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

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?

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.)

• Hm, I still can't see it :( – DesmondMiles Apr 26 at 11:50
• Expand the r.h.s. and see if it's possible to identify with the l.h.s. – Bernard Apr 26 at 11:52
• Alternatively, the cubic has degree $1$ in $z$ and so any factorization must be of the form $$y^2z-x^3=(a(x,y)z+b(x,y))\cdot c(x,y).$$ Then $c(x,y)$ divides both $y^2$ and $x^3$, so it is a unit and hence the cubic is irreducible. – Servaes Apr 26 at 12:52