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May 3, 2018 at 21:30 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Apr 3, 2018 at 19:11 answer added Xavier49 timeline score: 2
Apr 2, 2018 at 7:00 comment added Xavier49 @Mohan Thanks! Yes, but the "easily" part is not so easy for me, because I do not know how to tell when too singularities are 'the same', I mean I suppose there are some equivalence relations but I do not know which they are. By example one has the singularity $(x-y^a)(x+y^b)=0$, with say $a \leq b$. By expanding that polynomial, my intuition tells me that this is a $A_{2a}$ singularity, but I would like to understand rigorously why. More generally with e.g. $(x+vy^a)(x+uy^b)=0$, where $u,v$ are non equal constants.
Apr 1, 2018 at 18:35 comment added Mohan At any point $P\in C_1\cup C_2$, in the completion of the local ring of the surface, you can easily see that $C_1\cup C_2$ is given by $x(x-y^n)=0$ for suitable choices of parameters $x,y$. So, if characteristic is not 2, this can be changed to $x^2-y^{2n}=0$ which is what you are claiming.
Apr 1, 2018 at 17:04 comment added Xavier49 @Mohan : $A_n$ singularity has equation $y^2-x^{n+1}=0$
Apr 1, 2018 at 17:04 history edited Xavier49 CC BY-SA 3.0
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Apr 1, 2018 at 16:06 history edited Xavier49 CC BY-SA 3.0
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Apr 1, 2018 at 15:43 comment added Mohan What is $A_n$ singularity for you?
Apr 1, 2018 at 10:37 history asked Xavier49 CC BY-SA 3.0