If $M\otimes_S T$ is an $A$-module, is $M$ an $A$-module?

Question

Let $\mathbb{C}$ be the field of the complex numbers. Let $R=\mathbb{C}[x]$, $T=\mathbb{C}\langle x\rangle$ be the ring of entire series with convergence radius at least $1$, and let $S=\mathbb{C}\langle\langle x\rangle\rangle$ be the ring of entire series with infinite convergence radius. We have $R\subset S \subset T$. Let $A$ be an $R$-algebra which is flat and finitely generated as an $R$-module.

Let $M$ be a finitely generated $S$ module such that the $R$-module $M\otimes_S T$ admits an $A$-module structure extending the one of $R$-module. Can I conclude that the $A$-module structure descends to $M$, extending the $R$-module structure?

EDIT: I should state what I have in mind. Let $X$ be the affine algebraic curve over $\mathbb{C}$ given by the spectrum of $A$. The choice of a coordinate $x$ corresponds to the choice of a non constant morphism $f:X\to \mathbb{A}^1_{\mathbb{C}}$ of algebraic curves. By GAGA, we associate a morphism of Riemann surfaces $X^{an}\to (\mathbb{A}^1_{\mathbb{C}})^{an}$. Note that $\mathcal{O}_{\mathbb{A}^{1,an}}(\mathbb{A}^{1,an})\cong S$ and $\mathcal{O}_{X^{an}}(X^{an})\cong A\otimes_R S$. Let $U$ be open subset of $X$ given by the inverse image of the open unit ball via $f$. Again, note that $\mathcal{O}_{X^{an}}(U)\cong A\otimes_R T$. Now, suppose I have a coherent sheaf $\mathcal{F}$ on $U\subset X^{an}$. Assume further that $f_*\mathcal{F}$ can be analytically continuated to $(\mathbb{A}^1_{\mathbb{C}})^{an}$. Can $\mathcal{F}$ be analytically continuated to $X^{an}$?

Many thanks!

Answer 1

I may have a counter-example. Suppose $A$ is $\mathbb{C}[x,t]/(t^2=x+1)$, and let $M$ be $\mathbb{C}\langle \langle x\rangle\rangle $ as an $S$-module. Then, $M\otimes_S T=\mathbb{C}\langle x\rangle $ is given an $A$-module structure by defining the action of $t$ as the multiplication by $\sqrt{x+1}$ where $$\sqrt{x+1}=1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3-\frac{5}{128}x^4+...$$ But $\sqrt{1+x}$ does not defines an entire function and hence the $A$-module structure does not desend to $M$.