Show that $x$ and $y$ are irreducible in $K[x,y]/(y^2 - x(x+1)(x-1))$ where $K$ is an algebraically closed field

Question

I've already showed that for the affine smooth curve $X = V(y^2 - x(x+1)(x-1))$ the coordinate ring $K[X]$ is a domain and so what I've tried to show that the ideals $(x)$ and $(y)$ are prime ideals (then $x$ and $y$ are irreducible) but this hasn't worked as $$\frac{K[X]}{(y)} = \frac{K[x]}{(x(x+1)(x-1))}$$ which is clearly not a domain. I'm not sure what else to do.

Answer 1

As with many exercises involving a quadratic extension, there's a solution via the norm map. Any element of $f\in k[X]$ can be written $gy+h$ for $g,h\in k[x]$, and we define the norm of $f$ to be $N(f)=(gy+h)(gy-h)=g^2y^2-h^2=x(x+1)(x-1)g^2-h^2$, the determinant of the multiplication matrix by $f$ acting as a $k[x]$-automorphism of the rank-2 free $k[x]$-module $k[X]$ with basis $1,y$. It's a standard check that this is multiplicative and that $N(a)$ is a unit iff $a$ is, so if we have a factorization $a=bc$, then $N(a)=N(b)N(c)$, and if this is a non-trivial factorization then $N(b)$ and $N(c)$ must be proper divisors of $N(a)$.

In the case $a=x$, $N(x)=x^2$, and $N(-)$ cannot output anything of degree one, so $x$ is irreducible. In the case $a=y$, $N(y)=x(x+1)(x-1)$, so again the observation that $N(-)$ cannot output anything of degree one gives that $y$ is irreducible.