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Let $\textbf {X}$ be a noetherian scheme,

$\textbf {M(X)}$ be the categroy of coherent sheaf over the scheme $\textbf {X}$.

We denote $ \textbf {K$_0$(M(X))}$ to be $ \textbf {G$_0$(X)}$.

Now I know that since $\textbf {X}$ is noetherian it has a finite cover by Spec(A$_i$) for some $i$ = 1,2,..n. and A$_i$ for each $i$ is noetherian.

and there is an equivalence between the categories of M(Spec(A$_i$)) and fintely generated A$_i$ modules.

I have proved that G$_0$ (A$_i$$_{red}$) $\cong $ G$_0$ (A$_i$) where G$_0$ (R) is K$_0$(M(R));

Notation M(R) = finitely generated R module (R -noetherian in this case).

Now Can I say this, that since every open set in the cover of X has G$_0$ (A$_i$$_{red}$) $\cong $ G$_0$ (A$_i$) then

G$_0$(X) $\cong$ G$_0$(X$_{red}$) .

So my query is to prove the above mentioned statement does it suffice to prove it for affine noetherian scheme? As I know that every open affine subset of a noetherian scheme is noetherian.

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    $\begingroup$ This is an application of dévissage, see e.g. Weibel's K-book, Chapter 2. You can reduce to affine schemes but if you're not familiar with such reduction arguments, it's better you check in details that the abelian categories have same $K_0$ (of course at the end you use the result for affine schemes). This is also true for higher $G_n$, see Quillen's dévissage theorem in Higher algebraic K-theory I. $\endgroup$
    – François Brunault
    Commented Sep 18, 2019 at 8:10
  • $\begingroup$ So for reducing it to affine case is it sufficient to say every affine open set of noetherian scheme is noetherian? $\endgroup$
    – user139827
    Commented Sep 18, 2019 at 8:40
  • $\begingroup$ You need to consider the categories for the schemes not just for the affine cover. It depends how comfortable you are, but if not it's better to write the full proof. $\endgroup$
    – François Brunault
    Commented Sep 18, 2019 at 9:06
  • $\begingroup$ Okay thank you. $\endgroup$
    – user139827
    Commented Sep 18, 2019 at 9:14
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    $\begingroup$ No, you cannot deduce $G_0(X) = G_0(X_{red})$ just from the affine case. There is a Mayer-Vietoris sequence but it’s going to involve higher groups $G_i$ which you have to take into account. But you can apply devissage to M(X) directly and that works. $\endgroup$
    – crystalline
    Commented Sep 18, 2019 at 20:37

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