Convention. Throughout this series of postings, any ring is assumed to be unital and commutative (unless mentioned otherwise).
Unramified ring maps. We say that a ring map $A \rightarrow B$ is unramified if
Theorem. A finitely presented ring map $A \rightarrow B$ is unramified if and only if $\Omega^{1}_{B/A} = 0.$
Proof. Let $A \rightarrow B$ be unramified. Fix any $\mathfrak{q} \in \mathrm{Spec}(B)$ and write $\mathfrak{p} \in \mathrm{Spec}(A)$ to mean its contraction. We have $$B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{q}} \otimes_{B_{\mathfrak{q}}} \Omega^{1}_{B_{\mathfrak{q}}/A_{\mathfrak{p}}} \simeq \Omega^{1}_{(B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}})/(A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}})} = 0.$$ The vanishing occurred because $(B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}})/(A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}})$ is a finite separable field extension. The isomorphism is established because $$B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}} \simeq A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}} \otimes_{A_{\mathfrak{p}}} B_{\mathfrak{q}}.$$ Since $$(\Omega^{1}_{B/A})_{\mathfrak{q}} \simeq \Omega^{1}_{B_{\mathfrak{q}}/A_{\mathfrak{p}}}$$ and $A \rightarrow B$ is of finite type, we see that $\Omega^{1}_{B_{\mathfrak{q}}/A_{\mathfrak{p}}}$ is a finitely generated $B_{\mathfrak{q}}$-module. Thus, noting that $\mathfrak{p}B_{\mathfrak{q}} = \mathfrak{q}B_{\mathfrak{q}},$ by Nakayama's lemma, we must have $$(\Omega^{1}_{B/A})_{\mathfrak{q}} \simeq \Omega^{1}_{B_{\mathfrak{q}}/A_{\mathfrak{p}}} = 0,$$ and since $\mathfrak{q}$ was arbitrary, this implies that $\Omega^{1}_{B/A} = 0.$
Conversely, suppose that $\Omega^{1}_{B/A} = 0.$ Fix any $\mathfrak{q} \in \mathrm{Spec}(B)$ and denote again by $\mathfrak{p} \in \mathrm{Spec}(A)$ its contraction. Having $\Omega^{1}_{B/A} = 0$ implies that $\Omega^{1}_{(B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}})/(A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}})} = 0.$ As we have observed before, the map $A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}} \rightarrow B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}}$ is of finite type because $A \rightarrow B$ is. Since $B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}}$ is a local ring, our desired conclusion follows from the following lemma:
Lemma. Let $k$ be a field and $R$ a finite type $k$-algebra. Then the following are equivalent:
Proof of Lemma. It is immediate that the second implies the first applying the localization at each prime of $R,$ so we focus on proving the converse. Assume $\Omega^{1}_{R/k} = 0.$ We first claim that $\dim(R) = 0.$ We have $\dim(\bar{k} \otimes_{k} R) = \dim(R)$ (e.g., Vakil 11.1.G) so we may instead show that $\dim(\bar{k} \otimes_{k} R) = 0.$ For this, it is important to observe that $$\Omega^{1}_{(\bar{k} \otimes_{k} R)/\bar{k}} \simeq \bar{k} \otimes_{k} \Omega^{1}_{R/k} = 0.$$ Write $A := \bar{k} \otimes_{k} R$ for convenience, and note that $A$ is of finite type over $\bar{k}.$ Fix any maximal ideal $\mathfrak{m} \subset A.$ Since $\bar{k}$ is algebraically closed, we have $A/\mathfrak{m} \simeq \bar{k}.$ Therefore, we have $$\dim(A) = \dim(A_{\mathfrak{m}}) \leq \dim_{\bar{k}}(\mathfrak{m}/\mathfrak{m}^{2}) = \dim_{\bar{k}}(\Omega^{1}_{A/\bar{k}}) = 0.$$ This shows that $\dim(A) = 0$ as we claimed, which implies that $\dim(R) = 0,$ but it also shows that $\mathrm{Spec}(A)$ is regular at every closed point. Since $A$ is Noetherian, it follows that $A$ is isomorphic to a finite product of $\bar{k}$ as a $\bar{k}$-algebra. Since $$R \hookrightarrow \bar{k} \otimes_{k} R = A,$$ it follows that $R$ is a finite product of $0$-dimensional reduced Noetherian local rings. Each such local ring is regular because their base change over $\bar{k}$ is (applying Vakil 12.2.O), so it must be a field. Fix one of them and write $K$ to mean it. Since $K$ is finitely generated over $k,$ by Nullstellensatz (e.g., Vakil 3.2.5), the field $K$ must be a finite extension over $k.