Let $R$ be a local ring with maximal ideal $M$ and $I$ an ideal with $IM=(0)$. Let $A$ and $B$ be two $R$-algebras and $B$ flat over $R$. Let $\phi: A\to B$ be an $R$-morphism such that the induced homomorphism $\bar{\phi}: A/IA\to B /IB$ is an isomorphism. I want to show that $\phi$ is an isomorphism.
If $A,B$ were finite generated (e.g. noetherian) then the flat $R$-module would be free and therefore projective. this would imply injectivity. Lemma of Nakayama applied to cokernel of $\phi$ would imply injectivity. but neither the result "flat + fin. generated $\Rightarrow$ free", nor Nakayama lemma are available, because we not assumed that $A,B$ are finitely generated.
Surjectivity I think works as follows: Since $\bar{\phi}$ isomorphism, we find $b_1,…,b_k \in B$ and $l_1,…, l_n$ with $1+\sum_{j=1}^n b_j \cdot l_j \in \phi(A)$. $phi$ is a homomorphism of algebras and thus $(1+\sum_{j=1}^n b_j \cdot l_j)^2= 1+ 2 \sum_{j=1}^n b_j \cdot l_j \in \phi(A)$ (we used $I \subset M$, thus $I^2= I M =0$). this implies that $\sum_{j=1}^n b_j \cdot l_j \in \phi(A)$ and therefore $1_B \in \phi(A)$. this is surjectivity. Does it make sense?
I don't know how to prove injectivity.
Best Answer
I think we can generalize to the following claim:
Here is a lemma:
Proof of Claim: We show surjectivity first. It basically follows from (9) of Tag 00DV but I write out some details here. The condition that $\overline{\phi}$ is surjective is equivalent to $B/IB = (IB + \phi(A))/IB$; thus $B = IB + \phi(A)$; thus $B/\phi(A) = I(B/\phi(A))$; thus $B/\phi(A) = 0$ by the Lemma.
For injectivity, set $K := \ker \phi$ so that we have an exact sequence $0 \to K \to A \to B \to 0$ of $R$-modules; since $B$ is flat, this sequence stays exact when we tensor with $R/I$ (see Tag 00HL); thus $K/IK = 0$; thus $K = 0$ by the Lemma.