The classical book of Walker on algebraic curves described a method of computing the genus of curves with non-ordinary singularities by transforming such a curve into a birational one with only ordinary singularities. The result then follows from the degree-genus formula.
For positive integers $a > b \geq 2$, I have a curve $$F: Y^a+Z^a-X^bZ^{a-b}=0,$$ written this way so that $P=[1:0:0]$ is the only singularity on $F$, which happens to be non-ordinary, of multiplicity $a-b$. Also, we check that the points $[0:1:0],[0:0:1]$ do not lie on $F$.
The plane $Z=0$ intersects $F$ at $P$ with multiplicity $a$, and thus we make the change of variables $$X_0 = X, X_1=Y, X_2=Z-Y$$ to obtain the curve $$F: X_1^a+(X_1+X_2)^a-X_0^b(X_1+X_2)^{a-b}.$$ One easily verifies that both $X_1=0,X_2=0$ now intersects $F$ at $P$ with multiplicty $a-b,$ which is what we want. Now following the transformation of Walker, we have $$G(Y_0,Y_1,Y_2) = F(Y_1Y_2,Y_0Y_2,Y_0Y_1).$$ This has a factor of $Y_0^{a-b}$, which, upon factoring, gives us $$G(Y_0,Y_1,Y_2)=Y_0^{a-b}[Y_0^bY_2^a+Y_0^b(Y_1+Y_2)^a-(Y_1Y_2)^b(Y_1+Y_2)^{a-b}] = Y_0^{a-b}H(Y_0,Y_1,Y_2).$$ The new curve $H$ has degree $a+b$ and is birational to $F$, and it has ordinary singular points $[1:0:0],[0:1:0],[0:0:1]$ of multiplicities $a,b,b$ respectively. The intersection of the plane $Y_0=0$ with $H$ will yield $$(Y_1Y_2)^b(Y_1+Y_2)^{a-b} = 0,$$ which tells us that this plane is tangent to the singular point $Q=[0:1:-1]$ on $H$, and the multiplicity of the intersection is $a$ (with $a-b$ coming from the above equation, and $b$ coming from the multiplicity of $Q$ on $H$). By the method of Walker, it seems like the original non-ordinary singular point $P$ on $F$ only "splits" into this point, which is another non-ordinary singular point, and I am nowhere near performing any genus calculations.
Did I misunderstand something? How to I get around this phenomenon?
