I am looking at some notes on structure of complete local rings by Hochster.
Let $A$ be a Noetherian, complete local ring with maximal ideal $m$. I am looking at the ring of formal power series $R=A[[x]]$ and the finitely generated $R-$module, $M:=A[[x]]/\langle f\rangle$, where $f$ is regular of order $h$ in $A[[x]]$ (meaning the smallest order term in $f$ is $a_hx^h$ where $a_h$ is a unit in $A$).
I am told because $M$ is separable in the $m$-adic topology, $M/mM\cong K[[x]]/\langle\bar{f}\rangle$ where $\bar{f}$ is the image of $f$ under the map $A[[x]]\twoheadrightarrow K[[x]]$. I am trying to verify this claim.
I don't see at all why this follows. I know $M\otimes_A A[[x]]/m\cong M\otimes K[[x]]\cong M/mM$ from this post. I know $\hat{M}/\hat{M_n}\cong M/\mathfrak{m}^nM$ where $\hat{M_n}$ is the filtration in $\hat{M}$, whatever that is, but this has nothing to do with $M$ being separable. I am having a hard time getting my head around $M/mM$. Maybe this has something to do with associated graded rings, but I am not finding any theorems that help. But I have a feeling this follows very immediately from some fact I just forgot.
You have an exact sequence $ A[[X]]\stackrel{f}{\to} A[[X]]\to M\to 0$. Tensor it with $A[[X]]/m=K[[X]]$ and remember tensor product is right exact.