$ It remains to show that $K$ is separable over $k.$ To do so, fix any $\alpha \in K$ and denote by $f(x) \in k[x]$ the minimal polynomial of $\alpha$ over $k.$ We have $$\frac{\bar{k}[x]}{(f(x))} \simeq \bar{k} \otimes_{k} \frac{k[x]}{(f(x))} \simeq \bar{k} \otimes_{k} k[\alpha] \hookrightarrow \bar{k} \otimes_{k} R = A,$$ but since $A$ is reduced, the polynomial $f(x)$ must be square-free. This shows that $\alpha$ is separable over $K,$ as desired. $\Box$
- it is finitely presented, and
- for any $\mathfrak{q} \in \mathrm{Spec}(B),$ writing $\mathfrak{p} \in \mathrm{Spec}(A)$ for its contraction, the induced map $A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}} \rightarrow B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}}$ is a finite separable field extension.
Remark. Note that requiring that $B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}}$ is a field guarantees that $\mathfrak{p}B_{\mathfrak{q}}$ is a maximal ideal of $B_{\mathfrak{q}},$ which necessarily means that $\mathfrak{p}B_{\mathfrak{q}} = \mathfrak{q}B_{\mathfrak{q}}.$
Remark. It is often the case that $A_{\mathfrak{p}} \rightarrow B_{\mathfrak{q}}$ is not of finite type in this scenario. However, note that that $A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}} \rightarrow B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}}$ is the base change of the map $A \rightarrow B$ under $A \rightarrow A_{\mathfrak{p}} \rightarrow A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}},$ so (without any of the two conditions above)
Remark. It is often the case that $A_{\mathfrak{p}} \rightarrow B_{\mathfrak{q}}$ is not of finite type in this scenario. However, note that that $A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}} \rightarrow B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}}$ is the base change of the map $A \rightarrow B$ under $A \rightarrow A_{\mathfrak{p}} \rightarrow A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}},$ so (without any of the two conditions above)
- if $A \rightarrow B$ is of finite type, then so is $A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}} \rightarrow B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}}$;
- if $A \rightarrow B$ is finitely presented, then so is $A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}} \rightarrow B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}}.$
Theorem. A finitely presented ring map $A \rightarrow B$ is unramified if and only if $\Omega^{1}_{B/A} = 0.$
Proof. Let $A \rightarrow B$ be unramified. Fix any $\mathfrak{q} \in \mathrm{Spec}(B)$ and write $\mathfrak{p} \in \mathrm{Spec}(A)$ to mean its contraction. We have $$B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{q}} \otimes_{B_{\mathfrak{q}}} \Omega^{1}_{B_{\mathfrak{q}}/A_{\mathfrak{p}}} \simeq \Omega^{1}_{(B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}})/(A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}})} = 0.$$ The vanishing occurred because $(B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}})/(A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}})$ is a finite separable field extension. The isomorphism is established because $$B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}} \simeq A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}} \otimes_{A_{\mathfrak{p}}} B_{\mathfrak{q}}.$$ Since $$(\Omega^{1}_{B/A})_{\mathfrak{q}} \simeq \Omega^{1}_{B_{\mathfrak{q}}/A_{\mathfrak{p}}}$$ and $A \rightarrow B$ is of finite type, we see that $\Omega^{1}_{B_{\mathfrak{q}}/A_{\mathfrak{p}}}$ is a finitely generated $B_{\mathfrak{q}}$-module. Thus, noting that $\mathfrak{p}B_{\mathfrak{q}} = \mathfrak{q}B_{\mathfrak{q}},$ by Nakayama's lemma, we must have $$(\Omega^{1}_{B/A})_{\mathfrak{q}} \simeq \Omega^{1}_{B_{\mathfrak{q}}/A_{\mathfrak{p}}} = 0,$$ and since $\mathfrak{q}$ was arbitrary, this implies that $\Omega^{1}_{B/A} = 0.$
