How to prove $(s/t)\left(m/u+m'/u'\right)=(s/t)(m/u)+(s/t)(m'/u').$?

Question

Let $R$ be a commutative ring with unity 1. Let $T$ be nonempty multiplicative closed subset of $R$. Let $M$ be $R$-module. Define $$T^{-1}M=\{m/t\mid m\in M,t\neq 0\}.$$ Define the addition and multiplication on $T^{-1}M$ as below. $$m/t+m'/t'=(t'm+tm')/(tt')$$ $$(m/t)(m'/t')=(mm')/(tt').$$

Prove $$(s/t)\left(m/u+m'/u'\right)=(s/t)(m/u)+(s/t)(m'/u').$$ \begin{eqnarray*} (s/t)\left(m/u+m'/u'\right) &=& (s/t)\left((u'm+um')/(uu')\right)\\ &=& ((uu')s+t(u'm+um'))/(tuu')\\ &=& (uu's+tu'm+tum')/(tuu')\\ \text{I'm stuck here. I can't make the equation to this form.}\\ &=& ((tu'sm)+(tusm'))/(tutu')\\ &=& (sm)/(tu)+(sm')/(tu')\\ &=& (s/t)(m/u)+(s/t)(m'/u') \end{eqnarray*} So, I can't prove this question. Anyone can help me?

Answer 1

You have a typo in your line two; you need to multiply $(s/t)$ by $((u'm+um')/uu')$ and what you write is not that product. The argument goes as follows:

\begin{align*} (s/t)(m/u+m'/u')&=(s/t)\big((u'm+um')/uu'\big)\\ &=\big((s(u'm+um'))/tuu'\big)\\ &=\big((su'm+sum')/tuu'\big)\\ &=(t/t)\big((su'm+sum')/tuu'\big)\\ &=\big((tsu'm+tsum')/ttuu'\big)\\ &=\big((tu'sm+tusm')/tutu'\big)\\ &=(sm/tu)+(sm'/tu')\\ &=(s/t)(m/u)+(s/t)(m'/u'). \end{align*}