Let $R$ be a $\mathbb Z$ graded ring and let $f \in R_1$ where $f \ne 0$. Show that $R[f^{-1}]$ is a $\mathbb Z$ graded ring.

Question

I've defined $R[f^{-1}]_i = \{ \frac{r}{f^k} \mid r \in R_{i+k}\}$ and I've shown that each $R[f^{-1}]_i$ is an abelian group.

I've shown that $R[f^{-1}]_i R[f^{-1}]_j \subset R[f^{-1}]_{i+j}$.

I just need to show that $R[f^{-1}]$ is a direct sum of the $R[f^{-1}]_i$.

If $\frac{r}{f^k} \in R[f^{-1}]$, then $r \in R$ can be written uniquely $r=r_a + \cdots + r_{a+n}$ where $a \in \mathbb Z$ and $n \in \mathbb N$.

So, $\frac{r}{f^k} = \frac{r_a + \cdots + r_{a+n}}{f^k} = \frac{r_a}{f^k}+\cdots + \frac{r_{a+n}}{f^k} \in R[f^{-1}]_{a-k} + \cdots + R[f^{-1}]_{a+n-k}$.

So, $R[f^{-1}]$ is a sum of the $R[f^{-1}]_i$.

How can we show $R[f^{-1}]$ is a direct sum of the $R[f^{-1}]_i$?

Answer 1

Hint : multiply any relation $\sum_i \frac{r_i}{f^{k_i}} = 0$ in $R[f^{-1}]$ by a sufficiently large power of $f$ to get a relation in $R$.