Computing differential 1-forms of an algebra

We follow Chapter 2 of Bosch, Lütkebohmert, and Raynaud and notes by Hochster, as well as other references such as Vakil. By a ring, we will mean a commutative unital ring.

Let $R$ be a ring and $A$ an $R$-algebra. Given an $A$-module $M,$ an $R$-linear map $D : A \rightarrow M$ such that $$D(fg) = fD(g) + gD(f)$$ for all $f, g \in A$ is called an $R$-derivation of $A$ into $M$.

Remark. Note that $D(1) = D(1) + D(1)$ so that $D(1) = 0.$ Hence, for any $r \in R,$ we have $D(r) = D(r \cdot 1) = rD(1) = 0.$ That is, the derivative of any constant (i.e., element of $R$) is zero!

Warning. The only $R$-derivation $D : A \rightarrow M$ that is $A$-linear is the zero map, due to a similar argument as above. Whenever I get confused, I call $D$ an $R$-linear derivation instead.

We denote by $\mathrm{Der}_{R}(A, M)$ the set of all $R$-derivations of $A$ into $M,$ which is also an $A$-module given by the $A$-module structure of $M.$ The module of relative differential forms (of degree 1) of $A$ over $R$ is an $A$-module $\Omega^{1}_{A/R}$ with an $R$-derivation $d = d_{A/R} : A \rightarrow \Omega^{1}_{A/R},$ which is called the exterior differential, such that $$\mathrm{Hom}_{A}(\Omega^{1}_{A/R}, M) \simeq \mathrm{Der}_{R}(A, M)$$ in $\textbf{Set}$ given by $\phi \mapsto \phi \circ d.$ That is, the functor $\mathrm{Der}_{R}(A, -) : \textbf{Mod}_{A} \rightarrow \textbf{Set}$ is representable. Of course, this functor can be seen as $\textbf{Mod}_{A} \rightarrow \textbf{Mod}_{A},$ but we do not need this. It is abstract nonsense to show that such $d$ is unique, and one may construct it as follows.

Construction of the exterior differential. A formal construction of $d : A \rightarrow \Omega^{1}_{A/R}$ for the sake of existence is quite easy. That is, we take $$\Omega^{1}_{A/R} = \frac{\bigoplus_{a \in A} A da}{(d(a + b) - da - db, d(ab) - adb - bda, d(ra) - rda)_{a, b \in A, r \in R}}.$$ Taking $d : A \rightarrow \Omega^{1}_{A/R}$ by $a \mapsto da,$ we get what we need.

In practice, such a formal construction is quite useless, so we give better descriptions of $d : A \rightarrow \Omega^{1}_{A/R}.$

Lemma. Let $A := R[t_{i}]_{i \in I},$ a free $R$-algebra. Then we have $\Omega^{1}_{A/R} = \bigoplus_{i \in I} A dt_{i},$ where $d : A \rightarrow \bigoplus_{i \in I} A dt_{i}$ is given by $$df = \sum_{i \in I}\frac{\partial f}{\partial t_{i}} dt_{i}.$$ That is, the exterior derivative is given by taking the gradient.

Proof. Using the product rule for partial derivatives of polynomials, we check that $d$ is indeed an $R$-derivation. To check the universal property, consider any $R$-linear derivation $D : A \rightarrow M,$ the only way to define $A$-module map $\phi : \Omega^{1}_{A/R} \rightarrow M$ such that $D = \phi \circ d$ is to send $df$ to $Df,$ so in particular, we send $dt_{i}$ to $Dt_{i},$ which actually constructs such $\phi$ as $\Omega^{1}_{A/R}$ is free $A$-algebra over such elements $dt_{i}.$ This finishes the construction for this case. $\Box$

In general, an $R$-algebra $A$ is isomorphic to an $R$-algebra of the form $$R[t_{i}]_{i \in I}/\mathfrak{b},$$ where $\mathfrak{b} \subset R[t_{i}]_{i \in I}$ is an ideal. The next statement computes $d : A \rightarrow \Omega^{1}_{A/R}$ with respect to such a presentation. Now, the following, together with Lemma computes the exterior derivation in this case.

