Differential Ring (1): Some general result and interesting application

This blog aims to introduce the fundamental concept and property of differential rings (DR).

Define the integration ${\displaystyle \int (-)d_R}$$\displaystyle\int(-)d_R$​ on differential ring via the first isomorphism theorem.

Prove that $H(\mathrm{\Omega })$$H(\Omega)$ is integrally closed via the idea of DR, and consider the Lie bracket over $\mathrm{Der}(R).$$\mathrm{Der}(R).$

Basic definition and propertiesApplication to multivariable calculus IntegrationInitial object in Category of Differential Ring.Holomorphic function over connected domain is integrally closedLie Algebra over differential ring.Derivation on graded ringSquare matrix as a functor on Category of Differential Ring.

Basic definition and properties

Definition

Let $R$$R$ be a ring, a derivation on $R$$R$ is an operator $d:R\to R$$d: R \rightarrow R$ satisfied these two axioms.

For all $a,b\in R$$a, b \in R$, $d(a+b)=d(a)+d(b)$$d(a + b) = d(a) + d(b)$, $d(ab)=d(a)b+ad(b)$$d(ab) = d(a)b + a d(b)$.

The pair $(R,d)$$(R, d)$ is the differential ring.

Example

• $(C^{\mathrm{\infty }}(\mathbb{R}),\frac{d}{dx})$$(C^{\infty}(\mathbb{R}), \frac{d}{dx})$

• ${\displaystyle (C^{\mathrm{\infty }}({\mathbb{R}}^{\mathbb{n}}),\frac{\partial }{\partial x})}$$\displaystyle\left(C^{\infty}(\mathbb{R^n}), \frac{\partial}{\partial x}\right)$

• $(C^{\mathrm{\infty }}({\mathbb{R}}^2),{\partial }_x+i{\partial }_y)$$(C^{\infty}(\mathbb{R}^2), \partial_x + i\partial_y)$

• $(C^{\mathrm{\infty }}({\mathbb{R}}^3),f{\partial }_x+g{\partial }_y+h{\partial }_z)$$(C^{\infty}(\mathbb{R}^3), f\partial_x + g\partial_y + h\partial_z)$...

As you can see, vector fields are an important source of derivation. A trivial example is, for any ring $R$$R$, $0_R$$0_R$ is a derivation.

Proposition.

All the derivation of a commutative ring $R$$R$, i.e. $\text{Der}(R)$$\text{Der}(R)$ form a $R$$R$-Module.

Proof.

Let $d_1,d_2\in \text{Der}R$$d_1, d_2 \in \text{Der} R$

$(d_1+d_2)(a+b)=d_1(a+b)+d_2(a+b)=d_1(a)+d_1(b)+d_2(a)+d_2(b)$$(d_1 + d_2)(a + b) = d_1(a + b) + d_2(a + b) = d_1(a) + d_1(b) + d_2(a) + d_2(b)$
$=(d_1+d_2)(a)+(d_1+d_2)(b)$$= (d_1 + d_2)(a) + (d_1 + d_2)(b)$.

$(d_1+d_2)(ab)=d_1(ab)+d_2(ab)=d_1(a)b+a{d}_1(b)+d_2(a)b+a{d}_2(b)$$(d_1 + d_2)(ab) = d_1(ab) + d_2(ab) = d_1(a)b + a d_1(b) + d_2(a)b + ad_2(b)$
$=(d_1+d_2)(a)b+a(d_1+d_2)(b)$$= (d_1 + d_2)(a)b + a(d_1 + d_2)(b)$.

For any $r\in R$$r \in R$, $rd(a+b)=r(d(a)+d(b))=rd(a)+rd(b)$$rd(a + b) = r(d(a) + d(b)) = rd(a) + rd(b)$, $rd(ab)=r(d(a)b+ad(b))=rd(a)b+rad(b)$$rd(ab) = r(d(a)b + ad(b)) = rd(a)b + rad(b)$.