Conversely, suppose that $\Omega^{1}_{B/A} = 0.$ Fix any $\mathfrak{q} \in \mathrm{Spec}(B)$ and denote again by $\mathfrak{p} \in \mathrm{Spec}(A)$ its contraction. Having $\Omega^{1}_{B/A} = 0$ implies that $\Omega^{1}_{(B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}})/(A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}})} = 0.$ As we have observed before, the map $A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}} \rightarrow B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}}$ is of finite type because $A \rightarrow B$ is. Since $B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}}$ is a local ring, our desired conclusion follows from the following lemma:
Lemma. Let $k$ be a field and $R$ a finite type $k$-algebra. Then the following are equivalent:
- $\Omega^{1}_{R/k} = 0$;
- $R$ is a finite product of fields that are finite separable over $k.$
Proof of Lemma. It is immediate that the second implies the first applying the localization at each prime of $R,$ so we focus on proving the converse. Assume $\Omega^{1}_{R/k} = 0.$ We first claim that $\dim(R) = 0.$ We have $\dim(\bar{k} \otimes_{k} R) = \dim(R)$ (e.g., Vakil 11.1.G) so we may instead show that $\dim(\bar{k} \otimes_{k} R) = 0.$ For this, it is important to observe that $$\Omega^{1}_{(\bar{k} \otimes_{k} R)/\bar{k}} \simeq \bar{k} \otimes_{k} \Omega^{1}_{R/k} = 0.$$ Write $A := \bar{k} \otimes_{k} R$ for convenience, and note that $A$ is of finite type over $\bar{k}.$ Fix any maximal ideal $\mathfrak{m} \subset A.$ Since $\bar{k}$ is algebraically closed, we have $A/\mathfrak{m} \simeq \bar{k}.$ Therefore, we have $$\dim(A) = \dim(A_{\mathfrak{m}}) \leq \dim_{\bar{k}}(\mathfrak{m}/\mathfrak{m}^{2}) = \dim_{\bar{k}}(\Omega^{1}_{A/\bar{k}}) = 0.$$ This shows that $\dim(A) = 0$ as we claimed, which implies that $\dim(R) = 0,$ but it also shows that $\mathrm{Spec}(A)$ is regular at every closed point. Since $A$ is Noetherian, it follows that $A$ is isomorphic to a finite product of $\bar{k}$ as a $\bar{k}$-algebra. Since $$R \hookrightarrow \bar{k} \otimes_{k} R = A,$$ it follows that $R$ is a finite product of $0$-dimensional reduced Noetherian local rings. Each such local ring is regular because their base change over $\bar{k}$ is (applying Vakil 12.2.O), so it must be a field. Fix one of them and write $K$ to mean it. Since $K$ is finitely generated over $k,$ by Nullstellensatz (e.g., Vakil 3.2.5), the field $K$ must be a finite extension over $k.$ It remains to show that $K$ is separable over $k.$ To do so, fix any $\alpha \in K$ and denote by $f(x) \in k[x]$ the minimal polynomial of $\alpha$ over $k.$ We have $$\frac{\bar{k}[x]}{(f(x))} \simeq \bar{k} \otimes_{k} \frac{k[x]}{(f(x))} \simeq \bar{k} \otimes_{k} k[\alpha] \hookrightarrow \bar{k} \otimes_{k} R = A,$$ but since $A$ is reduced, the polynomial $f(x)$ must be square-free. This shows that $\alpha$ is separable over $K,$ as desired. $\Box$
Remark. Note that in the above proof, we have accidentally showed that for a finite type algebra $R$ over a field $k,$ the following are equivalent:
- $R$ is unramified over $k$;
- $R$ is a finite prodcut finite separable field extension of $k$;
- $\bar{k} \otimes_{k} R$ is a finite product of $\bar{k}.$
A ring map $A \rightarrow B$ is called unramified if it is of finite type and for every $\mathfrak{q} \in \mathrm{Spec}(B),$ writing $\mathfrak{p} \in \mathrm{Spec}(A)$ to mean its contraction, the induced local ring map $A_{\mathfrak{p}} \rightarrow B_{\mathfrak{q}}$ is unramified.
