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Jul 7, 2022 at 14:06 history edited Yuhang Chen CC BY-SA 4.0
added 277 characters in body
Jul 7, 2022 at 5:02 comment added naf Your computation is not correct if $p$ is $p_1$ or $p_2$: one only gets that $[p_1] + [p_2]$ is $0$ in $A^1(X)$, so $A^1(X) \cong \mathbb{Z}$.
Jul 7, 2022 at 1:47 history answered Yuhang Chen CC BY-SA 4.0