扶磊《代数几何》56页注释(3)

reversing the above argument的时候,遇到了一个拦路虎。

怎样搞定这一步
$S_{\bar{x_i}}\otimes_AA_p/pA_p={0}\Rightarrow S_{(\bar{x_i})}\otimes_AA_p/pA_p=0$

众所周知,$A\to A_p$是flat的,$A\to A_p/pA_p$貌似不一定是flat的啊,这可咋整呢?

突然灵感闪现,想起刘青的Algebraic Geometry and Arithmetic Curves第52页有一种将$S_{\bar{x_i}}$ grade的方法。

'Let us note that $B_f$ is a graded $B_{(f)}$-algebra, the homogeneous elements of degree $n(n\in \Bbb Z)$ being of the form $bf^{-N}$ with $b\in B$ and $\deg b=Nr+n$."

用这种方式将$S_{\bar{x_i}}$ grade以后,设$S_{\bar{x_i}}=\bigoplus_{n\in\Bbb Z}S^\prime_n$, 其中$S_{(\bar{x_i})}=S^\prime_0$。

由$S_{\bar{x_i}}\otimes_AA_p/pA_p=(\bigoplus_{n\in\Bbb Z}S^\prime_n)\otimes_AA_p/pA_p\cong \bigoplus_{n\in\Bbb Z}(S^\prime_n\otimes_AA_p/pA_p)$知$S_{(\bar{x_i})}\otimes_AA_p/pA_p$可以看作是$S_{\bar{x_i}}\otimes_AA_p/pA_p$的子模,而$S_{\bar{x_i}}\otimes_AA_p/pA_p=0$, 故$S_{(\bar{x_i})}\otimes_AA_p/pA_p=0$。

哥咋这么滴聪明捏!
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