Theorem.  Let $$A := B/\mathfrak{b},$$ where $B$ is an $R$-algebra and $\mathfrak{b} \subset B$ is an ideal. Then we can construct $$\Omega^{1}_{A/R} = \frac{\Omega^{1}_{B/R}}{\mathfrak{b}\Omega^{1}_{B/R} + Bd_{B/R}\mathfrak{b}},$$ where $d_{A/R} : A \rightarrow \Omega^{1}_{A/R}$ is given by $\bar{b} \mapsto \overline{d_{B/R}b}.$

Proof. Let $\Omega^{1}_{A/R}$ be as described in the statement. Then $\mathfrak{b}\Omega^{1}_{A/R} = 0,$ so $\Omega^{1}_{A/R}$ is a module over $A = B/\mathfrak{b}.$ The map $B \rightarrow \Omega^{1}_{A/R}$ given by $b \mapsto \overline{d_{B/R}b}$ is a $B$-linear map that kills $\mathfrak{b},$ so the induced map $d_{A/R} : \bar{b} \mapsto \overline{d_{B/R}b}$ is a well-defined module map over $A = B/\mathfrak{b}.$

To check the universal property, let $D : A \rightarrow M$ be any $R$-derivation. This is the same as saying that $M$ is a $B$-module such that $\mathfrak{b}M = 0$ and that we have an $R$-derivation $\tilde{D} : B \rightarrow M$ such that $\tilde{D}\mathfrak{b} = 0,$ given by $\tilde{D}b = D\bar{b}.$ Hence, there is a unique $B$-module map $\tilde{\phi} : \Omega^{1}_{B/R} \rightarrow M$ such that $d_{R/B}b \mapsto \tilde{D}b = D\bar{b}.$ This map kills $\mathfrak{b}\Omega^{1}_{B/R} + Bd_{B/R}\mathfrak{b},$ which gives an $A$-module map $\phi : \Omega^{1}_{A/R} \rightarrow M$ such that $\phi \circ d_{A/R} = D.$ The uniqueness of $\phi$ easily follows, which finishes the proof. $\Box$

Application. If $A = B/\mathfrak{b}$ with $B = R[x_{i}]_{i \in I},$ then we have $$\Omega^{1}_{B/R} = \bigoplus_{i \in I} B dx_{i}$$ so that $$\frac{\Omega^{1}_{B/R}}{\mathfrak{b}\Omega^{1}_{B/R}} = \frac{\bigoplus_{i \in I} B dx_{i}}{\bigoplus_{i \in I} \mathfrak{b}B dx_{i}} \simeq \bigoplus_{i \in I} A dx_{i}.$$ We can write $\mathfrak{b} = (f_{j})_{j \in J}$ for some family $f_{j} \in B$ so that $$d_{B/R}\mathfrak{b} = \{df_{j}\}_{j \in J} = \left\{\sum_{i \in I}\frac{\partial f_{j}}{\partial x_{i}} dx_{i}\right\}_{j \in J}.$$ This implies that under the above isomorphism, we have $$\frac{\mathfrak{b}\Omega^{1}_{B/R} + Bd_{B/R}\mathfrak{b}}{\mathfrak{b}\Omega^{1}_{B/R}} \simeq A\left\{\sum_{i \in I}\frac{\partial f_{j}}{\partial x_{i}} dx_{i}\right\}_{j \in J}.$$ Therefore, we may write $$\Omega^{1}_{A/R} = \bigoplus_{i \in I} A dx_{i} \Biggm/ A\left\{\sum_{i \in I}\frac{\partial f_{j}}{\partial x_{i}} dx_{i}\right\}_{j \in J}.$$ This is the content of Key fact 21.2.3 in Vakil. The author uses the last description to construct $\Omega^{1}_{A/R}.$

Jacobian description of differential $1$-forms. This is from 21.2.E in Vakil. Again, when $A = B/\mathfrak{b},$ where $B$ is an $R$-algebra and $\mathfrak{b} \subset B$ is an ideal, we have $$\Omega_{A/R}^{1} = \frac{\Omega^{1}_{B/R}}{\mathfrak{b}\Omega^{1}_{B/R} + Bd_{B/R}\mathfrak{b}},$$ and we saw this formally, but let's try to imbue some context now.

