Homogeneous localization. Let $S=\bigoplus_{d\ge 0} S_d$ be a graded ring with graduation given by the indices from $\mathbb{N}\cup\{0\}$. Let $T\subseteq S$ be a multiplicative subset consisting of homogeneous elements. Then we can define the usual localization $T^{-1} S$. We can give the graded $S$-module structure on $T^{-1}S$ as follows:
Define $S_d:=\{0\}$ for $d<0$. For any $n\in\mathbb{Z}$ define
$$(T^{-1}S)_n:=\bigg\{\frac{s}{t}\;:\; t\in T\cap S_d,\; s\in S_{n+d}\;\text{for some}\;d\ge 0\bigg\}$$
I have to prove the following:

Show that $$T^{-1}S =\bigoplus_{n\in\mathbb{Z}} \big(T^{-1}S\big)_n$$

I know the steps to take are as follows:
$(i)\quad T^{-1}S=\sum_{n\in\mathbb{Z}}\big(T^{-1}S\big)_n,$
$(ii)\quad\text{For any distinct choices}\;n_1,\dots, n_k, n_{k+1}\in\mathbb{Z}\;\text{we have}$
$$\big(T^{-1}S\big)_{n_{k+1}}\bigcap\bigg(\big(T^{-1}S)_{n_1}+\cdots +\big(T^{-1}S\big)_{n_k}\bigg)=\{0\}$$
Could anyone help me?
 A: For (i): If $s/t\in T^{-1}S$ with $\deg(t)=d$, writing $s=s_0+s_1+\cdots+s_n$ in $S$ with $s_i\in S_i$, view $$s_i/t\in (T^{-1}S)_{i-d}.$$ The other containment is easy.
This takes care of (i).
For (ii): For the 'direct sum' bit, enough to show that given  $s/t\in (T^{-1}S)_n$, if $$s/t\in\sum_{i=0}^{n-1} (T^{-1}S)_i,$$ then $s/t=0.$
Write $$s/t=s_0/t_0+\cdots+s_{n-1}/t_{n-1}$$ where for all $i$, $t_i$ is of degree $d_i$ and $\deg(s_i)=i+d_i.$ Let $\deg(t)=d$ so that $\deg(s)=d+n$.
Now clear the denominators to get
$$\dfrac{s}{t}=\dfrac{s_0t_1\cdots t_{n-1}+s_1t_0t_2\cdots t_{n-1}+\cdots+s_{n-1}t_0t_1\cdots t_{n-2}}{t_0t_1\cdots t_{n-1}}.$$
So $$t_0t_1\cdots t_{n-1}s=t(s_0t_1\cdots t_{n-1}+s_1t_0t_2\cdots t_{n-1}+\cdots+s_{n-1}t_0t_1\cdots t_{n-2})$$
in $S$.
The LHS is of degree $d+n+\sum d_i $ and the RHS is of degree $$max\{d+\sum d_i, d+1\sum d_i,\cdots , d+(n-1)+\sum d_i\}.$$
So the degrees don't match, hence $t_0t_1\cdots t_{n-1}s=0$ in $S$ hence, $s/t=0$ in $T^{-1}S$.
This finishes (ii).
