0

I have some question about the propositon(c) of thetheorem 3.16 in Qing Liu's book.

the statement is:

Let (A,m) be a Noetherian local ring, and $\hat{A}$ its m-adic completion.

Let (B,n) be a local ring such that $A\subseteq B \subseteq \hat{A}$ and mB =n. Then the n-adic completion $\hat{B}$ is isomorphic to $\hat{A}$.

the proof of c is:

Let n $\geq$ 1. We have $n^n=m^nB$.Since the composition $A/m^n \to B/m^nB \to \hat{A}/m^n\hat{A}$ is an isomorphism, $B/m^nB \to \hat{A}/m^n\hat{A}$ is surjective. It remains to show that it is injective; that is, that $m^n\hat{A}\cap B=m^nB$. We have $B=A+mB=A+m^2B=\dots=A+m^nB$.

I only want to know why B=A+mB. Can someone help me? [1]: https://i.stack.imgur.com/Cr1Hs.png [2]: https://i.stack.imgur.com/bsAtJ.png

梦尽缘灭
  • 1
  • https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference Please use mathjax to write your question. – Souparna Oct 12 '23 at 15:46
  • I Have edit my statement. @Souparna – 梦尽缘灭 Oct 12 '23 at 16:12
  • just i have an idea : because $ A/m,B/mB,\hat{A}/m\hat{A} $ is field, and field homomorphism is injective homomorphism. As $ B/mB \to \hat{A}/m\hat{A} $ is surjective, that is isomorphism, and $ A/m $ is isomorphism to $ \hat{A}/m\hat{A} $, so $A/m $ is isomorphism to $ B/mB $. As a result, B＝ A+mB. what i think is right? – 梦尽缘灭 Oct 12 '23 at 18:23

0 Answers0