= Plane quartic with exactly two nodes
{title2=$(X^2-Z^2)^2+XZY^2+Y^4=0$}
Over the complex numbers the displayed <projective plane curve> has precisely two singular points, $[1:0:1]$ and $[-1:0:1]$. At either, with $X=\pm1+u$, $Y=v$, $Z=1$, the leading quadratic term is $4u^2\pm v^2$, so they are <ordinary double points>. The derivative equations in the chart $Y=1$ have no solution: writing $d=X^2-Z^2$ forces both $d^2=-1/16$ and $d^2=1$. The <Bézout theorem> proves irreducibility, since a factorization into degrees $r,4-r$ would require at least three intersections, whereas the only possible component intersections are the two transverse <ordinary double points>.
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