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 $f\_1,f\_2,\ldots,f\_n$?
> 
> 
> 



How do I check that this is a Frobenius algebra?
Yes, this is discussed in page 659 of "Principles of Algebraic Geometry" by Griffiths and Harris. They call it *the local duality theorem*.


