Canonical divisor of an even-degree hyperelliptic curve

For the smooth curve $y^2=f(x)$ with square-free $f$ of degree $2n$, the differential $dx/y$ has divisorThus the canonical degree is $2n-4$, the genus is $n-1$, and $L(K)$ has basis $1,x,\dots ,x^{n-2}$.

The canonical map of an even-degree hyperelliptic curve isso it identifies the two points $(x,y)$ and $(x,-y)$ of a general fibre and is not an embedding.