About the canonical map $R\rightarrow \prod_{i=1}^{r}R[g_{i}^{-1}]$ Let $R$ be a complete and separated adic ring with a finitely generated ideal of
definition $I\subset R$. Then we say that $R$ does not have $I$-torsion, if the ideal $(I-\textrm{torsion})_{R}=\{r\in R\mid I^{n}r=0\textrm{ for some }n\in\mathbb{N}\}$ is trivial. Then how to prove that the canonical map $R\rightarrow \prod_{i=1}^{r}R[g_{i}^{-1}]$ (where $g_{i}$ generate $I$ as an ideal) is injective if and only if $R$ does not have $I$-torsion? In fact, I guess that the canonical map is the composition $R\rightarrow\prod_{i=1}^{r}R[X]\rightarrow \prod_{i=1}^{r}R[X]/(Xg_{i}-1)$.
 A: First, as hm2020 notes in their comment, the canonical map $R\to \prod_{i = 1}^r R[g_i^{-1}]$ can be viewed as the composition
$$
R\to\prod_{i = 1}^r R[X]\to\prod_{i = 1}^r R[X]/(Xg_i - 1).
$$
This is because the localization map $R\to R[g_i^{-1}]$ is identified with the composition $R\to R[X]\to R[X]/(Xg_i - 1)\cong R[g_i^{-1}]$.

Now, let $R$ and $I = (g_1,\dots, g_r)$ be as in your setup, and let $J = (I\textrm{-torsion})_R,$ for ease of notation.
Suppose the map is injective, and let $r\in J$ be an $I$-torsion element. This means that $I^n r = 0$ for some $n\in\Bbb{N},$ or more explicitly, that $r x = 0$ for any $x\in I^n.$ In particular, we must have $r g_i^n = 0$ for each $i,$ which implies that $r$ is sent to $0$ in each $R[g_i^{-1}].$ Since  we assumed the map to be injective, this means that $r = 0.$ Thus, $J = 0$.
Conversely, suppose that $R$ has no $I$-torsion, and suppose that there exists some $r\in R$ such that $r$ is sent to $0$ in $\prod_{i = 1}^r R[g_i^{-1}]$. This is equivalent to saying that $r$ is sent to $0$ in each $R[g_i^{-1}].$ Then for all $i,$ it follows that there exists $n_i\in\Bbb{N}$ such that $g_i^{n_i} r = 0.$ Let $N = r \max_i{n_i}$. Then $I^N$ is generated by products of the form $x_1\cdots x_N,$ where each $x_i\in I.$
Choose $a_{i,j}\in R$ such that each $x_j = a_{1,j}g_1 + \dots + a_{r,j} g_j.$ Expanding out the product $x_1\cdots x_N$ will yield a sum of elements of the form $b g_1^{e_1}\dots g_r^{e_r},$ where $b\in R$ and $e_1 + \dots + e_r = N.$ Then at least one of these $e_i$ must satisfy $e_j\geq N/r = \max_i\{n_i\}\geq n_j,$ which guarantees that $b g_1^{e_1}\dots g_r^{e_r}r = 0,$ so that $xr = 0.$ This implies that $r\in \{r\in R\mid I^N r = 0\}\subseteq J,$ which means that $r = 0,$ and we're done.

Notice that this did not require the assumption that $R$ is $I$-adically complete or separated. Compare this statement to the facts that $R\to S^{-1}R$ is injective if and only if $S$ does not contain any zero-divisors in $R,$ and that $R\to\prod_{i = 1}^r R[f_i^{-1}]$ is faithfully flat (hence injective) when $(f_1,\dots, f_r) = R.$