Example. Let $M$$M$ be a manifold, then $\text{Der}(C^{\mathrm{\infty }}(M))$$\text{Der}(C^\infty(M))$ form a $C^{\mathrm{\infty }}(M)$$C^\infty(M)$-Module.

Proposition. $d(1_R)=0$$d(1_R) = 0$​

Proof.

$$\begin{array}{}\text{(1)}& d(1_R)=d(1_R\cdot 1_R)=d(1_R)+1_{R}d(1_R)=2d(1_R)\Rightarrow d(1_R)=0◻\end{array}$$

Corolloary. For any $(R,d_R)$$(R,d_R)$, $\mathbb{Z}\subseteq \mathrm{Ker}{d}_R$$\mathbb Z\subseteq\mathrm{Ker}d_R$.

Proof. $d_R(n\cdot 1_R)=n{d}_R(1_R)=n\cdot 0=0,d((n-n)1_R)=d(n\cdot 1_R)+d(-n\cdot 1_R)$$d_R(n\cdot 1_R)=nd_R(1_R)=n\cdot0=0,d((n-n)1_R)=d(n\cdot 1_R)+d(-n\cdot 1_R)$.

Hence $0+d(-n\cdot 1_R)=0,d(-n\cdot 1_R)=0$$0+d(-n\cdot 1_R)=0,d(-n\cdot 1_R)=0$. $◻$$\square$

This tells us that $\mathrm{Der}(\mathbb{Z})=0$$\mathrm{Der}(\mathbb Z)=0$.

Proposition.

Let $a\in R,b\in R^{\ast }$$a \in R, b \in R^*$, and ${\displaystyle d\left(\frac{a}{b}\right)=\frac{d(a)b-ad(b)}{b^2}}$$\displaystyle d\left(\frac{a}{b}\right) = \frac{d(a)b - a d(b)}{b^2}$.

Proof.

Since $d(1_R)=d(b{b}^{-1})=d(b)b^{-1}+bd(b^{-1})=0,$$d(1_R) = d(b b^{-1}) = d(b)b^{-1} + bd(b^{-1}) = 0,$ hence $d(b^{-1})=-b^{-2}d(b)$$d(b^{-1}) = -b^{-2}d(b)$.

Apply the Leibniz Law,

${\displaystyle d\left(\frac{a}{b}\right)=\frac{d(a)}{b}+a(-\frac{d(b)}{b^2})=\frac{d(a)b-ad(b)}{b^2}◻}$$\displaystyle d\left(\frac{a}{b}\right) = \frac{d(a)}{b} + a \left(-\frac{d(b)}{b^2}\right) = \frac{d(a)b - a d(b)}{b^2} \square$

We will define the constant of $(R,d)$$(R, d)$, and prove that the constant in $R$$R$ form a subring of $R$$R$. There are lots of interesting examples of constants. For example, in $(C^{\mathrm{\infty }}({\mathbb{R}}^2),{\partial }_x+i{\partial }_y)$$(C^\infty(\mathbb{R}^2), \partial_x + i\partial_y)$, the constant is the ring of holomorphic function.

Definition.

The constant of $(R,d)$$(R, d)$ is the kernel of $d$$d$, in other words, $C:=\text{Ker}(d)$$C := \text{Ker}(d)$.

Example.

In $(C^{\mathrm{\infty }}({\mathbb{R}}^2),{\partial }_x+i{\partial }_y)$$(C^\infty(\mathbb{R}^2), \partial_x + i\partial_y)$, the constant is the holomorphic function ring by the C-R condition.

In $(C^{\mathrm{\infty }}({\mathbb{R}}^3),{\partial }_x)$$(C^{\infty}(\mathbb R^3),\partial_x)$, $f(y,z)\in \mathrm{Ker}({\partial }_x)$$f(y,z)\in\mathrm{Ker}(\partial_x)$.

Proposition.

