Show that a morphism \(f: X \to Y\) of schemes of finite type over \(k\) is étale if and only if the following condition is satisfied. For each \(x \in X\), let \(y = f(x)\). Let \(\hat{\mco}_x\) and \(\hat{\mco}_y\) be the completions of the local rings at \(x\) and \(y\). Choose fields of representatives (II, 8.25A) \(k(x) \subseteq \hat{\mco}_x\) and \(k(y) \subseteq \hat{\mco}_y\) so that \(k(y) \subseteq k(x)\) via the natural map \(\hat{\mco}_y \to \hat{\mco}_x\).
The condition is that for every \(x \in X\), the field \(k(x)\) is a separable algebraic extension of \(k(y)\), and the natural map \[ \hat{\mco}_y \tensor_{k(y)} k(x) \to \hat{\mco}_x \] is an isomorphism.
Solution
Fix \(x\in X\), put \(y=f(x)\), and write \[ A=\OO_{Y,y}, \qquad B=\OO_{X,x}, \qquad k_y=\kappa(y), \qquad k_x=\kappa(x). \] Let \(\mathfrak n\) and \(\mathfrak m\) be the maximal ideals of \(A\) and \(B\), respectively. Write \(\widehat A\) and \(\widehat B\) for their completions.
Choose compatible coefficient fields \[ k_y\subseteq\widehat A, \qquad k_x\subseteq\widehat B, \] as in the statement. They give the natural continuous homomorphism \[ \Phi:\widehat A\tensor_{k_y}k_x\longrightarrow\widehat B, \qquad a\tensor c\longmapsto f^\sharp(a)c. \]
1. If \(f\) is étale, then\[ \mathfrak m=\mathfrak nB \]and \(k_x/k_y\) is finite separable.
By Exercise III.10.3, an étale morphism is flat and unramified. Hartshorne’s definition of unramifiedness gives \[ \mathfrak nB=\mathfrak m \] and says that \(k_x/k_y\) is separable algebraic. Because \(B\) is essentially of finite type over \(A\), the residue-field extension is finitely generated. A finitely generated algebraic field extension is finite. Thus \(k_x/k_y\) is finite separable.
2. If \(f\) is étale, then for every \(r\ge0\) there is a natural isomorphism\[ \mathfrak m^r/\mathfrak m^{r+1} \cong (\mathfrak n^r/\mathfrak n^{r+1})\tensor_{k_y}k_x. \]
1. Tensor the exact sequence\[ 0\longrightarrow\mathfrak n^{r+1} \longrightarrow\mathfrak n^r \longrightarrow\mathfrak n^r/\mathfrak n^{r+1} \longrightarrow0 \]with \(B\). Flatness preserves exactness, so\[ (\mathfrak n^r/\mathfrak n^{r+1})\tensor_A B \cong \mathfrak n^rB/\mathfrak n^{r+1}B. \]By step 1,\[ \mathfrak n^rB=\mathfrak m^r. \]Moreover \(\mathfrak n\) annihilates\(\mathfrak n^r/\mathfrak n^{r+1}\), so tensoring with \(B\) factors through\[ B/\mathfrak nB=B/\mathfrak m=k_x. \]Hence\[ (\mathfrak n^r/\mathfrak n^{r+1})\tensor_A B \cong (\mathfrak n^r/\mathfrak n^{r+1})\tensor_{k_y}k_x, \]which proves the assertion.
3. If \(f\) is étale, the map\[ \Phi:\widehat A\tensor_{k_y}k_x\longrightarrow\widehat B \]is an isomorphism.
Because \(k_x/k_y\) is finite, the source is a finite free \(\widehat A\)-module. It is therefore complete and separated for the \(\mathfrak n\widehat A\)-adic filtration.
