# Irreducibility of polynomial $x^3-y^2$

I was told that

$$x^3-y^2$$ is irreducible in $$\Bbb C[x,y]$$.

But I cannot really give a sounding argument. I supposed that it may be factored as $$g_1, g_2$$, and considered those monomials divisible by $$y$$. But possible cancellations makes it hard.

EDIT: Thoughts:

Suppose $$(x^3-y^2)$$ is reducible in $$k[y][x] \cong k[x,y]$$, then it would factorize as a polynomial in $$x$$ over the ring $$k[y]$$. As the degree is $$3$$, there must exists a monomial factorization. Hence there is a polynomial solution $$x=P(y)$$, which is impossible.

• Your thoughts are correct. Another way is to factorize it in $k[x][y]$ and observe that $4x^3$ is not a square. – Saucy O'Path Oct 12 '18 at 23:04

## 1 Answer

A variant :

It is equivalent to show $$y^2-x^3$$ is irreducible. If it were not, it would factor as the product of two linear factors in $$y$$: $$(y-p(x))(y-q(x))$$. Expanding, we obtain by identification $$p(x)q(x)=-x^3, \qquad p(x)+q(x)=0,$$ whence $$p(x)^2=x^3$$, which is impossible for degree reasons.