Let $B = k[x_{1}, \dots, x_{n}],$ where $k$ is a field so that we may write $\mathfrak{b} = (f_{1}, \dots, f_{r})$ for some $f_{j} \in B.$ That is, we consider the case $$A = \frac{k[x_{1}, \dots, x_{n}]}{(f_{1}, \dots, f_{r})}$$ a finitely presented algebra over $k.$ We have $$\Omega_{B/R}^{1} = B dx_{1} \oplus \cdots \oplus B dx_{n},$$ so $$\begin{align*} \frac{\Omega_{B/R}^{1}}{\mathfrak{b}\Omega_{B/R}^{1}} &= \frac{B dx_{1} \oplus \cdots \oplus B dx_{n}}{\mathfrak{b} dx_{1} \oplus \cdots \oplus \mathfrak{b} dx_{n}} \\ &\simeq (B/\mathfrak{b}) dx_{1} \oplus \cdots \oplus (B/\mathfrak{b}) dx_{n} \\ &= A dx_{1} \oplus \cdots \oplus A dx_{n}. \end{align*}$$ Since $$\Omega_{A/R}^{1} = \frac{\Omega_{B/R}^{1}}{\mathfrak{b}\Omega_{B/R}^{1} + Bd_{B/R}\mathfrak{b}} \simeq \frac{\Omega_{B/R}^{1}/\mathfrak{b}\Omega_{B/R}^{1}}{d_{B/R}\mathfrak{b}(\Omega_{B/R}^{1}/\mathfrak{b}\Omega_{B/R}^{1})},$$ we may write $$\Omega_{A/R}^{1} = \frac{A dx_{1} \oplus \cdots \oplus A dx_{n}}{(df_{1}, \dots, df_{r})}.$$ Note that $$df_{j} = \frac{\partial f_{j}}{\partial x_{1}} dx_{1} + \cdots + \frac{\partial f_{j}}{\partial x_{n}} dx_{n}$$ in $A dx_{1} \oplus \cdots \oplus A dx_{n}.$

As a result, we have the exact sequence $$A dy_{1} \oplus \cdots \oplus A dy_{r} \xrightarrow{J} A dx_{1} \oplus \cdots \oplus A dx_{n} \rightarrow \Omega_{A/R}^{1} \rightarrow 0,$$ where $J$ can be described as an $A$-linear map $J : A^{\oplus r} \rightarrow A^{\oplus n}$ given by the following Jacobian matrix $$J = \left[ \frac{\partial f_{j}}{\partial x_{i}} \right]_{1 \leq i, j \leq n}$$ whose entries are in $A = k[x_{1}, \dots, x_{n}]/(f_{1}, \dots, f_{r}).$

Localization. If we have any ring map $B \rightarrow A$ and $S \subset A$ is a mulitplicative submonoid, then we have $$\Omega^{1}_{S^{-1}A/B} \simeq S^{-1}\Omega^{1}_{A/B},$$ as $S^{-1}A$-module. Moreover, one may check that the map $$S^{-1}A \rightarrow S^{-1}\Omega^{1}_{A/B}$$ given by $$\frac{a}{s} \mapsto \frac{sd_{A/B}(a) - ad_{A/B}(s)}{s^{2}}$$ is a well-defined $B$-linear derivation that is compatible with $$d_{S^{-1}A/B} : S^{-1}A \rightarrow \Omega^{1}_{S^{-1}A/B}$$ and the isomorphism given above.

Example. Let $k$ be a field. We have $$\Omega^{1}_{k(t_{1}, \dots, t_{n})/k} = k(t_{1}, \dots, t_{n})dt_{1} \oplus \cdots \oplus k(t_{1}, \dots, t_{n})dt_{n},$$ whose exterior derivative can be given by the quotient rule.

