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  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?

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Which book are you referring to? – A.P. Mar 16 '13 at 16:05

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 $$ B=A\oplus C $$

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