Proposition 21

The ring $S^{-1}A$ with the family $((S^{-1}A)_i)$ is a graded ring and $S^{-1}M$ with the family $((S^{-1}M)_i)$ is a graded module over the graded ring $S^{-1}A$. The canonical mappings $i^S_A$ and $i^S_M$ are homogeneous of degree 0.

Let $m \in M$, $s \in S_k$, $m' \in M_{j'}$, $s' \in S_{k'}$ and suppose that $j - k = j' - k' = i$; then $(m/s) - (m'/s') = (s'm - sm')/ss'$ and $s'm - sm' \in M_{j'+k'} = M_{j'+k'}$ and $ss' \in S_{k+k'}$, hence $(m/s) - (m'/s') \in (S^{-1}M)_i$ by definition; this shows that the $(S^{-1}M)_i$ are additive subgroups of $S^{-1}M$. The sum of these groups is the whole of $S^{-1}M$: for every $x \in S^{-1}M$ may be written as $m/s$ where $m \in M, s \in S; s$ is homogeneous by hypothesis and $m$ is a sum of homogeneous elements $m_j$; hence $x$ is the sum of the $m_j/s$, each of which belongs to a subgroup $(S^{-1}M)_i$. Finally the sum of the $(S^{-1}M)_i$ is direct; for let us consider a finite family of elements $x_q$ ($1 \leq q \leq n$) such that $x_q \in (S^{-1}M)_{i(q)}$, where the indices $i(q)$ are distinct, and suppose that $\sum_{q=1}^n x_q = 0$. Each $x_q$ may be written as $x_q = m_q/s$ where $s \in S_k$ and $m_q \in M_{i(q)+k}$; the hypothesis implies that there exists $s' \in S$ such that $s'\left( \sum_{q=1}^n m_q \right) = 0$; if the $s'm_q$ were not all zero, we would have a contradiction since, if $s' \in S_d$, then $s'm, \in M_{i(q)+d}$ and the $i(q) + d$ are all distinct. It follows that $x_q = 0$ for every index $q$.

It is immediately verified that, if $a \in (S^{-1}A)_i$ and $x \in (S^{-1}M)_j$, then $ax \in (S^{-1}M)_{i+j}$. Applying this result to the case $M = A$, we see first that $S^{-1}A$ is a ring graded by the $(S^{-1}A)_i$; then we see that $S^{-1}M$ is graded module over $S^{-1}A$. Finally, as $1 \in A_0$, $i_A^S$ and $i_M^S$ are homogeneous of degree 0.