$\DeclareMathOperator\Nil{\mathsf Nil}\DeclareMathOperator\ker{ker}$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?

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    $\begingroup$ Why do you feel that $K_0(Nil(\mathbb{F}))$ should be isomorphic to $\mathbb{Z}$? Can you turn that reason into a proof? $\endgroup$
    – Will Sawin
    Jan 11, 2021 at 19:11
  • $\begingroup$ Since all the elements are finite product of elements of the form $[\mathbb{F}^k , \nu]$ and their inverses. I felt that may be I can generate any element of $K_0Nil(\mathbb{F})$ by $[\mathbb{F}, \nu ']$, that is the extent of my progress here. $\endgroup$
    – user139827
    Jan 12, 2021 at 4:36

1 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).

  • $\begingroup$ I'm not sure I understand your last sentence. Indeed, it seems to me that you're not using that a submodule of a projective module is a summand, but rather that a submodule of a projective module is projective (in which case a PID would work just as well) $\endgroup$ Jan 12, 2021 at 9:39
  • $\begingroup$ @MaximeRamzi I'm using that the quotient of a projective module is projective, which is equivalent to the corresponding submodule being a summand :) $\endgroup$ Jan 12, 2021 at 10:06
  • $\begingroup$ For what it's worth, I believe the theorem $K_0(Nil(R))=K_0(R)$ is true for all regular Noetherian rings, but you need a more complicated proof. $\endgroup$ Jan 12, 2021 at 10:09
  • $\begingroup$ Great! Now I understand, and since we are using the filtration by kernel we should not need an algebraically closed field as well right? Thank you for your answer. $\endgroup$
    – user139827
    Jan 12, 2021 at 10:10
  • $\begingroup$ @DenisNardin wow! I will surely look for the proof for regular noetherian rings. $\endgroup$
    – user139827
    Jan 12, 2021 at 10:12

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