Let V be a set of zeroes of polynomials in ideal generated by $x^2 - yz$ and $x - xz$. I need to prove that it has three irreducible components.
My thoughts: $x - xz = 0$, then $x = 0$ or $z = 1$ so we have two planes intersecting zeroes of x^2 - yz.
If $x = 0$ then $x^2 - yz$ looks like $yz = 0$ and, again, $y = 0$ or $z = 0$
If $z = 1$ then $x^2 - yz$ looks like $x^2 - y = 0;\; y = x^2$
So, intuitively, we should have three following components:
Straight line $x = 0;y=0;z=t$
Straight line $x = 0; z=0; y = t$
Parabola $z=1; y = x^2; x=t$
But I have hard time understanding how this arguments works when we would write it with dual notation: showing the same fact via working with rings and ideas. And i would like to get help to understand it, because i feel that in more difficult situation such crude argument would not be enough.
