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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?

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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.)

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  • $\begingroup$ Hm, I still can't see it :( $\endgroup$ – DesmondMiles Apr 26 at 11:50
  • $\begingroup$ Expand the r.h.s. and see if it's possible to identify with the l.h.s. $\endgroup$ – Bernard Apr 26 at 11:52
  • $\begingroup$ 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. $\endgroup$ – Servaes Apr 26 at 12:52

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