Determining whether or not a projective curve is smooth over different fields $K$

Question

Let $n$ be a positive integer, and consider the projective plane cubic $\mathscr C_n$ given by the vanishing set of $$F(X,Y,Z):=X^3+Y^3+n^2Z^3 - 3n^2XZ^2$$ over a field $K$.

I want to determine for what fields $K$ this curve is smooth. Here is my progress so far.

We have \begin{align*} \frac{\partial F}{\partial X}=3(X^2 -n^2Z^2), \qquad\frac{\partial F}{\partial Y} = 3Y^2, \qquad \frac{\partial F}{\partial Z} = 3(n^2Z^2 - 2n^2XZ). \end{align*}

Now consider fields of characteristic zero first. If $(x:y:z)$ is a singular point, then the $Y$-derivative implies that we have $y=0$. The $X$-derivative implies we have $z=\pm\frac1nx$, which implies that $x\neq0\neq z$. Finally, then $Z$-derivative implies that we have $3x^2\pm6nx^2=0$, i.e., $1\pm 2n=0$, which has no solutions for integral $n$. Thus $\mathscr C_n$ is smooth for fields of characteristic zero.

Now suppose $K$ has characteristic $p$. If $p=3$, then clearly each derivative vanishes, so the curve is not smooth.

Is my reasoning so far correct? I'm not sure how to tackle the cases for $p\neq 3$.

Answer 1

Yes, this work looks fine so far. To deal with the positive characteristic case with $p\neq 3$, you'll need to split things up by whether $p|n$ or not.

If $p$ divides $n$, the curve is given by $X^3+Y^3$ which is singular at $[0:0:1]$.

If $p$ does not divide $n$, then the same logic from the characteristic zero case implies that any singular point has $Y=0$ and $Z=\pm\frac1nX$. Plugging this in to the $Z$-derivative, we see that we need to have $3(n^2Z^2 \pm 2n^3Z^2)=0$, or $3n^2Z^2(1 \pm 2n)=0$. As $3n$ is not divisible by $p$ and $X,Y$ can't simultaneously be zero because $Y$ is already zero, this means the only points where the Jacobian vanishes are $[\pm n:0:1]$ iff $1\pm 2n=0$.

Now we need to check if these points are on our curve. Plugging in, we get that our equation evaluates to $\pm n^3+n^2\mp 3n^3 = n^2(1\mp2n)$ As it's not possible for $1\pm 2n=0$ and $1\mp 2n$ to be zero simultaneously, we see that the Jacobian has positive rank at every point when $p$ and $n$ are coprime and $p\neq 3$.

To sum it up, the curve is singular iff $p=3$ or $p|n$.

Answer 2

Put all the partials equal to 0. If the characteristic p divides n, (0:0:1) is a singular point. Otherwise, all three partials vanish only at (1:0:2), provided 2n is congruent to 1 or -1 modulo p. But then (1:0:2) is not on the curve, so there are no singular points at all. Conclusion: no singular points unless p divides n, in which case (0:0:1) is the only singular point.