Remark. The proof for Theorem works if we replace "finitely presented" to "finite type". Some people (e.g., Vakil 21.6) use this more general definition for the definition of unramified maps. In (the first chapter of) Freitag and Kiehl, this is moot, because rings are assumed to be Noetherian.
Remark. The criterion for unramified ring maps using the module of differential $1$-forms is quite useful. For instance, let's show the following:
Lemma. The composition of two unramified ring maps is unramified.
Proof. Let $A \rightarrow B \rightarrow C$ be two unramified ring maps. Then consider the relative cotangent exact sequence (e.g., Vakil 21.2.9) $$C \otimes_{B} \Omega^{1}_{B/A} \rightarrow \Omega^{1}_{C/A} \rightarrow \Omega^{1}_{C/B} \rightarrow 0.$$ Our hypothesis guarantees that $\Omega^{1}_{B/A} = 0$ and $\Omega^{1}_{C/B} = 0,$ so it follows that $\Omega^{1}_{C/A} = 0.$ Since being finite type is preserved under taking compositions, the composition in question is unramified. $\Box$
Lemma. Let $A \rightarrow B$ be an unramified ring map. Given any ring map $A \rightarrow C,$ the induced map $C \rightarrow C \otimes_{A} B$ is unramified.
Proof. It is evident that $C \rightarrow C \otimes_{A} B$ is of finite type when $A \rightarrow B$ is. Since $$\Omega^{1}_{(C \otimes_{A} B)/C} \simeq (C \otimes_{A} B) \otimes_{B} \Omega^{1}_{B/A} = 0,$$ as $\Omega^{1}_{B/A} = 0$ by hypothesis, we are done. $\Box$
Lemma (unramified cancellation). Let $A \rightarrow B$ and $B \rightarrow C$ be finite type ring maps. If $A \rightarrow B \rightarrow C$ is unramified, then so is $B \rightarrow C.$
Proof. This immediately follows from the exact sequence $$C \otimes_{B} \Omega^{1}_{B/A} \rightarrow \Omega^{1}_{C/A} \rightarrow \Omega^{1}_{C/B} \rightarrow 0,$$ although we do not need the first term. $\Box$
We say a ring map $A \rightarrow B$ is étale if it satisfies the following two properties:
- it is flat;
- it is unramified.
It is immediate that the composition of two étale ring maps is étale.
Lemma. Let $A \rightarrow B$ be an étale ring map. Given any ring map $A \rightarrow C,$ the induced map $C \rightarrow C \otimes_{A} B$ is étale.
This follows from the following:
Lemma. Let $A \rightarrow B$ be an flat ring map. Given any ring map $A \rightarrow C,$ the induced map $C \rightarrow C \otimes_{A} B$ is flat.
Proof. This immediately follows from $M \otimes_{C} C \otimes_{A} B \simeq M \otimes_{A} B,$ for any $C$-module $M.$ $\Box$
Theorem (étale cancellation). Let $A \rightarrow B$ and $B \rightarrow C$ be finite type ring maps. If $A \rightarrow B$ and $A \rightarrow B \rightarrow C$ are étale, then so is $B \rightarrow C.$
Proof. This follows from the unramified cancellation and an analogous statement about flat ring maps. $\Box$
Remark. The statement about flat ring maps can be done painlessly with even greater generality by citing Vakil 10.1.19 and Vakil 21.2.F, the latter of which we will discuss in the third lecture.
Remark. The statement about flat ring maps can be done painlessly with even greater generality by citing Vakil 10.1.19 and Vakil 21.2.F, the latter of which we will discuss in the third lecture.
Étale algebras over a field. Let's classify all étale algebras $A$ over a field $k$ (i.e., all étale ring maps $k \rightarrow A$). For any $k$-algebra $A,$ we can see $A$ as a vector space over a field $k,$ which implies that it is free over $k,$ so it must be flat over $k.$ Hence, we immediately see that $A$ is unramified over $k$ if and only if $A$ is étale over $k.$ Thus, we have already shown the following above:
Theorem. Let $A$ be any algebra over a field $k.$ The following are equivalent:
- $A$ is étale over $k$;
- $A$ is unramified over $k$;
- $A$ is a finite product of finite separable field extensions of $k$;
- $\bar{k} \otimes_{k} A$ is a finite product of $\bar{k}.$
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