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A point on $F=0$ where all first partial derivatives vanish. For a reduced hypersurface this is the Jacobian criterion; a nondegenerate quadratic leading part defines an ordinary double point.

A nodal crossing locally consists of two complex branches meeting transversely and is not a one-dimensional complex manifold at the intersection.

For a plane algebraic curve this crossing is an ordinary double point, with two distinct tangent directions.

Take the plane quartic with exactly two nodesPut $d=X^2-Z^2$. Its partial derivatives areOn $Y=0$, the equation gives $X^2=Z^2$, producing exactly $P_{+}=[1:0:1]$ and $P_{-}=[-1:0:1]$, and their gradients vanish.

There are no singular points with $Y\ne 0$. Normalize $Y=1$. The derivative equations give $X=4dZ$ and $(16d^2+1)Z=0$. If $Z=0$, then $X=0$ and $F_Y=4\ne 0$, so $Z\ne 0$ and $d^2=-1/16$. The remaining equation gives $XZ=-2$, hence $Z^2=-1/(2d)$. But $d=X^2-Z^2=-2Z^2=1/d$, forcing $d^2=1$, a contradiction. Thus the singular set is exactly $\{P_{+},P_{-}\}$.

In the chart $Z=1$ near $P_{\epsilon }$, put $X=\epsilon +u$, $Y=v$ with $\epsilon =\pm 1$. The lowest-degree part is $4u^2+\epsilon v^2$, which splits into two distinct complex linear factors. Each singularity is therefore an ordinary double point, with two smooth transverse branches.

It remains to prove irreducibility. A repeated polynomial factor would make every point on that component singular, contradicting the finite singular set, so $F$ is reduced. If it were reducible, partition its irreducible factors into two coprime factors of degrees $r$ and $4-r$, $1\le r\le 3$. Every intersection of the two components would be singular on $C$, hence one of the two ordinary double points. At an ordinary double point shared by them, the two branches are transverse, so its intersection multiplicity is one. Their total intersection multiplicity would therefore be at most two. The Bézout theorem instead makes it $r(4-r)\ge 3$. This contradiction proves the displayed quartic is irreducible and has exactly two singular points.

Write $F=xyz+yzw+zwx+wxy$. A projective hypersurface singularity must have $F=0$ and all four first partial derivatives zero. If all coordinates were nonzero, writeOn $F=0$, differentiation gives $F_x=-xyzw/x^2\ne 0$, a contradiction. Thus some coordinate is zero; by symmetry take $w=0$. The surface equation gives $xyz=0$, so a second coordinate is zero, say $z=0$. Then $F_z=xy$, so a third coordinate is zero. The remaining four projective possibilities indeed have every first derivative zero. Therefore all singular points areOver the usual characteristic-zero ground field, each is an ordinary double point: near $[1:0:0:0]$, the affine equation has nondegenerate quadratic part $yz+zw+wy$, and symmetry treats the others.

Homogenization gives $F=X^4+X^2{Y}^2+Y^2{Z}^2+Z^4$, withIf the characteristic is not $2$, a singular point with $Y=0$ would force $X=Z=0$, impossible. If $Y\ne 0$, then $X^2+Z^2=0$. If either $X$ or $Z$ vanishes, both vanish, giving $[0:1:0]$. Otherwise the other two derivative equations require $Y^2=-2X^2=-2Z^2$, whence $X^2=Z^2$ and $2X^2=0$, again impossible. ThusIn the chart $Y=1$, the quadratic term is $X^2+Z^2$, the product of two distinct linear factors over $k$, so this is an ordinary double point.

In characteristic $2$,All first derivatives of this equation vanish. The equation defines a nonreduced scheme, singular everywhere; it no longer satisfies the irreducible/reduced hypersurface premise. If “curve” means the associated reduced variety, it is the union of two distinct lines, and only their intersection $\boxed{[1:0:1]}$ is singular. These two interpretations must not be conflated.

Over the complex numbers the displayed projective plane curve has precisely two singular points, $[1:0:1]$ and $[-1:0:1]$. At either, with $X=\pm 1+u$, $Y=v$, $Z=1$, 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.