Hopefully a quick one!
Let $R$ be a compact noetherian ring and $S$ be an $R$-module. If $T,T^\prime$ are compact submodules of $S$, does it then follow that the module $T+T^\prime$ is compact too (where $T+T^\prime$ is the module generated by $T \cup T^\prime$)?
I know that this is false for general modules, but I wonder if it works in this scenario.
Attempt: Suppose $T+T^\prime \subseteq \bigcup_{i \in I} U_i$ is an open cover and $R \subseteq \bigcup_{j=1}^n V_j$ is a finite (sub)cover (as $R$ is compact).
As $T,T^\prime \subseteq_{i \in I} U_i$ we can extract finite subcovers for each: $T=\bigcup_{i_1 \in I_1} U_{i_1}$ and $T^\prime = \bigcup_{i_2 \in I_2} U_{i_2}$ for $I_1, I_2 \subseteq I$ finite sets.
Note that then $T \cup T^\prime \subseteq \bigcup_{i_3 \in I_3} U_{i_3}$ where $I_3 = I_1 \cup I_2$ is also a finite set.
Then $T+T^\prime = R \cdot (T \cup T^\prime) \subseteq \bigcup_{j=1}^n V_j \cdot \left( \bigcup_{i_3 \in I_3} U_{i_3} \right)$ which is a finite union of a product of finitely-many open sets, hence a finite open cover of $T+T^\prime$.
Thanks in advance as ever,
M
Subcover of what? $U_i$ are subsets of $S$, not $R$. They do not cover $R$ at all, to pick a "subcover".
No, those subsets are hardly ever equal. The subset on the left consists of all elements of the form $t+t'$ for $t\in T$ and $t'\in T'$, while the subset on the right consists of all elements of the form $rx$ for $r\in R$ and $x\in T\cup T'$. In particular $R \cdot (T \cup T^\prime)$ is literally equal to $T\cup T'$.
A proper solution comes from the observation that $T+T'$ is a continuous image of $T\times T'$ which is compact (as a product of compact spaces). The surjective continuous mapping we are looking for is given by $(x,y)\mapsto x+y$. And this is regardless of what $R$ is (compact or not, noetherian or not), the only thing we need to know is that $T$ and $T'$ are compact.