Uniqueness of a local certain homomorphism (Etale Cohomology and the Weil Conjecture by Freitag, Kiehl)

Question

(All rings in the context are presumed to be commutative, unital and noetherian.)
Let $(A,n), (B_1, m_1),$ and $(B_2, m_2)$ be local rings with their maximal ideals, $k$ a field.
For $i=1,2$ there are local-etale maps $a_i: A \to B_i$ and local maps $p_i: B_i \to k$.

Recall that a map $\varphi: (R, n) \to (S,m)$ between local ring is called local if $\varphi(n) \subset n$;
it is local-etale if it is local, flat, unramified (meaning $\varphi(n) = m$ and $R/n \to S/m$ are finite separable) and $S$ arises as a localization of a finitely generated $R$-algebra.

Assume there exist two local maps $\phi_1, \phi_2: B_1 \to B_2$ compatible with $a_i$ and $p_i$, i.e. $\phi_j \circ a_1 = a_2$ and $p_2 \circ \phi_j = p_1$ for $j=1,2$.

Question: Why does this imply $\phi_1= \phi_2$?

Clearly, since $p_i$ factorizes as $B_i \to B_i/m_i \to k$ and field extensions $B_i/m_i \to k$ are always injective and so mono, the induced maps $\overline{\phi_1}, \overline{\phi_2}: B_1/m_1 \to B_2/m_2$ between residue fields coincide due to compatibility with $p_i$.

Why does this suffice to show that $\phi_1 $ and $ \phi_2$ already coincide?
Since $a_i$ unramified, we have equality $m_i = n \cdot B_i$ of maximal ideals, but note(!) that we cannot apply Nakayama lemma here to deduce from $B_1= \ker(\phi_1-\phi_2) + m_1 = \ker(\phi_1-\phi_2) + n \cdot B_1$ that $B_1= \ker(\phi_1-\phi_2)$, since $B_1$ is not a finite $A$-module.

Source: Etale Cohomology and the Weil Conjecture by Freitag, Kiehl, page 16, Construction of Henselian rings.

Answer 1

"First we reduce to the case where $B_1$ is actually etale $A$-algebra."

From the definition $$a_1: A\xrightarrow{} C_1\to S^{-1}C\simeq B_1$$ where $C_1$ is a etale A-algebra and $S$ a multiplicative closed subset of $C_1$.

Now from the remark (using quasi-finiteness and Zariski's main theorem: see paragraph after Def. 1.1 page 7) $C$ is actually a localization at $T$ of an etale $A$-algebra $D_1$ that is finite as $A$-module and $T$ a multiplicative closed subset. So we have a refined diagram

$$a_1: A\xrightarrow{\psi_1} D_1\to T^{-1}D_1 \simeq C_1\to S^{-1}C\simeq B_1.$$

Now apply the Nakayama argument to the composed maps as in your attempt $\phi_1, \phi_2: D_1\to B_2$, so we are done.

Answer 2

Here probably another way one can also prove it without Zariski main theorem as "deep input" at least if we assume $A, B_1, B_2$ to be noetherian.

Let $i_A: A \to \widehat{A}, i_{B_i}: B_i \to \widehat{B}_i$ the canonical maps to the completions and $\widehat{a}_i : \widehat{A} \to \widehat{B}_i$ the morphisms between completions. Noetherian condition implies $\cap_i m_2^i =0$ and therefore $i_{B_2}:B_2 \to \widehat{B}_2$ is mono.

Since $a_i $ are etale, $\widehat{a}_i$ are isomorphisms and the induced maps $\widehat{\phi}_j$ coinside, since both are $\widehat{a}_2 \circ \widehat{a}_1^{-1}$. Therefore $\widehat{\phi}_2 -\widehat{\phi}_1$ is zero morphism on $\widehat{B}_1$.

If we write a commutative diagram involving $A, B_i$ and their completions we see that $\widehat{\phi}_j \circ i_{B_1}= i_{B_2} \circ \phi_j$ and therefore we have also $ i_{B_2} \circ (\phi_2 -\phi_1)= (\widehat{\phi}_2 - \widehat{\phi}_1) \circ i_{B_1}=0$. Since $i_{B_2}$ mono, we get $\phi_2 -\phi_1=0$.