Computing singular points of curves, exercise 5.1 (Hartshorne) I am just trying to cross check my answer as it slightly differs from https://math.berkeley.edu/~reb/courses/256A/1.5.pdf to be sure of any mistake I am making. Here $k$ is an algebraically closed field with ${\mathrm{char}}~k \neq 2$.
5.1(a) The curve $Y \subseteq {\mathbb{A}}^2(k)$ is given by $F(x,y) = x^4 + y^4 - x^2$. The Jacobian
$$
J_P = \Big( ~{\frac {\partial F}{\partial x}}, {\frac {\partial F}{\partial y}}~ \Big) = \Big( 4x^3 - 2x, 4y^3 \Big)
$$
Surely, $y=0$ and if $x \neq 0$, then from $J_P$ we have $2x^2 = 1$ and from the curve $x^2 = 1$. This is absurd and hence $(0,0)$ is the only singular point.
5.1(c) $Y$ is given by $F(x,y) = x^4 + y^4 - x^3 + y^2$.
$$
J_P = \Big( 4x^3 - 3x^2, 4y^3 + 2y \Big).
$$
I checked that $(0,0)$ is always a singular point.
Now solving the equations in ${\mathrm{char}}~k = 7$, we get $(-1, \pm {\sqrt{3}})$ seems to be singular points, where ${\sqrt{3}} \in {\mathbb{F}}_7({\sqrt{3}}) \cong {\mathbb{F}}_{49} \subseteq k$.
Similarly, in ${\mathrm{char}}~k = 13$, we get $(4, \pm {\sqrt{6}})$ seems to be singular points, where ${\sqrt{6}} \in {\mathbb{F}}_{13}({\sqrt{6}}) \cong {\mathbb{F}}_{169} \subseteq k$.
I somehow can't see any mistake here.
 A: Your numbering is a little wacky: I think you might have made a typo when you say "5.4(c)".
Your solution to the first exercise is correct. You may alternately see that if $y=0$, then $x^4-x^2=0$, or $x^2(x^2-1)=0$, while at the same time we need $2x(2x^2-1)=0$, and the only root in common of these equations is $x=0$.
For the second, you are indeed correct that there are singular points which are not the origin in characteristics $7$ and $13$ (and no other characteristics). Here is a slightly different approach which verifies your solution. Take partial derivatives of $x^3=y^2+x^4+y^4$ and set them equal to zero, obtaining the equations $4x^3-3x^2=x^2(4x-3)=0$ and $2y+4y^3=2y(1+2y^2)=0$. Clearly $x=y=0$ satisfies all of these equations for every field. If $\operatorname{char} k\neq 2$, then we may have singular points which aren't the origin if $x=\frac34$ and $y^2=\frac12$. Plugging in to our original equation and simplifying, we see that such a point satisfies our equation iff $91=7\cdot13=0$ in $k$, or $k$ is of characteristic $7$ or $13$, as you have noted.
