Let $(A,\mathfrak{m})$ and $(B,\mathfrak{n})$ be local Noetherian rings. Suppose that $\phi : A\rightarrow B$ is a map such that $\phi(\mathfrak{m}) \subset \mathfrak{n}$ and suppose
- $A/\mathfrak{m} \cong B/\mathfrak{n}$;
- $\mathfrak{m} \rightarrow \mathfrak{n/n^2}$ is surjective;
- $B$ is a finitely generated $A$ module.
Show that $\phi$ is surjective.
Here is my approach so far:
Consider $0\rightarrow \mathfrak{m} \rightarrow A \rightarrow A/\mathfrak{m}\rightarrow 0$ and the exact sequence for $B$. Then we can establish an commutative diagram of the two exact sequences. (Sorry i do not know how to type that.) Then apply the snake's lemma, the isomorphism in condition 1 yields an isomophism $\mathfrak{n}/\phi(\mathfrak{m})\cong B/\phi(A)$. To see $\phi$ is surjective, it suffices to show $\phi(\mathfrak{m})=\mathfrak n$. Since $\mathfrak{n}$ is the maximal ideal of $B$, if $\phi(\mathfrak{m})+\mathfrak{n}^2 = \mathfrak{n}$, Nakayama will give the desired result. Now somehow I want to conclude $\phi(\mathfrak{m})+\mathfrak{n}^2 = \mathfrak{n}$ from conditions 2) and 3), which seems to be true but cannot really proceed.
Best Answer
It is clear that $\phi(\mathfrak{m}) + \mathfrak{n}^2 \subset \mathfrak{n}$. Take an element $a \in \mathfrak{n}$. Since $\mathfrak{m} \rightarrow \mathfrak{n}/\mathfrak{n}^2$ is surjective, there exists $x \in \mathfrak{m}$ such that $a - \phi(x) \in \mathfrak{n}^2$. Thus, $a = \phi(x) + b$ for some $b \in \mathfrak{n}^2$. This shows that $\mathfrak{n} \subset \phi(\mathfrak{m}) + \mathfrak{n}^2$, and we are done.
I hope I am not misunderstanding what the map $\mathfrak{m} \rightarrow \mathfrak{n}/\mathfrak{n}^2$ is. I also want to say that you did all the difficult parts already.
Also take a look at Exercise 10 from Atiyah-Macdonald Chapter 2. While your proof using Snake Lemma is very nice, one can prove this statement by using Nakayama twice. The first time you use Nakayama is to deduce from $\phi(\mathfrak{m}) + \mathfrak{n}^2 = \mathfrak{n}$ that $\phi(\mathfrak{m}) = \mathfrak{n}$. This gives you that $\mathfrak{n} = \mathfrak{m}B$, i.e., the extension of $\mathfrak{m}$ in $B$ is $\mathfrak{n}$. Now, we know that the map $A\mathfrak{m} \rightarrow B/\mathfrak{m}B$ is surjective. Hence, by an argument similar to the one I gave above, you can show that $im(\phi) + \mathfrak{m}B = B$. Since $B$ is a finitely generated $A$-module, you use Nakayama again to deduce that $im(\phi) = B$.
This also shows you that you do not need $A$ Noetherian, and it suffices to only assume that the maximal ideal of $B$ is finitely generated.