Example. Again, let $k$ be a field and let $A := k(t_{1}, \dots, t_{n})[x_{1}, \dots, x_{m}].$ We want to compute $\Omega^{1}_{A/k}.$ We have $$A = k(t_{1}, \dots, t_{n})[x_{1}, \dots, x_{m}] = S^{-1}k[t_{1}, \dots, t_{n}, x_{1}, \dots, x_{m}],$$ where $$S = k[t_{1}, \dots, t_{n}] \setminus (0) \subset k[t_{1}, \dots, t_{n}] \subset k[t_{1}, \dots, t_{n}, x_{1}, \dots, x_{m}].$$ Hence, we have $$\begin{align*}\Omega^{1}_{A/k} &= S^{-1}\Omega^{1}_{k[t_{1}, \dots, t_{n}, x_{1}, \dots, x_{m}]/k} \\ &= S^{-1}\left(\bigoplus_{i=1}^{m}k[t_{1}, \dots, t_{n}, x_{1}, \dots, x_{m}]dt_{i} \oplus \bigoplus_{j=1}^{n}k[t_{1}, \dots, t_{n}, x_{1}, \dots, x_{m}]dx_{j}\right) \\ &= \bigoplus_{i=1}^{m}k(t_{1}, \dots, t_{n})[x_{1}, \dots, x_{m}]dt_{i} \oplus \bigoplus_{j=1}^{n}k(t_{1}, \dots, t_{n})[x_{1}, \dots, x_{m}]dx_{j}, \end{align*}$$ which is quite concrete.

Example. We keep using $k$ to denote the base field. Then consider $$A = \frac{k(t_{1}, \dots, t_{n})[x_{1}, \dots, x_{n}]}{(f_{1}(\boldsymbol{t}, x_{1}, \dots, x_{n}), \cdots, f_{r}(\boldsymbol{t}, x_{1}, \dots, x_{n}))}.$$ Write $$B = k(t_{1}, \dots, t_{n})[x_{1}, \dots, x_{n}]$$ and $$\mathfrak{b} = (f_{1}(\boldsymbol{t}, x_{1}, \dots, x_{n}), \cdots, f_{r}(\boldsymbol{t}, x_{1}, \dots, x_{n})).$$ We have $$\Omega^{1}_{B/k} = \bigoplus_{i=1}^{m}k(\boldsymbol{t})[\boldsymbol{x}]dt_{i} \oplus \bigoplus_{j=1}^{n}k(\boldsymbol{t})[\boldsymbol{x}]dx_{j}$$ and $$\begin{align*}d_{B/k}\mathfrak{b} &= \{df_{1}, \dots, df_{r}\} \\ &= \left\{ \sum_{i=1}^{m}(f_{l})_{t_{i}} dt_{i} + \sum_{j=1}^{n}(f_{l})_{x_{j}} dx_{i} \right\}_{l=1}^{r}\end{align*}.$$ This lets us compute $$\begin{align*} \Omega^{1}_{A/k} &= \frac{\Omega^{1}_{B/k}}{\mathfrak{b}\Omega^{1}_{B/k} + Bd_{B/k}\mathfrak{b}} \\ &\simeq \frac{\Omega^{1}_{B/k}/\mathfrak{b}\Omega^{1}_{B/k}}{(\mathfrak{b}\Omega^{1}_{B/k} + Bd_{B/k}\mathfrak{b})/\mathfrak{b}\Omega^{1}_{B/k}} \\ &\simeq \frac{\bigoplus_{i=1}^{m}Adt_{i} \oplus \bigoplus_{j=1}^{n}Adx_{j} }{A\left\{ \sum_{i=1}^{m}(f_{l})_{t_{i}} dt_{i} + \sum_{j=1}^{n}(f_{l})_{x_{j}} dx_{j} \right\}_{l=1}^{r}}, \end{align*}$$ which is very explicit.

Diagonal description. Consider a scheme map $\mathrm{Spec}(A) \rightarrow \mathrm{Spec}(R)$ between affine schemes. The diagonal map $\Delta_{A/R} : \mathrm{Spec}(A) \rightarrow \mathrm{Spec}(A) \times_{R} \mathrm{Spec}(A)$ is the scheme map induced by the ring map $A \otimes_{R} A \rightarrow A$ given by $a \otimes a' \mapsto aa'$ (as also remarked the proof of Proposition 10.1.3 in Vakil.) Denote by $I$ the kernel of this ring map, and consider the map $d : A \rightarrow I/I^{2}$ given by $f \mapsto 1 \otimes f - f \otimes 1$ modulo $I^{2}.$

Remark. The map $A \otimes A \rightarrow A$ given above is called the multiplication map of $A$ over $R,$ because it is the $R$-linear map corresponding to the actual multiplication map $A \times A \rightarrow A.$ Indeed, we have $1 \otimes f - f \otimes 1 \in I$ because $$1 \otimes f - f \otimes 1 \mapsto f - f = 0$$ under the multiplication map.

