I have to prove that in $\mathbb{C}[x, y]$ we have the equality
$$ (x^3-x^2, x^2y-x^2, xy-y, y^2-y)=(x^2, y) \cap (x-1, y-1).$$
I have proven the inclusion $\subset$ proving that every generator of the LHS can be written as a linear combination both of elements of $(x^2, y)$ and of elements of $(x-1, y-1)$ (e.g.: $x^2y-x^2=x^2\cdot(y-1)$), but I am stuck with the reverse inclusion. Any suggestion?
Take an arbitrary element of the RHS and subtract linear combinations of the generators on the left hand side to simplify your element: the first and last generator on the LHS are enough to reduce your element to something of the form: $$ax^2+by+cxy+dx^2y,$$ as being an element of the RHS implies there is no constant or $\lambda x$ term.
Then the third generator gets rid of the $cxy$ term and the second gets rid of the $dx^2y$ term, leaving $$a'x^2+b'y.$$
Substituting in $x=1, y=1$ we get $0$ as we have an element of the RHS. Thus $a'+b'=0$, so we are left with a scalar multiple of $x^2-y$.
But $$x^2-y=(x+1)(xy-y)-(x^2y-x^2) $$
We conclude that any element of the RHS can be broken down to a linear combination of the 4 generators on the LHS.