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

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!

• Extending along which homomorphism $R\to A$? Commented Mar 4, 2021 at 19:08
• Sorry, I still don't see it. To explain what does it mean to extend an $R$-module structure to an $A$-module structure, you must have an $R$-algebra structure on $A$, what is it? Commented Mar 4, 2021 at 19:15
• @მამუკაჯიბლაძე Thank you, my question was more than confusing without the condition that $A$ is an $R$-algebra. Hope this is fine now. Commented Mar 4, 2021 at 19:25
• But do you mean $A$ finitely generated as an $R$-module, or as $R$-algebra? does $A:=R[T]$ satisfies your hypothesis? I think not, otherwise it seems easy to find a counter-example.
– Joël
Commented Mar 5, 2021 at 5:47
• Interesting question.
– Joël
Commented Mar 5, 2021 at 13:12

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 $$M$$.