How do I compute this Milnor number

Question

I need to compute $\mu (x^5+y^5)=5$ on the point $p=(0,0)\in\mathbb{C}^2$. By definition, for $f\in\mathbb{C}[x,y]$, I have $$ \mu(f)=\dim\dfrac{\mathcal{O}_{(0,0)}}{<\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y}>} $$ We have $\dfrac{\partial f}{\partial x} = 4x^4$ and $\dfrac{\partial f}{\partial y} = 4y^4$. Therefore I just need to find a basis for the above quotient space. My guess was $$ \dfrac{\mathcal{O}_{(0,0)}}{<\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y}>} = <1,x,y,x^2y,xy^2> $$ but I'm pretty sure this is wrong. I'm having a real hard time figuring out how to compute those bases so any help would be appreciated. I could also use some reference to any method of computign those, if such a thing exists.

Answer 1

Your polynomial is of Brieskorn-type, i.e., $f = \sum_{i = 1}^n z^{a_i}_i$, so by a formula due to Pham, $\mu(f) = \prod_{i = 1}^n (a_i - 1)$. Thus, the answer to your question is 16 not 5. To compute the Milnor number using the local algebra approach, as you mention, observe \begin{align} \langle \partial_x f, \partial_y f \rangle = \langle 1, x, x^2, x^3, y, xy, x^2y, x^3 y, y^2, x y^2, x^2 y^2, x^3 y^2, y^3, xy^3, x^2y^3, x^3 y^3 \rangle. \end{align} These are the basis monomials which are not annihilated by the quotient.