This question was part of my class notes in commutative algebra and it has been left to solve by itself.
Let $K[X,Y]$ be the polynomial ring over a field $K$, $A'=\left< X^2 ,XY\right> \subseteq K[X,Y]$ and $S=K[X,Y]\setminus\left<X\right>$. Then show that $A'\nsubseteq A'S^{-1} (K[X,Y]) \cap K[X,Y]$.
I think $A'$ should always be a subset of RHS as $A'$ is an ideal but I am asked to prove otherwise.
Can, you please tell what's wrong with my argument and help with proof?
Thanks!
As pointed out by @user26857, the question was probably supposed to be:
Show that $$A'\subsetneq A'S^{-1} (K[X,Y]) \cap K[X,Y].$$
We know $$A'\subseteq A'S^{-1}(K[X,Y]),$$ as $A'S^{-1}(K[X,Y])$ is simply the ideal in the localisation $S^{-1}(K[X,Y])$, generated by the elements of $A'$.
We also know $A'\subseteq K[X,Y]$, as we were given that $A'$ is an ideal in $K[X,Y]$ generated by certain elements (namely $X^2,XY$).
Thus $$A'\subseteq A'S^{-1} (K[X,Y]) \cap K[X,Y].$$
On the other hand $X$ is a linear combination of $X^2$ and $XY$ in the localisation $S^{-1}(K[X,Y])$: $$(X^2+XY)(X+Y)^{−1}=X$$
Thus $X\in A'S^{-1} (K[X,Y])$. Clearly $X\in K[X,Y]$. Combining these statements we have: $$X\in A'S^{-1} (K[X,Y]) \cap K[X,Y].$$
However $X\notin A'$ as it is a degree one polynomial in $X,Y$. We can conclude that:
$$A'\subsetneq A'S^{-1} (K[X,Y]) \cap K[X,Y].$$