Check if singularity of curve is node in positive characteristic Given a plane curve it is easy to check whether a point is singular by using the Jacobi criterion. However, I am stuck with checking whether it is a node or worse, especially in the case of positive characteristic. Is there are general way to check that without computing the completed local ring every time, we can assume the field is algebraically closed, if this helps. Two concrete examples in characteristic 2 that appeared in my work are:

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*$y^2-x(x-1)(x-\lambda)$ and the singularity at $x=\sqrt\lambda$, which is smooth in characteristic $0$

*$y(y-x^2+1)$ and the singularity at $(1,0)$, which has two nodal singularities $(1,0)$ and $(-1,0)$ in characteristic $0$.

If there is no general easy way, maybe one can illustrate helpful techniques for these examples?
 A: At least for plane curves, you can tell based on examining the lowest-degree term of the defining polynomial at that point. To be specific, if $(a,b)$ is a point on the curve $V(f)$, write $f=\sum f_r$ for $f_r$ homogeneous of degree $r$ in $(x-a)$ and $(y-b)$. We wish to examine the smallest $r$ so that $f_r$ is nonzero: the case of a node is when $r=2$ and $f_2$ splits as a product of distinct linear factors. (This is equivalent to what Sergey suggests in the comments, but dressed up slightly differently.)
Let's investigate this with your examples:

*

*$y^2-x(x-1)(x-\lambda)$: expand $x(x-1)(x-\lambda)$ as $x^3-(1+\lambda)x^2+\lambda x$, so our equation can be written as $(y-?)^2+(x-\sqrt{\lambda})^3+(1+\sqrt{\lambda})(x-\sqrt{\lambda})^2$. The degree-two portion is of the form $c(y-?)^2+d(x-?)^2$, which is a square and therefore not a node.


*$y(y-x^2+1)$: writing $-x^2+1=(x+1)^2$ the equation becomes $y^2-(x+1)^2y$ which is again not a node. (Compare in characteristic not two, where we get $x^2-1=(x-1)^2+2(x-1)$ and therefore $y(y-x^2+1)=y(y-(x-1)^2-2(x-1))=y^2-2(x-1)y-(x-1)^2y$, which is a node as $y^2-2(x-1)y=y(y-2(x-1))$ is a product of distinct linear factors.)
