TITLE: Irreducible components of $I = (x^3 - y^3, x^2 - y^2 + 2x - 2y)$.
QUESTION [5 upvotes]: I am trying to find the irreducible components of $V(I),$ where $I = (x^3 - y^3, x^2 - y^2 + 2x - 2y)$.

So far I've factorized as follows: 
 $$x^2 - y^2 + 2x - 2y = (x-y)(x + y + 2)$$
$$x^3 - y^3 = (x -y)(x - \omega y)(x - \omega^2 y)$$
From this I've got that we must have the following equal to zero: 
$$x + y +2, -\omega^2 y + 2 $$
$$x + y + 2, -\omega y+ 2$$
or 
$$x-y.$$
However I'm not sure if this is everything I need to do. Do I now just express them as subsets of $V = V(I)$, such that the union of them is equal to $V$?

REPLY [2 votes]: We have
$$x^2 - y^2 + 2x - 2y = (x-y)(x + y + 2), \quad x^3 - y^3= (x -y)(x^2+xy+y^2)$$
So, $(x,y) \in V(I)$ iff $x=y$ or $x + y + 2=0=x^2+xy+y^2$.
The last equation has two solutions: $x=-1 \pm \sqrt3i, y=-1 \mp \sqrt3i$.
Therefore, $V(I)$ has three irreducible components: a line and two isolated points.