Show that $y-x^2 \notin \langle x(x-1), (x-1)y\rangle$ in $\mathbb{C}[x,y]$

Question

I am in the middle of an exercise and I am stuck in a final step in which I want to show that $y-x^2 \notin \langle x(x-1), (x-1)y\rangle$ in $\mathbb{C}[x,y]$. My first thought was that $x-1$ does not divide $y-x^2$, but I have no idea how to prove this properly.

Answer 1

You can formalize this intuition by using that $$\langle x(x-1),(x-1)y\rangle$$ is the set of polynomials that can be written as $$x(x-1)P(x,y)+(x-1)yQ(x,y)$$ for some polynomials $P,Q$. However, any such polynomial can be written as $$(x-1)\big(xP(x,y)+yQ(x,y)\big).$$ Can you finish from here?

Answer 2

I will use $(~~~~)$ to denote an ideal.

If $y-x^2$ were in the ideal, then $V(y-x^2) \supset V(x(x-1), (x-1)y)$. However, a point such as $(1,0)$ is in $V(x(x-1), (x-1)y)$ but not in $V(y-x^2)$. This is a contradiction.