Affine Varieties

Question

I came across a task from an old exam I don't know how to solve. The task is the following. Let $V=\{(x,y) \in \mathbb{C}^2 : x^4=y^4\}$. 1) Find the irreducible Components of $V$. 2) What is the dimension of $V$? I struggle a lot with affine varieties actually, so any help is greatly appreciated. What I know is that $V=Z(x+iy)\cup Z(x-iy)\cup Z(x+y)\cup Z(x-y)$ and I think that a subset $A \subseteq k^n $ with a field $k$ is irreducible exactly when $I(A)$ is a prime ideal with $I(A)$ being the set of all polynomials that have every $a \in A$ as a root. Thank you in advance for any help!

Answer 1

All of the ideals $x \pm y$ and $x \pm iy$ are prime. For example, $\mathbb C[x, y] / (x - y) \cong \mathbb C[t]$ via the isomorphism $x \mapsto t$ and $y \mapsto t$. This gives us (a).

For (b), we have the following chain of ideals, $$ 0 \subset (x^4 - y^4) \subset (x - y) \subset (x, y). $$

Since the $\dim \mathbb C[x, y] = 2$, we have $1 \le \dim V \le 2$. The dimension of $V$ cannot be $2$, for otherwise we would have a chain of ideals in which each $\mathfrak p_i$ is prime, $$ 0 \subset (x^4 - y^4) \subset \mathfrak p_1 \subset \mathfrak p_2 \subset \mathfrak p_3. $$

This would contradict the fact that $\dim \mathbb C[x, y] = 2$. We conclude that $\dim V = 1$.