$K_0(\mathsf{Nil}(R))$ when $R$ is a field

Question

$\DeclareMathOperator\Nil{\mathsf Nil}\DeclareMathOperator\ker{ker}$I was reading through The $K$- book by Charles A. Weibel. There I found a very interesting category $\Nil(R)$, which consists of pairs like $(P , \nu)$, where $P$ is a finitely generated projective module and $ \nu : P \rightarrow P$ is a nilpotent endomorphism. Now if I define a forgetful functor $F : \Nil(R) \rightarrow P(R)$ (that forgets the nilpotent) where $P(R)$ is the category of finitely generated projective $R$-module induces a split surjective group homomorphism $F'$ on $K_0$ of the said categories such that $$K_0(\Nil(R)) \cong K_0(R) \oplus \ker(F').$$ I have understood that the $\ker(F')$ is generated by elements of the form $[R^n ,\nu] - n[R,0]$, now I was thinking that what happens if I replace $R$ by a field $\mathbb{F}$ then I have $$K_0(\Nil(\mathbb{F})) \cong \mathbb{Z} \oplus \ker(F').$$ Where now $\ker(F')$ is generated by elements of the form $[\mathbb{F^n} ,\nu] - n[\mathbb{F},0]$, now I have a feeling that $K_0(\Nil(\mathbb{F}))$ should be isomorphic to $\mathbb{Z}$, but then $\ker(F')$ should be trivial so my question is in the case of field is $\ker(F')$ trivial?

Answer 1

The answer is yes, and this follows essentially from the Jordan decomposition of nilpotent endomorphisms.

Let $(F^n,\nu)$ be an $n$-dimensional vector space and a nilpotent endomorphism. Then $\nu^n=0$ and we can write a filtration $$ F^n=\ker\nu^n\supseteq \ker \nu^{n-1} \supseteq \ker \nu^{n-2} \supseteq \cdots \supseteq \ker \nu \supseteq 0\,.$$ Since $\nu(\ker\nu^i)\subseteq \ker \nu^{i-1}$, we obtain an identity in $K_0(Nil(F))$ $$ [F^n,\nu] \cong \left[\bigoplus_{i=0}^{n-1} \ker \nu^{i+1}/\ker\nu^i,0\right]\cong [F^n,0]\,.$$ Therefore $K_0(Nil(F))=\mathbb{Z}$, as requested. Note that here we have used that $F$ is a field to prove that any submodule of a projective module is a summand (or, equivalently that the quotient of a projective module is projective).