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.



Faithfully flat ring homomorphism and annihilator
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 $A\otimes\_{\mathbb{Z}}S$.