Let $c\in \text{Ker}(d)$$c \in \text{Ker}(d)$, then $d(ca)=cd(a),d(ac)=d(a)c$$d(ca) = c d(a), d(ac) = d(a)c$.

Proof.

It follows from the Leibniz Law directly. $d(ca)=d(c)a+cd(a)=d(a)c+cd(a)=d(a)c+d(ac)=d(a)c◻$$d(ca) = d(c)a + c d(a) = d(a)c + c d(a) = d(a)c + d(ac) = d(a)c \square$​

Remark. Differential ring is a particular case of differential $R-$$R-$algebra. In general, for a $R$$R$-algebra, we could let $d$$d$ be $R-$$R-$ linear. For more general case, see this blog Math Essays: Derivations from Rings to Modules: A Categorical Perspective Exploration (marco-yuze-zheng.blogspot.com)

Application to multivariable calculus

Consider $(C^{\mathrm{\infty }}({\mathbb{R}}^2),{\partial }_x+i{\partial }_y)$$(C^\infty(\mathbb{R}^2), \partial_x + i\partial_y)$, then $({\partial }_x+i{\partial }_y)(\cos z\cdot f)=\cos z({\partial }_{x}f+i{\partial }_{y}f)=\cos z({\partial }_{x}f+i{\partial }_{y}f)$$(\partial_x + i\partial_y)(\cos z \cdot f) = \cos z(\partial_x f + i\partial_y f) = \cos z(\partial_x f + i\partial_y f)$​.

In multi-varible calculus, when you deal with sth like ${\partial }_x(e^x\sin (y+z))$$\partial_x(e^x\sin(y+z))$, some teacher will tell you that you should pretend that $\sin (y+z)$$\sin(y+z)$ is a constant. But what do you mean pretend? We are mathematician, we do not need to pretend.

Consider the differential ring $(C^{\mathrm{\infty }}({\mathbb{R}}^3),{\partial }_x)$$(C^{\infty}(\mathbb R^3),\partial_x)$, oberve that $\sin (y+z)\in \mathrm{Ker}({\partial }_x)$$\sin(y+z)\in\mathrm{Ker}(\partial_x)$, we have

$$\begin{array}{}\text{(2)}& {\partial }_x(e^x\sin (y+z))=\sin (y+z){\partial }_x(e^x).\end{array}$$

Proposition.

$\text{Ker}(d)$$\text{Ker}(d)$ is a subring of $(R,d)$$(R, d)$.

Proof.

We already know that $1_R\in \text{Ker}(d)$$1_R \in \text{Ker}(d)$, and $\text{Ker}(d)$$\text{Ker}(d)$ is an abelian group. We need to prove that $\text{Ker}(d)$$\text{Ker}(d)$ is closed for multiplication.

Let $a,b\in \text{Ker}(d)$$a, b \in \text{Ker}(d)$, then $d(a\cdot b)=d(a)b+ad(b)=0$$d(a \cdot b) = d(a)b + ad(b) = 0$, $d(ba)=d(b)a+bd(a)=0$$d(ba) = d(b)a + bd(a) = 0$.

Observe that $\text{Ker}(d)$$\text{Ker}(d)$ is also a DR. Since $0_R$$0_R$ is also a derivation. $◻$$\square$

Remark

$(R,d)$$(R, d)$ is an $\text{Ker}(d)$$\text{Ker}(d)$-Algebra, via the inclusion map $i:\text{Ker}(d)\to R$$i: \text{Ker}(d) \rightarrow R$, $d$$d$ is the $\text{Ker}(d)$$\text{Ker}(d)$-Module endomorphism.

For example,

• $f:(C^{\mathrm{\infty }}(\mathbb{R}),\frac{d}{dx})$$f: (C^\infty(\mathbb{R}), \frac{d}{dx})$ is an $\mathbb{R}$$\mathbb R$-v ector space, and ${\displaystyle \frac{d}{dx}}$$\displaystyle\frac{d}{dx}$ is the $R$$R$ linear map.

