Homogenization gives $F=X^4+X^2Y^2+Y^2Z^2+Z^4$, with
$$
F_X=4X^3+2XY^2,\qquad F_Y=2Y(X^2+Z^2),\qquad F_Z=2ZY^2+4Z^3.
$$
If the characteristic is not $2$, a singular point with $Y=0$ would force $X=Z=0$, impossible. If $Y\ne0$, 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. Thus
$$
\boxed{\operatorname{Sing}(X)=\{[0:1:0]\}\quad(\operatorname{char}k\ne2).}
$$
In 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 $2$,
$$
\boxed{F=(X^2+XY+YZ+Z^2)^2=((X+Z)(X+Y+Z))^2.}
$$
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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 2 / 22I / c / Solution

= Node of a plane algebraic curve
{synonym}

For a reduced plane <algebraic curve>, an ordinary double point is a point whose lowest nonzero local homogeneous term has degree $2$ and factors into two distinct linear factors over an <algebraic closure>. The two factors give distinct tangent directions. In characteristic not $2$, $x^2+y^2+\text{higher terms}$ is an example.

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 Incoming links: Ordinary double point