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Let $X$ be a non-reduced Noetherian scheme. We define $K^0(X)$ to be the Grothendieck group of the derived category $Perf(X)$ and $K_0(X)$ to be the Grothendieck group of the derived category $D^b_{coh}(X)$. Then we have a natural Cartan homomorphism $$ K^0(X)\to K_0(X). $$ given by inclusion of the derived categories $Perf(X)\subset D^b_{coh}(X)$.

Next we consider the associated reduced scheme $X_{red}$ with $f: X_{red}\to X$ and we have $K^0(X_{red})$ and $K_0(X_{red})$ also. Moreover there are natural maps $$ f^*:K^0(X)\to K^0(X_{red}) $$ and $$ f_*:K_0(X_{red})\to K_0(X). $$

By the devissage theorem in K-theory, $f_*:K_0(X_{red})\to K_0(X)$ is an isomorphism but this is not true for $f^*$.

My question is: do we have the following commutative diagram $$\begin{array}[c]{ccc} K^0(X)&{\rightarrow}&K_0(X)\\ \downarrow\scriptstyle{f^*}&&\uparrow\scriptstyle{\cong}\\ K^0(X_{red})&{\rightarrow}&K_0(X_{red})? \end{array}$$

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  • $\begingroup$ That square is not commutative. Just consider the images of the element $1$ in $K^0(X)$. $\endgroup$
    – Jason Starr
    Commented May 1, 2015 at 15:56
  • $\begingroup$ @JasonStarr Thank you Jason! Actually this question is inspired by your answer to my previous MO question mathoverflow.net/questions/203145/…, where you mentioned that the composition $Pic(X)\hookrightarrow K^0(X)\to K_0(X)$ factors through the pull back $Pic(X)\to Pic(X_{red})$. I think the latter goes to $K^0(X_{red})$ and then to $K_0(X_{red})$. That's why I thought there is a commutative diagram in K-theory. Could you explain a little bit more why the map from the Picard group factors through $Pic(X_{red})$? $\endgroup$
    – Zhaoting Wei
    Commented May 1, 2015 at 21:49
  • $\begingroup$ The map from $K_0(X_{\text{red}})$ to $K_0(X)$ that completes that commutative diagram is not the pushforward map. It is a $K^0(X)$-module homomorphism that sends the class $[\mathcal{O}_{X_{\text{red}}}]$ to $[\mathcal{O}_X]$. $\endgroup$
    – Jason Starr
    Commented May 1, 2015 at 22:40
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    $\begingroup$ "You mean ..." No, that is not what I am saying. Here is another way of saying it: since the pushforward map $K_0(f)$ is an isomorphism of $K^0(X)$-modules, and since the action of $K^0(X)$ on $K_0(X_{\text{red}})$ is induced by the ring homomorphism $K^0(f):K^0(X)\to K^0(X_{\text{red}})$, it follows that the ideal $\text{Ker}(K^0(f))$ annihilates $K_0(X)$. In particular, if this ideal is nonzero (which can easily be arranged), then the induced homomorphism $K^0(X)\to K_0(X)$ is not injective. $\endgroup$
    – Jason Starr
    Commented May 2, 2015 at 0:15
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    $\begingroup$ You may be right. At any rate, the point of my answer to your original question is that $\text{Ker}(K^0(f))$ annihilates $K_0(X)$. However, now I am not certain whether or not there is a homomorphism from $K_0(X_{\text{red}})$ to $K_0(X)$ making the diagram commute. $\endgroup$
    – Jason Starr
    Commented May 2, 2015 at 14:52

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