Artinianness detected by tensoring with a module-finite extension

Determine whether an R-module A must be Artinian as an R-module whenever A\otimes_R S is Artinian as an R-module, for a module-finite extension of Noetherian local rings R\hookrightarrow S.

Background

Theorem 1(b) proves that if A is an Artinian R-module, then A\otimes_R S is Artinian as an S-module and hence as an R-module. The unresolved converse asks whether Artinianness can be recovered from the Artinianness of the tensor product when no flatness, direct-summand, or annihilation hypothesis is imposed on the module-finite extension R\hookrightarrow S.

The paper notes that A\otimes_R S is a quotient of a finite direct sum At, where t=\ell_R(S/\mathfrak mS), but the Artinianness of A\otimes_R S does not immediately imply that of At. Lemma 5 establishes a positive answer under additional assumptions: the extension is flat, R is an R-module direct summand of S, or \mathfrak mA=0.

References

From the statement of Theorem \ref{T:1}(b), it is natural to ask the following question: \textit{Let $A$ be an $R$-module such that $A\otimes_RS$ is an Artinian $R$-module. Is $A$ an Artinian $R$-module?} It should be mentioned that, for a submodule $A'$ of $A$, the natural homomorphisms $A\to A\otimes_RS$ and $A'\otimes_RS\to A\otimes_RS$ are not necessarily injective, see Example \ref{E:1}. Although $A\otimes_RS$ is Artinian and it is a quotient of $At$ (where $t=\ell_R(S/\frak mS))$, we do not know whether $At$ is Artinian.

— The transfer of Artinian and Cohen-Macaulay properties under module-finite extensions  (2609.11003 - Chau, 10 Sep 2026) in Section 3, immediately after Theorem 1, before Lemma 5