• $f:(C^{\mathrm{\infty }}({\mathbb{R}}^2),{\partial }_x+i{\partial }_y)$$f: (C^\infty(\mathbb{R}^2), \partial_x + i\partial_y)$ is the $H({\mathbb{R}}^2)$$H(\mathbb{R}^2)$-Module.

What about ${\displaystyle (C^{\mathrm{\infty }}(-2,-1)\cup (1,2),\frac{d}{dx})}$$\displaystyle \left(C^\infty(-2, -1) \cup (1, 2), \frac{d}{dx}\right)$? The ${\displaystyle \text{Ker}\left(\frac{d}{dx}\right)}$$\displaystyle\text{Ker}\left(\frac{d}{dx}\right)$ is generated by ${\chi }_{(-2,-1)},{\chi }_{(1,2)}$$\chi_{(-2,-1)}, \chi_{(1,2)}$, where $\chi$$\chi$ is the characteristic function.

As an $\text{IR}$$\text{IR}$ vector space, ${\displaystyle \text{dim Ker}\left(\frac{d}{dx}\right)=2}$$\displaystyle\text{dim Ker}\left(\frac{d}{dx}\right) = 2$ tell us how many path-connected components the domain has.

As we mentioned before, the vector field is an important source of derivation.

Then the kernel of the derivation is a ring! In other words, the solution of PDE like $(f{\partial }_x+g{\partial }_y+h{\partial }_z)f=0$$(f\partial_x + g\partial_y + h\partial_z)f = 0$ form a ring!

Integration

We can define the integration on a differential ring by the first isomorphism theorem.

Consider $d_R:R\to d_R(R)$$d_R: R\to d_R(R)$ as a $\mathrm{Ker}(d_R)$$\mathrm{Ker}(d_R)$-module homomorphism, which is surjective. By first isomorphism theorem, we get that:

$$\begin{array}{}\text{(3)}& \tilde{d}:R/\mathrm{Ker}(d_R)\cong d_R(R)\end{array}$$

Then we can define the integration as ${\displaystyle \int (-)d_R:={\tilde{d}}^{-1}}$$\displaystyle \int (-)d_R:=\tilde{d}^{-1}$, whcih is a $\mathrm{Ker}(d_R)$$\mathrm{Ker}(d_R)$​-module homomorphism ad well.

Let $d(a)\in d_R(R)$$d(a)\in d_R(R)$, then ${\displaystyle \int d(a)d_R=a+\mathrm{Ker}(d_R)}$$\displaystyle\int d(a)d_R=a+\mathrm{Ker}(d_R)$.

Think about the high school calculus.

$$\begin{array}{}\text{(4)}& \int f(t)dt=F(t)+C\end{array}$$

Here $F(t)+C\in C^{\mathrm{\infty }}(\mathbb{R})/\mathbb{R}$$F(t)+C\in C^{\infty}(\mathbb R)/\mathbb R$.

Now use this idea to consider ${\displaystyle \int e^x\sin (y+z)dx}$$\displaystyle\int e^{x}\sin (y+z)dx$. It should be equal to ${\displaystyle \sin (y+z)\int e^{x}dx=\sin (y+z)e^x+C(y,z)}$$\displaystyle \sin (y+z)\int e^{x}dx=\sin(y+z)e^x+C(y,z)$.

Proposition.

If $(F,d)$$(F, d)$ is a differential field, then $\text{Ker}(d)$$\text{Ker}(d)$ is a field as well.

Proof.

We only need to prove that $a\in \text{Ker}(d)⟹a^{-1}\in \text{Ker}(d)$$a \in \text{Ker}(d) \implies a^{-1} \in \text{Ker}(d)$.

It follows from ${\displaystyle d(a^{-1})=-\frac{d(a)}{a^2}=0}$$\displaystyle d(a^{-1}) = -\frac{d(a)}{a^2} = 0$. $◻$$\square$

Proposition.

