(hyper) elliptic curve in characteristic two and the Jacobian criterion Let $k$ be a field of characteristic two and let $E$ be a curve given by 
$$
y^2=x*(x+1)*(x^2+x+1)*(x^3+x+1)\quad\text{or}\quad y^2=f(x)
$$
Now we have $dy^2/dy=2y=0$ and consider the Jacobian criterion:
$$
J=(f'(x),\:\:\:0\:)
$$
what does this imply for $E$? Where is this curve regular?
I thought that $E$ is regular except at the zeros of $f'$. Is this true?
 A: One must be careful here: first, one has to distinguish between regular points of the curve and smooth points. $x\in E$ is regular, if the local ring $O_{E,x}$ is regular, that is a discrete valuation ring. $x\in E$ is smooth if $x$ is regular and remains regular after every purely inseparable base change $E\times_k\ell$ of the curve. So regularity and smoothness coincide if $k$ is a perfect field.
The Jacobian criterion (as You apply it) gives the smooth points of $E$, provided $k$ is a perfect field. Under this additional assumption you are right that the regular points $(a,b)\in E$ are precisely the ones satisfying $f^\prime(a)=0$.
Consider the point $(0,0)\in E$: its local ring is $k[x,y]_{(x,y)}$, which is a local extension of the local ring $k[x]_{(x)}$. Except $x$ itself all the polynomial factors appearing on the right hand side of the defining equation of $E$ are units in $k[x]_{(x)}$ and therefore in $k[x,y]_{(x,y)}$. Hence $(x,y)k[x,y]_{(x,y)}=yk[x,y]_{(x,y)}$, because $y^2=ux$ with a unit $u$ of $k[x,y]_{(x,y)}$. We have shown that the maximal ideal $(x,y)k[x,y]_{(x,y)}$ of the local ring in $(0,0)$ is generated by one element, thus it is a regular local ring and $(0,0)$ is a regular point.
