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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 $f_1,f_2,\ldots,f_n$?

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

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  • $\begingroup$ Ok. I'll check it out. Is it easy to say what kind of ingredients come into the proof? $\endgroup$ – André Henriques Oct 23 '14 at 21:52
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    $\begingroup$ @AndréHenriques The main steps are roughly: 1) Reduce the global duality theorem (the statement in your question) to the local case around a single zero. 2) Prove it for the case $f_i=z_i^{k_i}$. 3) Deduce the general form by using the transformation formula (which says that if you take the residue with respect to $Af$ then the residue changes by factor of $\det (A)$.) $\endgroup$ – Gjergji Zaimi Oct 23 '14 at 22:14

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