Localization at $\mathfrak p$, extension and contraction of an ideal 
Consider $R= K[X,Y]/(XY)$ and $I$ be the ideal generated by $[X^n]$ in $R$ for  $n>2$. Then prove $I^{ec} \neq I$ where extensions and contractions of $I$ are taken considering the map $R \to R_{\mathfrak p}$ where $\mathfrak p=([X])$ and $R_{\mathfrak p}$ denotes localization of $R$ at $\mathfrak p$. 

My try: We've $I=\{[f(X)] \mid \deg(f)>2\}$ and $R^c=S= \{[g(Y)]\}$. So, $I^e=\{\frac{[f(X)]}{[g(Y)]} \mid \deg(f)>2\}$. But then clearly $I^{ec}=I$, isn't it ?
 A: The ideal $I^{ec}$ is bigger than the ideal $I$. This is why:
Take an element $f \in I$. This element is by definition of $I$ divisible by $X^3$, which means it's of the form $X^3g$ for some $g\in R$. Now, in $I^e$, the element $f$ becomes $\frac{f}{1} = \frac{X^3g}{1}$. I claim that this is equal (in $R_p$) to $\frac{0}{1}$. By definition of localization, two elements $\frac ab$ and $\frac cd$ are equal if there is an element $s$ in the multiplicative system such that $s(ad - bc) = 0$ In our case, we're looking for an $s \in R\setminus [X]$ such that
$$
s(f\cdot 1 - 1 \cdot 0) = 0\\
sf = 0\\
sX^3g = 0
$$
But $Y$ is in the complement of $(X)$, so it's a valid $s$. We see that since $XY = 0$ in $R$, this makes $YX^3g = 0$. Since this works for any $f \in I$, we've shown that $I^e = \left(\frac 01\right)$ is the zero ideal in $R_p$. Consequently, the ideal $I^{ec}$ is the kernel of the localization, which is $(X)$.
The moral is: Whenever your localization inverts a zero divisor (in this case $Y$), all elements that are killed by that zero divisor becomes $0$ in the localization. It is, in fact, the only way localization can fail to be injective.
