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

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

• Why do you feel that $K_0(Nil(\mathbb{F}))$ should be isomorphic to $\mathbb{Z}$? Can you turn that reason into a proof? Jan 11, 2021 at 19:11
• 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.
– user139827
Jan 12, 2021 at 4:36

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).

• 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) Jan 12, 2021 at 9:39
• @MaximeRamzi I'm using that the quotient of a projective module is projective, which is equivalent to the corresponding submodule being a summand :) Jan 12, 2021 at 10:06
• 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. Jan 12, 2021 at 10:09
• 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.
– user139827
Jan 12, 2021 at 10:10
• @DenisNardin wow! I will surely look for the proof for regular noetherian rings.
– user139827
Jan 12, 2021 at 10:12