Let $(R,d)$$(R, d)$ be a commutative DR, then $d(a^n)=n{a}^{n-1}d(a)$$d(a^n) = na^{n-1}d(a)$.

Proof.

We will prove it via mathematical induction.

Obviously it is true for $k=0,1$$k = 0, 1$, now suppose $d(a^k)=k{a}^{k-1}d(a)$$d(a^k) = ka^{k-1}d(a)$.

Proposition.

Let $(R,d)$$(R, d)$ be a DR, let $I\subset R$$I \subset R$ be an ideal. Then ${\displaystyle (\frac{R}{I},\overline{d})}$$\displaystyle\left(\frac{R}{I}, \bar{d}\right)$ is a DR iff $d(I)\subset I$$d(I) \subset I$.

Proof.

If ${\displaystyle (\frac{R}{I},\overline{d})}$$\displaystyle\left(\frac{R}{I}, \bar{d}\right)$ is DR, then $d(I)=d(0+I)=0$$d(I) = d(0 + I) = 0$, hence $d(I)\subset I$$d(I) \subset I$ in $R$$R$.

If $d(I)\subset I$$d(I) \subset I$, firstly, we need to check $\overline{d}$$\bar{d}$ is well defined. If $a-b\in I$$a - b \in I$, then $d(a)-d(b)=d(a-b)\in d(I)=0$$d(a) - d(b) = d(a - b) \in d(I) = 0$.

Then ${\displaystyle (\frac{R}{I},\overline{d})}$$\displaystyle\left(\frac{R}{I}, \bar{d}\right)$ is a derivation ring since ${\displaystyle d\left(a+b+I\right)=\overline{d}(a)+\overline{d}(b).d(ab+I)=\overline{d}(ab)◻}$$\displaystyle d\left({a + b + I}\right) = \bar{d}(a) + \bar{d}(b). d(ab + I) = \bar{d}(ab) \square$

Remark.

We call the ideal that satisfies $dI\subset I$$dI \subset I$ differential ideal.

Example.

Consider $(\mathbb{Z}[X,Y],X{\partial }_X+{\partial }_Y)$$(\mathbb{Z}[X, Y], X\partial_X + \partial_Y)$ and the quotient map $\pi :\mathbb{Z}[X,Y]\to \mathbb{Z}[Y]$$\pi: \mathbb{Z}[X, Y] \rightarrow \mathbb{Z}[Y]$, the ideal is $(X)$$(X)$.

Obviously $(X{\partial }_X+{\partial }_Y)(X)=X{\partial }_X(X)\subset (X)$$(X\partial_X + \partial_Y)(X) = X\partial_X(X) \subset (X)$. Hence $(\mathbb{Z}[Y],X{\partial }_X+{\partial }_Y)\simeq (\mathbb{Z}[Y],{\partial }_Y)$$(\mathbb{Z}[Y], X\partial_X + \partial_Y) \simeq (\mathbb{Z}[Y], \partial_Y)$ is a derivation ring.

Well, as you see, I use the notation $\simeq$$\simeq$ for isomorphism, but what is homomorphism between two DR?

Definition.

Let $f:R\to R^{\prime }$$f : R \rightarrow R'$ be a ring homomorphism, if it satisfies this property $d^{\prime }\circ f=f\circ d$$d' \circ f = f \circ d$, then $f:(R,d)\to (R^{\prime },d^{\prime })$$f : (R, d) \rightarrow (R', d')$ is a differential ring homomorphism.

Remark.

The definition tells us that $d^{\prime }\circ f(a)=f\circ d(a)$$d' \circ f(a) = f \circ d(a)$. Hence $d^{\prime }\circ f(a+b)=f(d(a))+f(d(b))$$d' \circ f(a + b) = f(d(a)) + f(d(b))$, $d^{\prime }\circ f(ab)=f(d(a)b+ad(b))=f(d(a))f(b)+f(a)d^{\prime }(f(b))$$d' \circ f(ab) = f(d(a)b + ad(b)) = f(d(a))f(b) + f(a)d'(f(b))$. $f$$f$ preserve the derivation structure.

