This question is based on an example from Mel Hochster's notes on commutative algebra.
Let $k$ be a field. Consider $R=k[X(1-X),Y,XY]\subset S=k[X,Y]$. It is straightforward to show that this is an integral extension of integral domains.
Now consider the ideals $(1-X,Y),(X)\in \mathrm{Spec}(S).$ It is remarked that the ideal $(1-X,Y)$ contracts to the ideal $(X(1-X),Y,XY)$ and the ideal $(X)$ contracts to $(X(1-X),XY)$.
But I have trouble convincing myself why this is the case. Any hint/help will be greatly appreciated. Thanks in advance.
In the first case notice that $(x(1-x),y,xy)$ is a maximal ideal of $R$ and $(x(1-x),y,xy)\subseteq(1-x,y)S\cap R$, so they must be equal.
For the second part I would do the following.
Let $xf(x,y)=g(x(1-x),y,xy)\in xS\cap R$. Then write $g(x(1-x),y,xy)=\sum a_{ijk}[x(1-x)]^iy^j(xy)^k$. Now rewrite $$g(x(1-x),y,xy)=\sum_{i>0}a_{ijk}[x(1-x)]^iy^j(xy)^k+\sum_{k>0}a_{0jk}y^j(xy)^k+\sum a_{0j0}y^j.$$ One can see that the first two sums belong to the ideal of $R$ generated by $x(1-x)$ and $xy$, while the third sum must be zero since in $S$ it must be a multiple of $x$.