# A question from Otto Forster's book on Riemann surfaces

I am reading section 14, A Finiteness Theorem of Otto Forster's book Lectures on Riemann Surfaces, and come across a problem on Theorem 14.15 on page 117. In the proof Forster introduces a function

$$F=\det(f\delta_{\nu\mu}-c_{\nu\mu})_{\nu\mu}$$

which is holomorphic, where $f$ is holomorphic, but I don't know why it follows that $F\xi_\nu\mid_Y=0$. I am wondering whether we should replace $F$ with the matrix $(f\delta_{\nu\mu}-c_{\nu\mu})$, but since the proof relies heavily on this claim, I get puzzled. Is there something wrong or am I misunderstanding some stuff? How should I understand this theorem?

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The argument is similar to the proof of Nakayama's lemma .Take everything on (1) to one side and multiply by the adjugate matrix. t –  Mohan Ramachandran Apr 8 '13 at 18:34

I found that argument confusing too. If you choose a basis $\xi_{\mu}$ of eigenvectors of $C=(c_{\mu \nu})$, you can arrange that $C$ is in Jordan normal form. Then we get that $F=\det(fI-C)$ is the product of all determinants of the various blocks, and so multiplied by any generalized eigenvector of $C$ gives us $0$ because it applies $f-\lambda$ (where $\lambda$ is the eigenvalue) enough times to kill the generalized eigenvector: $(f-\lambda)^k \xi_{\nu} = (C-\lambda I)^k \xi_{\nu}=0$ if $k$ is as large as the size of the Jordan block that $\xi_{\nu}$ belongs to. Check this carefully, because I haven't thought about Forster's book in a long time (and because my first answer was wrong).