Moreover, it is not hard to see that $f$$f$ preserve constants.

Example.

Consider $\pi :\mathbb{Z}[X,Y]\to \mathbb{Z}[Y]$$π : \mathbb{Z}[X, Y] \rightarrow \mathbb{Z}[Y]$ again.

$\pi \circ ((X{\partial }_X+{\partial }_Y)(a))=\pi ((X{\partial }_X(a)+{\partial }_Y(a)))=\pi (0)=0\text{or}\pi (({\partial }_Y(a))={\partial }_Y(\pi (a))$$\pi \circ ((X\partial_X + \partial_Y)(a)) = \pi((X\partial_X(a) + \partial_Y(a))) = \pi(0) = 0 \quad \text{or} \quad \pi((\partial_Y(a)) = \partial_Y(\pi(a))$

Consider the $(C^{\mathrm{\infty }}(M),V)$$(C^\infty(M), V)$, where $V$$V$ is a vector field, the restriction map $re{s}_V^U:C^{\mathrm{\infty }}(U)\to C^{\mathrm{\infty }}(U^{\prime })$$res^U_V : C^\infty(U) \rightarrow C^\infty(U')$​​ is a differential ring homomorphism.

Initial object in Category of Differential Ring.

As we already see, the only derivation on $\mathbb{Z}$$\mathbb Z$ is $0$$0$. As we know, $\mathbb{Z}$$\mathbb Z$ is the initial object in category of ring.

We will prove that $(\mathbb{Z},0)$$(\mathbb Z,0)$​ is the initial object in category of differential ring.

We know the only ring homomorphism from $\mathbb{Z}$$\mathbb Z$ to $R$$R$ is defined as:

$$\begin{array}{}\text{(5)}& f:\mathbb{Z}\to R,n⟼n\cdot 1_R\end{array}$$

And it is easy to see that $f$$f$​ preserve the derivation.

$$\begin{array}{}\text{(6)}& d\circ f(n)=d(n\cdot 1_R)=0=f\circ 0(n)\end{array}$$

Holomorphic function over connected domain is integrally closed

Definition.

Let $(K,d_K)\subset (L,d_L)$$(K, d_K) \subset (L, d_L)$ be two integral differential domains, if $\mathrm{\forall }k\in K,d_L(k)=d_K(k)$$\forall k \in K, d_L(k) = d_K(k)$, then $(K,d_K)\subset (L,d_L)$$(K, d_K) \subset (L, d_L)$ is a integral differential extension.

Proposition.

Let $(K,d_K)\subset (L,d_L)$$(K, d_K) \subset (L, d_L)$ be an integral differential extension, if $\alpha$$\alpha$ is algebraic over $\text{Ker}(d)$$\text{Ker}(d)$, then $d(\alpha )=0$$d(\alpha) = 0$.

Proof.

Let $p(x)\in \text{Ker}(d)[X]$$p(x) \in \text{Ker}(d)[X]$ be the minimal polynomial of $\alpha$$\alpha$.

Then $p(\alpha )=0\Rightarrow d_L(p(\alpha ))=p^{\prime }(\alpha )d_L(\alpha )=0$$p(\alpha) = 0 \Rightarrow d_L(p(\alpha)) = p'(\alpha)d_L(\alpha) = 0$. $L$$L$ is integral domain implies that $p^{\prime }(\alpha )=0$$p'(\alpha) = 0$ or $d_L(\alpha )=0$$d_L(\alpha) = 0$

Since $\mathrm{deg}{p}^{\prime }(X)<\mathrm{deg}p(X)$$\deg p'(X) < \deg p(X)$ and $p(X)$$p(X)$ is the minimal polynomial, $p^{\prime }(x)\ne 0$$p'(x) \neq 0$. Hence $d(\alpha )=0◻$$d(\alpha) = 0 \square$

Corollary.

