category RMod as grothendieck category
1. Proposition
Let \(R\) be a ring and \(\mathrm{RMod}\) the category RMod. Then \(\mathrm{RMod}\) is a grothendieck category
2. Proof
2.1. a)
2.2. b)
follows from universal property of a ring as module as for different module-homomorphism \(f,g: M \rightarrow N\) there exists an element \(m \in M\) with \(f(m) \neq g(m)\) Hence
\begin{align*} \varphi: R \rightarrow& M \\ 1 \mapsto& m \end{align*}suffices to show that \(R\) is a seperator