Timeline for Do we have the following "devissage commutative diagram" in K-theory?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| May 2, 2015 at 19:57 | comment | added | Zhaoting Wei | Let us continue this discussion in chat. | |
| May 2, 2015 at 14:52 | comment | added | Jason Starr | 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. | |
| May 2, 2015 at 3:10 | comment | added | Zhaoting Wei | @JasonStarr I got your last reasoning that $K^0(X)\to K_0(X)$ is not injective. Nevertheless I'm also interested in your comment "The map from $K_0(X_{red})$ 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 $[\mathcal{O}_{X_{red}}]$ to $[\mathcal{O}_X]$". Is this map well-defined and an isomorphism? | |
| May 2, 2015 at 0:15 | comment | added | Jason Starr | "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. | |
| May 1, 2015 at 22:53 | comment | added | Zhaoting Wei | @JasonStarr You mean the isomorphism $K_0(X)\overset{\cong}{\to}K_0(X_{red})$ is given by the pull back $f^*$, isn't it? | |
| May 1, 2015 at 22:40 | comment | added | Jason Starr | 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 $[\mathcal{O}_X]$. | |
| May 1, 2015 at 21:49 | comment | added | Zhaoting Wei | @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 $Pic(X_{red})$? | |
| May 1, 2015 at 15:56 | comment | added | Jason Starr | That square is not commutative. Just consider the images of the element $1$ in $K^0(X)$. | |
| May 1, 2015 at 14:23 | history | asked | Zhaoting Wei | CC BY-SA 3.0 |