ag.algebraic geometry - A necessary and sufficient condition for a curve to have an $A_k$ singularity.

Although the above answer involving the local algebra and the Milnor number is correct, it is often very hard to apply in real situations. Especially if you have a general function with arbitrary coefficients. You can perform a rather messy iterative process to check for an $A_k.$ In general the condition is far too ugly to want to, or be able to, write down.

You have a curve in the plane given by $f(x,y) = 0.$ The Taylor series, with respect to $x$ and $y$ is what you're really interested in. Let's assume we are only interested in the origin. If the linear terms vanish then you know that you have a singular point (a critical point of $f$). In that case you consider the quadratic part. If the quadratic part is non-degenerate, i.e. not a perfect square, then you have Morse singularity. These are $mathscr{A}$-equivalent to $x^2 pm y^2,$ and give the so-called $A_1^{pm}$-singularity types.

(Notice that $mathscr{A}$-equivalence has no relevance to the $A$ in $A_k.$ $mathscr{A}$-equivalence is also called $mathscr{RL}$-equivalence. You allow diffeomorphic changes of coordinate in the source and target (right and left sides of the commutativity diagram.)

If $f$ has a zero linear part and a degenerate quadratic part, we complete the square on the quadratic part. Then take a change of coordinates that turns the quadratic part into $tilde{x}^2$. The condition for exactly an $A_2$ is that $tilde{x}$ does not divide the new, post-coordinate change, cubic term. If not then $f$ is $mathscr{A}$-equivalent to $tilde{x}^2 + tilde{y}^3.$ (There is no $pm$ because $(x,y) mapsto (x,-y)$ changes the sign of the cubic term.

If $tilde{x}$ does divide the new cubic term, then you can complete the square on the three jet, i.e. on the quadratic and cubic terms as a whole. You take a change of coordinates so that this completed square become, say $X^2$. The condition for an $A_3^{pm}$ is that $X$ does not divide the new, post-coordinate change, quadric terms. If not then $f$ is $mathscr{A}$-equivalent to $X^2 pm Y^4.$

In general you follow the same pattern. Complete the square, take a formal power series change of coordinates so that the perfect square becomes $x_{new}^2.$ Check if $x_{new}$ divides the next set of fixed order terms. If not then stop. If $x_{new}$ didn't divide the order $n$-terms then $f$ is $mathscr{A}$-equivalent to $x_{new}^2 pm y_{new}^n.$

You just repeat the pattern: Is the quadratic part degenerate? If so then change coordinates (by a formal power series of low, but sufficient order) so that the degenerate part becomes $x_{new}^2 + O(3).$ Check if $x_{new}$ divides the cubic terms. If not then you have $x^2 + y^3.$ If so then complete the square on the new 3-jet and change coordinates so that you have $x_{new}^2 + O(4)$. Does $x_{new}^2$ divide the quartic terms? If not then you have $x^2 pm y^4.$ If so then complete the square on the new 4-jet and change coordinates. Just keep completing the square, checking divisibility, changing coordinates.

The conditions on the coefficients soon spiral out of control. To check an $A_6$ you'll need a computer program. For a general polynomial it's impossible without a computer. (Except for very special cases!) I wrote a program in Maple to calculate the conditions up to $A_k$ once, but the output was so messy then I gave up. Having said that, for an explicit polynomial it's child's play.

• Next Avoiding Minkowski's theorem in algebraic number theory.
• Previous at.algebraic topology - relationship between borromean rings and hanging-a-picture-from-three-nails puzzle?