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Suppose that the ring homomorphism $R\rightarrow S$ is faithfully flat ($R$ and $S$ are Noetherian commutative rings). Let $A$ be an Artinian $R$-module. Do we have $0:_S(A\bigotimes_R S)=(0:_RA)S$.

I know this the case when $A$ is Noehterian. Also by purity I know that $\big(0:_S(A\bigotimes_R S)\big)\bigcap R=0:_RA$ for an arbitrary $R$-module A.

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No. Let $A=\mathbb{Z}[1/2]/\mathbb{Z}$. Let $R\to S$ be $\mathbb{Z}\to \mathbb{Z}_2\times \mathbb{Z}_3$, where $\mathbb{Z}_n$ denotes $\mathbb{Z}[1/n]$.

Then $Ann_{\mathbb{Z}}A=0$. But $(1,0)\in S$ kills $A\otimes_{\mathbb{Z}}S$.

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The following counterexample might be interesting too. There exists an unmixed d-dimensional commutative Noetherian local domain $(R,\mathfrak{m})$ such that its completion is not unmixed. Then $H^d_\mathfrak{m}(R)$ is a counterexample since the annihilator of $H^d_\mathfrak{m}(R)$ is the unmixed part of the zero ideal.

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