Let $A = k[x, y]$ for a field $k$ and let $M$ be the $A$-module $k[x, y]/(x)$. Given a maximal ideal $m$ in $A$, we write $M_m$ for the localization of $M$ at $m$.
Show that
- $M_m=0$ for $m=(x-1,y)$ and
- $M_m \neq 0$ for $m = (x, y - 1)$.
Facts: $M=k[x, y]/(x) \cong k[y]$, and $M_m=S^{-1}M$.
Why is the localization $M_m$ given with $S=A-m$ and not $S=M-m$? In A&M (page 38) it defines the $S^{-1}A$-module as $S^{-1}M$ with $M$ an A-module in place of the ring A.
For Q1:
$S=A-m=k[x,y]-(x-1, y)$. Any ideal contains $0$ and the only $Ann_m(s)$, $s \in S$ is elements of $k$, but $k$ is a field. Also we have the fact: "Some properties of the localization $R^* = S^{−1}R =$ {$0$} if and only if $S$ contains $0$" from wikipedia. Since $0 \notin S$ how can $M_m=0$? These elements are in the localization $\frac{1}{a}, \frac{1}{x}, \frac{y}{x} \in S^{-1}M = M_m$
For Q2: $S=A-m=k[x,y]-(x, y-1)$. Again $0$ is in $m$ and $k$ is a field, and $k[y]$ is a field extension with no zero-divisors - is that enough to say that $M_m \neq 0$? E.g. $1, y, y^2 \in S$ so $\frac{1}{y} \in S^{-1}M = M_m$
edit: replace $S^{-1}M$ for $S^{-1}A$ in Q1