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

Hello pricey customer to our community We will proffer you an answer to this query ac.commutative algebra – \$K_0(mathsf{Nil}(R))\$ when \$R\$ is a bailiwick ,and the respond will breathe typical via documented info sources, We welcome you and proffer you fresh questions and solutions, Many customer are questioning in regards to the respond to this query.

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).

we are going to proffer you the answer to ac.commutative algebra – \$K_0(mathsf{Nil}(R))\$ when \$R\$ is a bailiwick query through our community which brings all of the solutions from a number of dependable sources.