# 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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(TeX comment) Insert a missing dollar sign to make the question readable. –  P Vanchinathan Apr 8 '13 at 9:57
It's not a missing dollar sign. Actually I can fix it by escaping (with a backslash) two of the underscores, but I do NOT understand why that ought to work, hence I'm leaving the question as is. Someone more familiar with the inner workings of jsMath should look at it. –  José Figueroa-O'Farrill Apr 8 '13 at 10:23
It can also be fixed by enclosing a few of the pieces of MathJax with backticks, e.g. (dollar sign)...(dollar sign), but since this hack will not work on the Stack Exchange network, I'm hesitant to add to what is already a daunting problem (given how commonly this has been used in the past on MO). See here for an example of how math inside backticks appears on the SE network: math.stackexchange.com/questions/354677/… –  Zev Chonoles Apr 8 '13 at 12:12
Actually I asked the very same question again, and corrected the terrible math writing by simply adding "`" in front of every dollar sign just as suggested. I don't know why, and it seems to happen when you type subscripts。 –  xuxuzhu Apr 8 '13 at 16:15
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).