Given $I=\langle xy, xz+z(y^2-z^2)\rangle$, prove that $I=\langle x, z(y^2-z^2)\rangle \cap \langle y, xz-z^3)\rangle $.

Question

This is Exercise 3c. from Chapter 9, Section 7 of Ideals, Varieties, and Algorithms by Cox et al.

Given $I=\langle xy, xz+z(y^2-z^2)\rangle$, prove that $I=\langle x, z(y^2-z^2)\rangle \cap \langle y, xz-z^3)\rangle $.

I proved it using Groebner basis algorithm.

But the problem also says that there is an elementary argument. So I tried:

The direction $\subset$ is trivial. To show the other direction,

$$\text{Let } f=p_1 x+p_2 z(y^2-z^2)=q_1 y+q_2(xz-z^3)\\ f=(p_1-p_2z+p_2z)x+p_2z(y^2-z^2)\\ =(p_1-p_2z)x+p_2(xz+z(y^2-z^2))$$

For this to be in $I$, we need that $p_1-p_2z$ has a factor $y$.

Or from the other expression,

$$f=(q_1-q_2yz+q_2yz)y+q_2(xz-z^3)\\ =(q_1-q_2yz)y+q_2(xz+z(y^2-z^2))$$

We need that $q_1-q_2yz$ has a factor $x$.

I don't know how to continue. Any help will be greatly appreciated!

Edit

Using $q_1=xu+zv$ and $p_1=sy+tz$, we get $$f=uxy+vyz+q_2(xz-z^3)\\ =sxy+txz+p_2z(y^2-z^2)$$

I don't see how to put this into my argument above. Also since we don't know the nature of $u,v,s,t,p,q$, we cannot equate anything. So how to continue?

Edit 2 Now $uxy-sxy\in \langle z\rangle$, so we let $u-s=wz$. This leads to $$f=(s+wz)xy+vyz+q_2(xz-z^3)\\ =sxy+txz+p_2z(y^2-z^2)$$ So $$wxy+vy+q_2(x-z^2)=tx+p_2(y^2-z^2)$$

I believe we want to show either $wx+v=yq_2$ or $t=p_2$.

Following the same idea, we can let $q_2(x-z^2)-tx+p_2z^2=Ay$. This gives us $wx+v+A=p_2y$.

Answer 1

We prove that $$\langle xy, xz+z(y^2-z^2)\rangle=\langle x, xz+z(y^2-z^2)\rangle\cap\langle y, xz+z(y^2-z^2)\rangle.$$

This reduces easily to show that $xp\in\langle y, xz+z(y^2-z^2)\rangle\implies p\in\langle y, xz+z(y^2-z^2)\rangle$.
Notice that $\langle y, xz+z(y^2-z^2)\rangle=\langle y, z(x-z^2)\rangle$.
Now write $xp=yu+z(x-z^2)v$, and since $xp\in\langle y, z\rangle$ we get $p\in\langle y, z\rangle$. Thus $p=ya+zb$, and from $xp=yu+z(x-z^2)v$ we get $xya+xzb=yu+z(x-z^2)v$, that is, $y(xa-u)=z[(x-z^2)v-xb]$. Then $z\mid xa-u$, and therefore $u=xa+zw$. Now we get $xp=xya+yzw+z(x-z^2)v$, so $x(p-ya)\in\langle yz, z(x-z^2)\rangle$. Since $p-ya=zq$ we get $xq\in\langle y, x-z^2\rangle$. The ideal $\langle y, x-z^2\rangle$ is prime and this shows that $q\in\langle y, x-z^2\rangle$, and finally we get $p\in\langle y, z(x-z^2)\rangle$.