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ADDED TO ANSWER THE COMMENT BELOW. I do not know any explicit expression for the Milnor number, I think that in general you cannot avoid to compute the $\mathbb{C}$-basis for the Milnor algebra. I agree that these computations are tedious by hand, however you can use a Computer Algebra software like SINGULAR (which is free) to do this quickly and easily.

And yes, there are similar conditions for $D_k$ and $E_6$, $E_7$, $E_8$. Let me state the condition for $D_k$.

Let $f \in \boldsymbol{m}^3 \subset \mathcal{O}_o$ and $k \geq 4$. Denote by $f^{(3)}$ the $3$-jet of $f$. Then the following are equivalent:

• $f^{(3)}$ factors into at least two different factors and $\mu(f, o)=k$;
• $f$ is of type $D_k$.
• Moreover, $f^{(3)}$ factors into three different factors if and only if $f$ is of type $D_4$.

The conditions for $E_6$, $E_7$, $E_8$ are a bit more complicate and I will not state them here. You will find them in the book of GREUEL, LOSSEN and SHUSTIN, p. 154.

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At least over $\mathbb{C}$, there is a simple answer.

A plane curve $f(x,y)=0$ has a singularity of type $A_k$ in $o=(0,0)$ if and only if

• $o$ is a $double$ $point$, that is all first partial derivatives of $f$ vanish in $o$ but there is at least one second partial derivative which is not zero;

• the $Milnor$ $number$

$\mu(f, o):= \dim_{\mathbb{C}}\mathcal{O}_{o}/(f_x, f_y)$

is equal to $k$. Here $\mathcal{O}_{o}$ denotes the ring of convergent power series.

This can be generalized in higher dimensions. In fact, one proves that a (germ of) complex hypersurface singularity $f(x_1, ...,x_n)=0$ is of type $A_k$ if and only if

• the corank

$\textrm{crk}(f):=n-\textrm{rank}(Hessian)(f)(o)$ \textrm{crk}(f):=n-\textrm{rank}(\textrm{Hessian}(f))(o)$is$ \leq 1$; • the Milnor number$\mu(f, o):= \dim_{\mathbb{C}}\mathcal{O}_{o}/(J_f)$is equal to$k\$.

This follows from a sort of generalized Morse Lemma. See the book GREUEL - LOSSEN - SHUSTIN "Introduction to singularities and deformations" p. 150 for the proof.

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