Primary decomposition of $I=\langle xy,x^3-x^2,x^2y-xy\rangle$

Question

I should prove that:

$I=\langle xy,x^3-x^2,x^2y-xy\rangle=\langle x\rangle \cap \langle x-1,y\rangle \cap \langle x^2,y\rangle$, in $k[x,y]$, where $k$ is a field.

Well, $$xy=y\cdot x =x\cdot y\in \langle x\rangle \cap \langle x-1,y\rangle \cap \langle x^2,y\rangle. $$

In the same way, $$x^3-x^2=(x^2-x)\cdot x=x^2\cdot (x-1)=(x-1)\cdot x^2\in \langle x\rangle \cap \langle x-1,y\rangle \cap \langle x^2,y\rangle $$ and $$x^2y-xy=(xy-y)\cdot x =(x^2-x)\cdot y\in \langle x\rangle \cap \langle x-1,y\rangle \cap \langle x^2,y\rangle.$$

So, $I\subseteq \langle x\rangle \cap \langle x-1,y\rangle \cap \langle x^2,y\rangle $.

However, I am not sure how to proceed at reverse.

Many thanks in advance!

Answer 1

Suppose $f$ is an element of $\langle x\rangle \cap \langle x-1,y\rangle \cap \langle x^2,y\rangle$.

Since $f \in \langle x\rangle$, we have $f = xf'$ for some $f'$.

Since $xf' = (x - 1)f' + f' \in \langle x - 1, y\rangle$, we have $f' \in \langle x - 1, y\rangle$. Thus write $f' = (x - 1)g + yh$.

Since $f = xf' = x(x - 1)g + xyh \in \langle x^2, y\rangle$, we have $xg \in \langle x^2, y\rangle$.

Hence there are $u, v$ such that $xg = x^2u + yv$. Rewrite this as $x(g - xu) = yv$.

Since $k[x, y]$ is a UFD, we know that there exists $p$ such that $g - xu = yp$ and $v = xp$.

Thus in the end, $g = xu + yp$ and $f = x(x - 1)g + xyh = (x^3 - x^2)u + (x^2y - xy)p + xyh\in\langle xy,x^3-x^2,x^2y-xy\rangle$.