Can the following sum be counted or expressed in terms of special functions?

ac.commutative algebra – $K_0(mathsf{Nil}(R))$ when $R$ is a bailiwick Answer

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ac.commutative algebra – $K_0(mathsf{Nil}(R))$ when $R$ is a bailiwick

The respond is sure, and this follows basically from the Jordan decomposition of nilpotent endomorphisms.

Let $(F^n,nu)$ breathe an $n$-dimensional vector area and a nilpotent endomorphism. Then $nu^n=0$ and we are able to write a filtration
$$ F^n=kernu^nsupseteq ker nu^{n-1} supseteq ker nu^{n-2} supseteq cdots supseteq ker nu supseteq 0,.$$
Since $nu(kernu^i)subseteq ker nu^{i-1}$, we secure an id in $K_0(Nil(F))$
$$ [F^n,nu] cong left[bigoplus_{i=0}^{n-1} ker nu^{i+1}/kernu^i,0right]cong [F^n,0],.$$
Therefore $K_0(Nil(F))=mathbb{Z}$, as requested. Note that right here we’ve used that $F$ is a bailiwick to show that any submodule of a projective module is a summand (or, equivalently that the quotient of a projective module is projective).

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