$y^2z - x^3$ is irreducible and with one singularity

Question

Give an example for an irreducible cubic curve in $\mathbb{C}\mathbb{P}^2$ with exactly one singular point.

It is easy to check that $y^2z - x^3$ has only [0,0,1] as a singularity. But how to show it is irreducible?

Answer 1

Hint: elementary method.

If it factors, it is a product of homogeneous polynomials. Necessarily, one can obtain a factorisation as a product of a polynomial of (total) degree 2 and a linear polynomial.

On the other hand, it has degree $2$ in $y$, so a factorisation, if any, can be found in the form $$y^2z-x^3=(y+\ell(x,z))(yz+q(x,z)), $$ where $\ell(x,z)$ is linear and $q(x,z)$ is quadratic. Can you deduce a contradiction?

With Eisenstein's criterion:

Consider this polynomial as being in $\mathbf C[y,z][x]$. Then in $y^z-x^3$, $z$ divides all coefficients but the leading coefficient, and $z^2$ doesn't divide the constant term. Hence the polynomial is irreducible in the U.F.D. $\mathbf C[x,y,z]$. (Actually, this is valid for any field of coefficients.)