Since $I$ kills $I/I^{2},$ we see that $I/I^{2}$ is a module over $(A \otimes_{R} A)/I.$ Since the map $A \otimes_{R} A \rightarrow A$ is surjective, we have $(A \otimes_{R} A)/I \simeq A.$ This makes $I/I^{2}$ an $A$-module. Note that $a \otimes 1 = 1 \otimes a$ in $(A \otimes_{R} A)/I$ both corresponding to $a \in A.$ Hence, we have $$a \cdot (1 \otimes f - f \otimes 1) = a \otimes f - (af) \otimes 1 =  1 \otimes (af) - f \otimes a$$ in $I/I^{2},$ as we admittedly omitted bar notations. The map $d : A \rightarrow I/I^{2}$ is an $R$-module map because $$\begin{align*}d(rf) &= 1 \otimes (rf) - (rf) \otimes 1 \\ &= r \otimes f - (rf) \otimes 1 \\ &= r \cdot (1 \otimes f) - r \cdot (f \otimes 1) \\ &= r \cdot (1 \otimes f - f \otimes 1) \\ &= r \cdot df\end{align*}$$ in $I/I^{2}.$

Moreover, we note that $d : A \rightarrow I/I^{2}$ is an $R$-derivation as we can check $$\begin{align*}d(fg) &= 1 \otimes (fg) - (fg) \otimes 1 \\ &= (1 \otimes f)(1 \otimes g) - (f \otimes 1)(g \otimes 1) \\ &= (1 \otimes f)(1 \otimes g) - (1 \otimes f)(g \otimes 1) + (g \otimes 1)(1 \otimes f) - (g \otimes 1)(f \otimes 1) \\ &= f \cdot (1 \otimes g -  g \otimes 1) + g \cdot (1 \otimes f - f \otimes 1) \\ &= f \cdot d(g) + g \cdot d(f), \end{align*}$$ in $I/I^{2},$ where again, we omitted many bars in the middle.

Theorem. The map $d : A \rightarrow I/I^{2}$ also describes the exterior derivative $d_{A/R} : A \rightarrow \Omega^{1}_{A/R}.$

Proof. We may assume $A = R[x_{i}]_{i \in S}/(f_{j})_{j \in T}.$ Then $$A \otimes_{R} A \simeq \frac{R[x_{i}, y_{i}]_{i \in S}}{(f_{j}(\boldsymbol{x}), f_{j}(\boldsymbol{y}))_{j \in T}} = \frac{R[x_{i}, \Delta_{i}]_{i \in S}}{(f_{j}(\boldsymbol{x}), f_{j}(\boldsymbol{x} + \boldsymbol{\Delta}))_{j \in T}},$$ where we used the following change of variables: $\Delta_{i} = y_{i} - x_{i}.$ In this presentation, the multiplication map $A \otimes_{R} A \rightarrow A$ is given by the map $$\frac{R[x_{i}, \Delta_{i}]_{i \in S}}{(f_{j}(\boldsymbol{x}), f_{j}(\boldsymbol{x} + \boldsymbol{\Delta}))_{j \in T}} \rightarrow \frac{R[x_{i}]_{i \in S}}{(f_{j}(\boldsymbol{x}))_{j \in T}}$$ by $x_{i} \mapsto x_{i}$ and $\Delta_{i} \mapsto 0.$ Hence, by inspection, we can see that the kernel $I$ can be written as $$I = (\overline{\boldsymbol{\Delta}}) = \frac{(\boldsymbol{\Delta}, f_{j}(\boldsymbol{x}), f_{j}(\boldsymbol{x} + \boldsymbol{\Delta}))_{j \in T}}{(f_{j}(\boldsymbol{x}), f_{j}(\boldsymbol{x} + \boldsymbol{\Delta}))_{j \in T}},$$ so $$I/I^{2} \simeq \frac{(\boldsymbol{\Delta}, f_{j}(\boldsymbol{x}), f_{j}(\boldsymbol{x} + \boldsymbol{\Delta}))_{j \in T}}{({\boldsymbol{\Delta}}^{2}, f_{j}(\boldsymbol{x}), f_{j}(\boldsymbol{x} + \boldsymbol{\Delta}))_{j \in T}}.$$ Now, note that for any $f(\boldsymbol{x}) \in R[x_{i}]_{i \in S},$ we have $$f(\boldsymbol{x} + \boldsymbol{\Delta}) = f(\boldsymbol{x}) + \sum_{i \in S}\frac{\partial f(\boldsymbol{x})}{\partial x_{i}} \Delta_{i} + \sum_{i, i' \in S}\Delta_{i}\Delta_{i'}g_{i,i'}(\boldsymbol{x}, \boldsymbol{\Delta})$$ in $R[x_{i}, \Delta_{i}]_{i \in S}$ for suitable $g_{i,i'}(\boldsymbol{x}, \boldsymbol{\Delta}),$ which are in fact zero for all but finitely many $(i,i') \in S^{2}.$ In particular, this implies that $$({\boldsymbol{\Delta}}^{2}, f_{j}(\boldsymbol{x}), f_{j}(\boldsymbol{x} + \boldsymbol{\Delta}))_{j \in T} = \left({\boldsymbol{\Delta}}^{2}, f_{j}(\boldsymbol{x}), \sum_{i \in I} \frac{\partial f_{j}(\boldsymbol{x})}{\partial x_{i}} \Delta_{i}\right)_{j \in T}.$$ Write $B = R[x_{i}]_{i \in S} = R[\boldsymbol{x}]$ and $\mathfrak{b} = (f_{j}(\boldsymbol{x}))_{j \in T}.$ Then we can consider the surjective $B$-linear map $$\Omega^{1}_{B/R} = \bigoplus_{i \in I} B dx_{i} \twoheadrightarrow I/I^{2}$$ defined by $dx_{i} \mapsto \overline{\Delta_{i}}.$

