Let $A$ be a commutative ring. Show that if $A$ is Noetherian, then $A[[x_1,...,x_n]]$ is a faithfully flat $A$-algebra.
I've already shown that $A[[x_1,...,x_n]]$ is a flat $A$-algebra (by showing that $A[[x_1,...,x_n]]$ is a flat $A[x_1,...,x_n]$-algebra and $A[x_1,...,x_n]$ is a flat $A$-algebra). How can I take this further to show that $A[[x_1,...,x_n]]$ is a faithfully flat $A$-algebra?
A separate exercise (shown below) that I've completed provides different equivalent conditions for a flat $R$-algebra $S$ to be faithfully flat, where $R$ and $S$ are commutative rings. Could I use one of these conditions?
Let $S$ be a flat $R$-algebra. Then the following conditions are equivalent:
i) $a^{ec} = a$ for all ideals $a$ of $R$.
ii) $\text{Spec}(S) \rightarrow \text{Spec}(R)$ is surjective.
iii) For every maximal ideal $m$ of $R$ we have $m^e \neq (1)$.
iv) If $M$ is any non-zero $R$-module, then $M_S \neq 0$.
v) For every $R$-module $M$, the mapping $x \mapsto 1 \otimes x$ of $R$ into $M_S$ is injective.
Thanks!