If $\delta \colon R[x_1, \dotsc, x_r] \to M$ is an $R$-derivation, then $\delta P = \sum_{i = 1}^r \frac{\partial P}{\partial x_i} dx_i$

Question

Let $\phi \colon R \rightarrow S$ be a ring homomorphism and $M$ an $S$-module. An $R$-derivation $\delta \colon S \rightarrow M$ is a $R$-module homomorphism such that:

1. $\phi(R) \subset \ker(\delta)$,

2. (Leibniz’ rule:) $\delta(s s') = s \delta s' + s' \delta s$ for all $s, s' \in S$.

Then, according to the exercises on my lecture notes, if $\delta \colon R[x_1, \dotsc, x_r] \rightarrow M$ is an $R$-derivation, then $\delta P = \sum_{i = 1}^r \frac{\partial P}{\partial x_i} dx_i$. I’m not totally sure what they mean by $dx_i$ here, since I think they don’t have defined the operator $d$ as far as I know. I guess it should be just the same as $\delta x_i$?

I’d like to prove this. I thought of doing it by induction: for the case $r = 1$, it is fairly easy to see using the above properties that $\delta P = \frac{\partial P}{\partial x} \delta x$. Then, if we write a multivariable polynomial as $\sum_{j = 0}^k a_j x^j_{n + 1}$, where $a_j \in R[x_1, \dotsc, x_n]$, we may assume that the result holds for $n$ and try to prove it for $n + 1$: $$ \delta P = \delta\Biggl( \sum_{j = 0}^k a_j x^j_{n+1} \Biggr) = \sum_{j = 0}^k \Bigl( a_j \delta(x_{n+1}^j) + \delta(a_j) x_{n+1}^j \Bigr) $$ … where we may use that $\delta(x_{n+1}^j) = j x_{n+1}^{j-1} \delta(x_{n+1})$ and that $\delta(a_j) = \sum_{i = 1}^n \frac{\partial{a_j}}{\partial{x_i}}dx_i$ (because of the induction hypothesis), but I don’t really know where to go from there.

I guess it’s a matter of manipulating expressions with polynomials in several variables, but I’m also not entirely sure what the exercise expects me to prove (as I don’t know what “$dx_i$” is).

Answer 1

Let $$ be a commutative ring, let $A$ be a commutative $$-algebra and let $M$ be an $A$-module. We abbreviate “$$-derivation” as derivation”.

Some general observations.

1. For any two derivations $δ, δ' \colon A \to M$ their equalizer $\{ a ∈ A \mid δ(a) = δ'(a) \}$ is a $$-subalgebra of $A$. Consequently, any derivation from $A$ to $M$ is uniquely determined by its action on a $$-algebra generating set of $A$.

2. Let us denote by $\mathrm{Der}_(A, M)$ the subspace of $\mathrm{Hom}_(A, M)$ consisting of derivations. The $$-module $\mathrm{Hom}_(A, M)$ is actually an $A$-module via $(a f)(a') ≔ a f(a')$, and $\mathrm{Der}_(A, M)$ is an $A$-submodule of $\mathrm{Hom}_(A, M)$. More explicitly, $A$-linear combination of derivations are again derivations.

3. If $N$ in another $A$-module and $f \colon M \to N$ is a homomorphism of $A$-modules, then for every derivation $δ \colon A \to M$ the composite $f ∘ δ \colon A \to N$ is again a derivation.

Let now $A = [x_1, \dotsc, x_n]$. Every $$-derivation $A \to M$ is uniquely determined by the values $δ(x_1), \dotsc, δ(x_n)$ thanks to observation (1).

The special case $M = A$.

Let us consider the special case $M = A$ for a moment. The partial derivative maps $∂_i ≔ ∂ / ∂ x_i \colon A \to A$ are derivations with $$ ∂_i( x_j ) = δ_{ij} $$ for all $i, j = 1, \dotsc, n$. It follows from these equations that for every choice of coefficients $a_1, \dotsc, a_n ∈ A$ the derivation $\sum_{i = 1}^n a_i ∂_i$ satisfies $$ \biggl( \sum_{i = 1}^n a_i ∂_i \biggr)( x_j ) = \sum_{i = 1}^n a_i ∂_i( x_j ) = \sum_{i = 1}^n a_i δ_{ij} = a_j $$ for every $j = 1, \dotsc, n$. Since every derivation $δ \colon A \to A$ is uniquely determined by its values $δ(x_1), \dotsc, δ(x_n)$, it further follows that every element of $\mathrm{Der}_(A, A)$ is a unique $A$-linear combination of $∂_1, \dotsc, ∂_n$. In other words, $\mathrm{Der}_(A, A)$ is free as an $A$-module with basis $∂_1, \dotsc, ∂_n$.

As formulas, every derivation $δ \colon A \to A$ is given by $δ = \sum_{i = 1}^n δ(x_i) ∂_i$, or more explicitly by $$ δ(p) = \sum_{i = 1}^n \frac{∂ p}{∂ x_i} δ(x_i) $$ for every polynomial $p ∈ A$.

The general case.

Let now $M$ be an arbitrary $A$-module. Every element $m$ of $M$ corresponds to a homomorphism of $A$-modules $$ f_m \colon A \longrightarrow M \,, \quad a \longmapsto am \,. $$ According to observation (3) we can use the homomorphism $f_m$ to push forward derivations $δ \colon A \to A$ to derivations $f_m ∘ δ \colon A \to M$. We denote the derivation $f_m ∘ δ$ by $δ m$, so that $$ (δ m)(a) = δ(a) m \,. $$ We find in particular that $$ (∂_i m)(x_j) = ∂_i(x_j) m = δ_{ij} m \,. $$ Since every derivation $δ \colon A \to M$ is uniquely determined by its values $δ(x_1), \dotsc, δ(x_n)$, we find that every such derivation $δ$ can be uniquely written as a linear combination $$ δ = \sum_{i = 1}^n ∂_i m_i $$ with $m_1, \dotsc, m_n ∈ M$. More explicitly, the “coefficients” $m_i$ are given by $m_i = δ(x_i)$. This entails the following:

• Every derivation $δ \colon A \to M$ is of the form $δ = \sum_{i = 1}^n ∂_i δ(x_i)$, or more explicitly $$ δ(p) = \sum_{i = 1}^n \frac{∂ p}{∂ x_i} δ(x_i) $$ for every polynomial $p$.

• The isomorphism of $A$-modules $\mathrm{Der}_(A, A) ≅ A^{⊕ n}$ generalizes to an isomorphism of $A$-modules $\mathrm{Der}_(A, M) ≅ M^{⊕ n}$.