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

Nat
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3 Answers3

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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 $a_i=0, \forall i.$

Krish
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    (And of course it's a prime ideal since the homomorphic image is a domain.) – anon Mar 27 '15 at 13:44
  • @anon: Yes! I deliberately left out that point for OP to find out by himself/herself. Well, this is not a complete answer, just a hint!! ;) If OP has any more question, I'll be happy to help as much as possible. – Krish Mar 27 '15 at 13:48
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    My initial problem was indeed to prove that $Ker(\phi)=<y^3-x^2>$, in particular I have trouble showing Kerϕ⊆<y^3−x^2> (the other inclusion is fine...) – Nat Mar 27 '15 at 16:23
  • @Nat: I've added an explanation. See if it's fine. Please feel free to ask if you have any more doubts. – Krish Mar 27 '15 at 17:03
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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 $by$.

Attila Smith
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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.

vudu vucu
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    And don't forget to mention that having no root is equivalent to being irreducible for cubic polynomials. That's important to mention since it's not true in general for polynomials; the same reasoning cannot be used for your generalization (even though I suspect the conclusion is still correct). Also since this fact holds for quadratics you might as well have considered $\Bbb C(x)$ instead. By the way, do you know why having no root and being irreducible are equivalent for quadratics and cubics (but not higher degrees)? – anon Mar 27 '15 at 13:41
  • you are right.. – vudu vucu Mar 27 '15 at 13:43
  • How can we justify that there are no roots in $C(y)$? – Anakhand Dec 05 '23 at 18:49