{-1}(U)$. Then $rm=f(r,m)\in U$. If $rm=0$, then there exists open submodule $N$ contains $m$ (since $\tau$ in $M$ is discrete topology) and open ideal $Ann(m)$ contains $r$ such that $f(Ann(m),N)\subset U$. But, if $rm\neq 0$, how can we find open ideal that contains $r$? For information, $R$ is Hausdorff topological ring. How can we prove this? Thank you for any help.

Answer 1

It doesn't have to be an open ideal! It has to be an open subset! So note that the subset you are looking for is simply

$$r+Ann(m)=\{r+s\ |\ s\in Ann(m)\}$$

which is open since $x\mapsto r+x$ is a homeomorphism $R\to R$.

Now you can pick $N$ to be simply $\{m\}$, it doesn't have to be a submodule as well, in fact in general it cannot be for $f(Ann(m),N)\subseteq U$ to hold. With that it is straightforward that $$f\big((r+Ann(m))\times\{m\}\big)=\{rm\}$$ is a subset of $U$.