Irreducible Decomposition of a Vareity

Question

How would one decompose the variety $V=(x+y)^s$ into irreducible components? Would it be just $V=(x+y) \cup (x+y) \cup ... \cup (x+y)$, s number of times? And this is over $\mathbb{C}$.

Answer 1

If you mean $V=V((x+y)^s) \subseteq \mathbb C^2$, then we alyways have $V(I)=V(\sqrt I)$ (since $f(x)^s=0 \Rightarrow f(x)=0$), so that actually $V=V(x+y)$. So as a variety, $V$ is in fact irreducible.

However, as a scheme, $V$ is still irreducible, but it is "fat". For a scheme theorist, $V$ is like the line $y=-x$ $s$ times.