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Consider the following question: If two nodes collide what do you get? First of all it can not be a strict $A_2$ node, because the delta invariant of that is $1$. So it has to be more singular than an $A_2$ node. It can be an $A_3$ node because the delta invariant of that is $2$.

Is there any simple argument to show that if an $A_2$ node and an $A_1$ node collide, then we can not get a strict $A_3$ node? The delta invariant doesn't help. Is there some other invariant that can answer this question? Note that I am NOT asking what do we actually get when an $A_2$ node and an $A_1$ node collide. I merely want to show that we can not get a strict $A_3$ node.

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up vote 5 down vote accepted

I am not sure if the following counts as a simple argument:

A miniversal deformation of the $A_3$ singularity is given by the family $y^2 = a + bx + cx^2 + x^4$. There is no member in this family with nodes of type $A_1$ and $A_2$ so it follows that we cannot get an $A_3$ node from collisions of two nodes of type $A_1$ and $A_2$.

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Thank you. Is the miniversal deformation of any ADE singularity known? I wanted to show that two nodes and one cusp can not collide to a strict A_4 node. And is the miniversal deformation of E_7 node known? –  Ritwik Nov 17 '11 at 14:53
    
By the way, how do we know that the family you wrote has no singularities of type A_1 and A_2? –  Ritwik Nov 17 '11 at 14:56
    
The miniversal deformation of any planar singularity is easy to write down. See for example Hartshorne's "Deformation theory", section 14. –  ulrich Nov 17 '11 at 15:12
    
The singularities in the family correspond to multiple roots of the polynomial $a + bx + cx^2 + x^4$. Since this is of degree $4$, it can never have both a double root and a triple root. –  ulrich Nov 17 '11 at 15:16
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