Computation of $\mathrm{Ext}_R(R[x^{-1}],M)$ Let $x$ be an element of a commutative ring $R$. It seems that the following statements are true:

Let us write $R[x^{-1}]$ as the direct limit $$R[x^{-1}]\simeq \varinjlim(R\xrightarrow{x} R\cdots ).$$
Let $M$ be an $R$-module. Let $TM$ denote the tower
$$( \cdots \rightarrow M \xrightarrow{x} M \xrightarrow{x} M). $$


Then


*

*


$$\mathrm{Ext}^1_R(R[x^{-1}],M) \simeq {\lim}^1 TM$$



*For $i\ge 2$, we have
$$ \mathrm{Ext}^i_R(R[x^{-1}],M) \simeq 0 $$


How does one prove each of these statements? For the 2nd point I wonder if there is a simple projective resolution.
 A: Coming from $R[x^{-1}]=R[U]/(1-Ux),$ we have a short exact sequence
$$0\rightarrow (1-Ux)R[U] \stackrel{\subseteq}{\rightarrow} R[U] \rightarrow R[x^{-1}]\rightarrow 0,$$
in which the first two $R$-modules on the left are free of countable rank (upon observing that $1-Ux$ is a non-zero divisor on $R[U]$): In fact, the short exact sequence is isomorphic to
$$0 \rightarrow R^{\oplus \mathbb{N}}\stackrel{\alpha}{\rightarrow} R^{\oplus \mathbb{N}} \rightarrow R[x^{-1}]\rightarrow 0,$$
where the map $\alpha$ is given by $$\alpha:(r_0, r_1, r_2,\dots)\mapsto (r_0, r_1-xr_0, r_2-xr_1,\dots).$$
Since this is a two-term projective resolution of $R[x^{-1}]$, this in particular shows that $\mathrm{Ext}_R^{\geq 2}(R[x^{-1}], M)=0$, showing (2).
In fact, applying $\mathrm{Hom}_R(-, M)$ to $\alpha$ yields, up to the canonical isomorphisms $\mathrm{Hom}_R( R^{\oplus \mathbb{N}}, M) \simeq M^{\times \mathbb{N}}$, the map
$$ \beta: M^{\times \mathbb{N}} \rightarrow M^{\times \mathbb{N}},\;\;  (m_0, m_1, m_2, \dots)\mapsto (m_0-xm_1, m_1-xm_2, \dots),$$
Then the cokernel $\beta$ is on one hand equal to $\mathrm{Ext}_R^1(R[x^{-1}], M),$ and on the other hand, equal to $\varprojlim^{1}(\dots \stackrel{x}{\rightarrow} M \stackrel{x}{\rightarrow} M)$ by definition (e.g. the one in nLab). This shows (1).
(Terminological remark (and some related references): If it happens that this $\mathrm{Ext}^1$ also vanishes, $M$ is called $x$-contraadjusted by Positselski. If additionally also $(\varprojlim TM \simeq )\mathrm{Hom}_R(R[x^{-1}], M)=0$, $M$ is called "$x$-contramodule" by Positselski,"$x$-adically derived complete" by Stacksproject, and "analytically $x$-complete" by Rezk.)
