# 扶磊《代数几何》56页注释（2）

We may find a natural number $N$ such that $S_k\otimes_AA_p/pA_p=0$ for any $k\ge N$. By Nakayama's lemma, we have $S_k\otimes_AA_p=0$ for any $k\ge N$.

$A_p$只有一个极大理想$pA_p$，它的大根就是$pA_p$。如果能证明$pA_p(S_k\otimes_AA_p)=S_k\otimes_AA_p$就解决问题了。