Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am currently trying to understand the very basics of complex algebraic curves and I came across the following statement in the book by F. Kirwan (Definition 2.9):

The multiplicity of the complex algebraic curve $C$ defined by the polynomial $P(x,y)$ at a point $(a,b) \in C$ is the smalles positive integer $m$ such that $$ \frac{\partial^m P}{\partial x^i \partial y^j}(a,b) \ne 0 $$ for some $i \ge 0$, $j \ge 0$ such that $i + j = m$.

The polynomial $$ \sum_{i+j = m} \frac{\partial^m P}{\partial x^i \partial y^j}(a,b) \frac{(x-a)^i(y-b)^j}{i!j!} $$ is then homogeneous of degree $m$. Here is where I get stuck - it is clear that it is homgeneous in the variables $X = x - a$, $Y = y-b$, but how can I see that $$ \sum_{i+j = m} \frac{\partial^m P}{\partial x^i \partial y^j}(a,b) \frac{(\lambda x-a)^i(\lambda y-b)^j}{i!j!} = \lambda^m \sum_{i+j = m} \frac{\partial^m P}{\partial x^i \partial y^j}(a,b) \frac{(x-a)^i(y-b)^j}{i!j!} \quad ? $$

share|improve this question
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.