Let $k$ be an algebraically closed field. Let $A = k[x,y,z,w]$, $\mathfrak p = (x, w)$, $\mathfrak a = (x^d, w)$, for some $d \in \mathbb N$.
How do I compute $(A/\mathfrak a)_\mathfrak p$?
How do I compute the length of $(A/\mathfrak a)_\mathfrak p$ as an $A_\mathfrak p$-module?
EDIT: Is the following procedure correct?
My basic idea is to use the exact sequence
$$ 0 \longrightarrow \mathfrak p/\mathfrak a \longrightarrow A/\mathfrak a \longrightarrow A/\mathfrak p \longrightarrow 0, $$
and localize it at $\mathfrak p$. Since $w = 0$ in all three quotients, we may replace $A, \mathfrak p, \mathfrak a$ as follows:
$$ A = k[x,y,z], \qquad \qquad \mathfrak p = (x), \qquad \qquad \mathfrak a = (x^d) = \mathfrak p^d. $$
Then the aforementioned exact sequence becomes
$$ 0 \longrightarrow \mathfrak p/\mathfrak p^d \longrightarrow A/\mathfrak p^d \longrightarrow A/\mathfrak p \longrightarrow 0. $$
Since $A$ is an integral domain and $\mathfrak p = xA$ is a nonzero principal ideal, the $A$-homomorphism $\varphi : A \to \mathfrak p$ defined by $\varphi(f) = xf$ is an isomorphism. Then $\mathfrak p/\mathfrak p^d$ is isomorphic to $A/\mathfrak p^{d-1}$.
After localizing at $\mathfrak p$, the exact sequence becomes
$$ 0 \longrightarrow (\mathfrak p/\mathfrak p^n)_\mathfrak p \longrightarrow (A/\mathfrak p^n)_\mathfrak p \longrightarrow (A/\mathfrak p)_\mathfrak p \longrightarrow 0. $$
Inductively, $(\mathfrak p/\mathfrak p^d)_\mathfrak p \cong (A/\mathfrak p^{d-1})_\mathfrak p$ has length $d-1$. Since $(A/\mathfrak p)_\mathfrak p$ is the residue field of $A_\mathfrak p$, it has length $1$. Therefore, $(A/\mathfrak p^d)_\mathfrak p$ has length $(d-1) + 1 = d$.
