Showing that two ideals are equivalent.

Question

For some polynomial ring $k[x,y]$ I want to show that the ideal $I=(x^2-x,y^2-y)$ is equal to the ideal $J=(x,y)\cap(x,y-1)\cap(x-1,y)\cap(x-1,y-1)$.

So far I have shown that $k[x,y]/I$ is isomorphic to $k[x,y]/J$, If I can show that $I\subset J$ them I am done.

Attempt:

An arbitrary element in $I$ is $h=h_1(x^2-x)+h_2(y^2-y)$ where $h_1,h_2 \in k[x,y]$. This gives that $h=h_1(x-1)(x)+h_2(y-1)(y)$.

Now I feel like this should be an element in $J$ since we have the generators of the ideals appearing in it, but I'm not sure how to argue that this is enough to say its an element of $J$.

Answer 1

Set $I_1:=(x,y), I_2:=(x,y-1), I_3:=(x-1,y)$ and $I_4:=(x-1,y-1)$. Clearly $I_i'$s are pairwise comaximal, that is, for each $i\not= j$, $I_i+I_j=R$. Thus, $I_1I_2I_3I_4=I_1\cap I_2\cap I_3\cap I_4=J$. Clearly, $I_1I_2I_3I_4\subseteq(x^2-x,y^2-y)$, and so $J\subseteq I$. Also for each $i$, $I\subseteq I_i$. Thus $I\subseteq I_1\cap I_2\cap I_3\cap I_4=J$, and so $(x^2-x,y^2-y)=(x,y)\cap(x,y-1)\cap(x-1,y)\cap(x-1,y-1)$.

Answer 2

The ideals are the same by the equation $$g(x^2-x) + h(y^2-y) = g(x-1)x + h(y-1)y.$$