It's time to compute its kernel. Consider a general element $$\sum_{i \in I}g_{i}(\boldsymbol{x}) dx_{i} \mapsto 0$$ under this map. This implies that $$\sum_{i \in I}g_{i}(\boldsymbol{x}) \Delta_{i} \in \left({\boldsymbol{\Delta}}^{2}, f_{j}(\boldsymbol{x}), \sum_{i \in S} \frac{\partial f_{j}(\boldsymbol{x})}{\partial x_{i}} \Delta_{i}\right)_{j \in T}$$ in $R[\boldsymbol{x}, \boldsymbol{\Delta}],$ so we may write $$\sum_{i \in S}g_{i}(\boldsymbol{x}) \Delta_{i} = \sum_{i,i' \in S}h_{i,i'}(\boldsymbol{x}, \boldsymbol{\Delta}) \Delta_{i}\Delta_{i'} + \sum_{j \in S}h_{j}(\boldsymbol{x}, \boldsymbol{\Delta})f_{j}(\boldsymbol{x}) + \sum_{j \in S}\sum_{i \in S} \frac{\partial f_{j}(\boldsymbol{x})}{\partial x_{i}} \Delta_{i},$$ so $$g_{i}(\boldsymbol{x}) = \frac{\partial f_{j}(\boldsymbol{x})}{\partial x_{i}} + \text{ some element in } (f_{j}(\boldsymbol{x}))_{j \in T} = \mathfrak{b}$$ in $R[\boldsymbol{x}] = B.$ Conversely, for any such $g_{i}(\boldsymbol{x}),$ the sum $\sum_{i \in S}g_{i}(\boldsymbol{x}) dx_{i}$ is in the kernel. This implies that the kernel is precisely $$\mathfrak{b}\Omega_{B/R} + Bd_{B/R}\mathfrak{b},$$ so we now have the $B$-lienar isomorphism $$\Omega^{1}_{A/R} = \frac{\Omega^{1}_{B/R}}{\mathfrak{b}\Omega_{B/R} + Bd_{B/R}\mathfrak{b}} \simeq I/I^{2}.$$ We note that the $R$-linear derivation $A = R[\boldsymbol{x}]/(f_{j}(\boldsymbol{x}))_{j \in T} \rightarrow I/I^{2}$ is given by $g \mapsto 1 \otimes g - g \otimes 1$ can be explicitly described as $$g(\boldsymbol{x}) \mapsto g(\boldsymbol{x} + \boldsymbol{\Delta}) - g(\boldsymbol{x}) = \sum_{i \in I}\frac{\partial g(\boldsymbol{x})}{\partial x_{i}} \Delta_{i},$$ which shows that it must be the exterior derivative. $\Box$