Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
  1. $\ker(f)$ direct sum with $\operatorname{im}(g)$, does it mean that number of polynomials in $\ker(f)$ must be the same as the number of polynomials in $\operatorname{im}(g)$?

In other words, for example $\mathbb Q[y_1,y_2,y_3] \rightarrow \mathbb Q[x_1,x_2,x_3]$, number of variables in domain must be the same as number of variables in codomain?

If not, how can they be direct sum?

  1. a:${}\qquad S \rightarrow G$

    b: ${}\qquad S \rightarrow H$

    c: ${}\qquad G \rightarrow H$

    $b = c \circ a$

    where $S$ is semigroup, $G$ is Grothendieck.

I guess to get $G$ is to find $d: G \rightarrow S$ from $a: S \rightarrow G$ and then $G = \ker(d)$ direct sum $\operatorname{im}(a)$, however in a book algebraic K-theory and its application page 4 define above.

How to express $K_0(R)$ = Grothendieck group of semi group $\operatorname{Proj}(R)$ of isomorphism classes of finitely generated projective modules over $R$ in maple code?

share|improve this question
    
Which book are you referring to? –  A.P. Mar 16 '13 at 16:05

1 Answer 1

Given a monoid $(M,*)$, its Grothendieck group $G(M)$ is the smallest group containing $M$ as a submonoid. You can think of it as $M$ with the missing inverses added. For example, when you consider $(M,*)=(\operatorname{Proj}(R),\oplus)$ (with identity the $0$ module), then for every $P\in \operatorname{Proj}(R)$ you add objects $P'$ such that $P\oplus P' = 0$.

In this special case, you actually have another way of defining $K_0(R)=G(\operatorname{Proj}(R))$, which can be proven to be equivalent (for example, see Rotman's Advanced Modern Algebra). Namely, by taking the free abelian group generated by the elements of $\operatorname{Proj}(R)$ and quotienting out the relations $$ B-A-C $$ whenever $$ 0 \to A \to B \to C \to 0 $$ is an exact sequence, i.e. if there are an injective map $f:A\to B$ and a surjective map $g:B\to C$ such that $\operatorname{im}(f)=\ker(g)$ (it is somewhat more than requiring that $g\circ f=0$). Note that since $C$ is projective, there is such a sequence if and only if $$ B=A\oplus C $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.