Why is $y^2-x^3\in \mathbb{C}[x,y]$ irreducible?

Question

How can I prove that $y^2-x^3\in \mathbb{C}[x,y]$ is irreducible?

Answer 1

It is in $C(y)[x]$ and has no root in $C(y)$.

Note that: $C(y)$ is a field and since it is cubic, we just need to show it has no root.

Answer 2

Hint: Define a map $\phi : \mathbb C[x,y] \to \mathbb C[t]$ by $x \mapsto t^2, y \mapsto t^3.$ Show that this is a ring homomorphism and the kernel is the ideal $\langle y^2-x^3 \rangle.$ Now this is a prime ideal and hence the element $y^2-x^3$ is a prime element.

Added: Let $f(x,y) \in \ker \phi.$ Then we can write $$f(x,y)=f_1(x,y)(y^2-x^3) + f_2(x,y)$$ where the highest degree of $x$ in $f_2(x,y)$ is atmost 2 and that of $y$ is atmost 1 (Why?). So we can write $$ f_2(x,y)=a_0+a_1x+a_2y+a_3xy+a_4x^2+a_5x^2y$$ for some $a_0, a_1, \dots , a_5 \in \mathbb C.$ Now $f_2(t^2,t^3)=0.$ This shows that $a_i=0, \forall i.$

Answer 3

Grade the ring $ \mathbb{C}[x,y]$ by $ \operatorname {deg} x=2,\operatorname {deg}y=3$.
A factorization of $y^2-x^3$ would have homogeneous factors and one of them must have degree $1,2$ or $3$.
This is impossible because the only homogeneous polynomials of these degrees are $ax$ or $by$.