Completion does not change the associated graded pieces of a noetherian local ring: \[ (\mathfrak n\widehat A)^r/(\mathfrak n\widehat A)^{r+1} \cong \mathfrak n^r/\mathfrak n^{r+1}, \] and similarly for \(B\). Thus the map induced by \(\Phi\) on the \(r\)th associated graded piece is exactly the isomorphism of step 2.
It follows inductively that for every \(N\ge1\), \(\Phi\) induces an isomorphism \[ (\widehat A\tensor_{k_y}k_x)/(\mathfrak n\widehat A)^N \xrightarrow{\sim} \widehat B/(\mathfrak m\widehat B)^N. \] Indeed, pass from \(N\) to \(N+1\) using the short exact sequences whose kernels are the \(N\)th associated graded pieces. Taking inverse limits and using completeness and separatedness gives \[ \boxed{ \widehat A\tensor_{k_y}k_x\xrightarrow{\sim}\widehat B. } \] This proves the required completed-local-ring condition in the forward direction.
4. Conversely, assume that \(k_x/k_y\) is separable algebraic and that \(\Phi\) is an isomorphism. Then\[ \mathfrak m\widehat B=\mathfrak n\widehat B. \]
Again \(k_x/k_y\) is finite because it is algebraic and finitely generated. Modulo \(\mathfrak n\widehat A\), the source of \(\Phi\) is \[ (\widehat A/\mathfrak n\widehat A)\tensor_{k_y}k_x \cong k_y\tensor_{k_y}k_x \cong k_x, \] a field. Therefore \[ \mathfrak n(\widehat A\tensor_{k_y}k_x) \] is the maximal ideal of the source. Under the local-ring isomorphism \(\Phi\) it maps to the maximal ideal \(\mathfrak m\widehat B\) of \(\widehat B\). Hence \[ \boxed{\mathfrak n\widehat B=\mathfrak m\widehat B}. \]
5. Under the hypotheses of step 4, the local map \(A\to B\) is flat and satisfies\[ \mathfrak nB=\mathfrak m. \]
The completion \(\widehat A\) is flat over the noetherian local ring \(A\). Since \(k_x\) is a vector space over \(k_y\), \[ \widehat A\tensor_{k_y}k_x \] is also flat as an \(A\)-module. By the assumed isomorphism \(\Phi\), \(\widehat B\) is therefore flat over \(A\).
The completion map \[ B\longrightarrow\widehat B \] is faithfully flat. To test flatness of \(B\) over \(A\), tensor an exact sequence of \(A\)-modules first with \(B\) and then with the faithfully flat \(B\)-module \(\widehat B\). The result is exact because \(\widehat B\) is \(A\)-flat; faithful flatness then shows that the sequence after tensoring with \(B\) was already exact. Thus \(B\) is flat over \(A\).
Now step 4 gives \[ (\mathfrak m/\mathfrak nB)\tensor_B\widehat B=0. \] Faithful flatness of \(B\to\widehat B\) implies \[ \mathfrak m/\mathfrak nB=0. \] Hence \[ \boxed{\mathfrak nB=\mathfrak m}. \]
6. The completed-local-ring condition implies that \(f\) is étale at \(x\).
Step 5 proves that \(A\to B\) is flat and that \[ \mathfrak m_y\OO_{X,x}=\mathfrak m_x. \] By hypothesis \(k_x/k_y\) is separable algebraic. Thus \(f\) is unramified at \(x\) in Hartshorne’s sense. Exercise III.10.3 says that flat plus unramified is equivalent to étale. Therefore \(f\) is étale at \(x\).
7. The two conditions are equivalent globally.
If \(f\) is étale, steps <1>1–<1>3 establish the stated completed-local-ring condition for every \(x\in X\). Conversely, if that condition holds for every \(x\), steps <1>4–<1>6 show that \(f\) is étale at every point of \(X\). Hence \(f\) is étale.
8. Q.E.D.
Steps <1>1–<1>3 prove the forward implication by comparing associated graded rings, and steps <1>4–<1>7 recover flatness and unramifiedness from the completed isomorphism and descend them to the original local rings.