TITLE: How do I check that this is a Frobenius algebra?
QUESTION [12 upvotes]: Let $f_1,f_2,\ldots,f_n\in \mathbb C[z_1,\ldots, z_n]$ be such that the quotient ring
$$A:=\mathbb C[z_1,\ldots, z_n]/(f_1,f_2,\ldots,f_n)$$
is finite dimensional (in other words, it's a zero-dimensional complete intersection).
I've heard that such a ring is always a Frobenius algbera when equipped with the counit
$$
\varepsilon(g):= \int_{|f_1|=\ldots=|f_n|=1} \frac{g(z_1,\ldots,z_n)}{\prod_i f_i(z_1,\ldots,z_n)}dz_1\ldots dz_n,
$$
and I'd like to check that claim. By definition, $A$ is a Frobenius algebra if the bilinear form $\langle g,h\rangle:=\varepsilon(gh)$ is non-degenerate. 
So I would like to know whether:

If $g\in \mathbb C[z_1,\ldots, z_n]$ is such that
  $$\int_{|f_1|=\ldots=|f_n|=1} \frac{g(z_1,\ldots,z_n)h(z_1,\ldots,z_n)}{\prod_i f_i(z_1,\ldots,z_n)}dz_1\ldots dz_n=0$$
  for every $h\in \mathbb C[z_1,\ldots, z_n]$, does is then follow that $g$ is in the ideal generated by $f_1,f_2,\ldots,f_n$?

REPLY [14 votes]: Yes, this is discussed in page 659 of "Principles of Algebraic Geometry" by Griffiths and Harris. They call it the local duality theorem.