$H(\mathrm{\Omega })$$H(\Omega)$ is integrally closed. Where $\mathrm{\Omega }$$\Omega$ is connected.

Remark.

So what is integrally closed? Let $R$$R$ be an integral domain, $R\subset Q(R)$$R \subset Q(R)$, where $Q(R)$$Q(R)$ is the fraction field of $R$$R$.

If $R$$R$ satisfies the following property, we call $R$$R$ integrally closed.

If $r\in Q(R)$$r \in Q(R)$ is a root of a monic polynomial $m(X)\in R[X]$$m(X) \in R[X]$, then $r\in R$$r \in R$.

Proof.

Consider the differential extension $(H(\mathrm{\Omega }),{\partial }_X+i{\partial }_Y)\subset (M(\mathrm{\Omega }),{\partial }_X+i{\partial }_Y)$$(H(\Omega), \partial_X + i\partial_Y) \subset (M(\Omega), \partial_X + i\partial_Y)$. Where $H$$H$ means holomorphic and $M$$M$ means meromorphic.

$m(f)=0\Rightarrow ({\partial }_X+i{\partial }_Y)f=0$$m(f) = 0 \Rightarrow (\partial_X + i\partial_Y)f = 0$, hence $f$$f$ is holomorphic as well. Therefore, $\text{Hol}(\mathrm{\Omega })$$\text{Hol}(\Omega)$ is integrally closed. $◻$$\square$

Lie Algebra over differential ring.

Consider the $\mathrm{Der}(R)$$\mathrm{Der}(R)$, whcih is a $R-$$R-$module. Define the Lie bracket $[d,d^{\prime }]=d{d}^{\prime }-d^{\prime }d$$[d,d']=dd'-d'd$​ on it, we can get an Lie algebra.

Obviously $[d,d^{\prime }]$$[d,d']$ is $\mathbb{Z}-$$\mathbb Z-$linear map, and $[-,-]$$[-,-]$ is bilinear.

We only need to check Leibniz Law.

$$\begin{array}{}\text{(7)}& [d,d^{\prime }](fg)=(d{d}^{\prime }-d^{\prime }d)(fg)=d(d^{\prime }(f)g+f{d}^{\prime }(g))-d^{\prime }(d(f)g+fd(g))\end{array}$$

expand it we get:

$$\begin{array}{}\text{(8)}& d{d}^{\prime }(f)g+d^{\prime }(f)d(g)+d(f)d^{\prime }(g)+fd{d}^{\prime }(g)-d^{\prime }d(f)g-d(f)d^{\prime }(g)-d^{\prime }(f)d(g)-f{d}^{\prime }d(g)\end{array}$$

Simplify it we get:

$$\begin{array}{}\text{(9)}& (d{d}^{\prime }-d^{\prime }d)(f)g+f(d{d}^{\prime }-d^{\prime }d)(g)=[d,d^{\prime }](f)g+f[d,d^{\prime }](g)\end{array}$$

For the proof of Jacobi identity, see Math Essays: Introduction to Weyl Algebra (marco-yuze-zheng.blogspot.com)

Derivation on graded ring

Let ${\displaystyle R=\underset{i=0}{\overset{\mathrm{\infty }}{⨁}}{R}_i}$$\displaystyle R=\bigoplus_{i=0}^\infty R_i$ be a graded ring. Let ${\displaystyle r=\sum _{k=0}^n{r}_k}$$\displaystyle r=\sum_{k=0}^n r_k$, define ${\displaystyle d(r)=\sum _{k=0}^{n}k{r}_k}$$\displaystyle d(r)=\sum_{k=0}^n kr_k$.

Easy to see it is $R-$$R-$linear. Now we only neeed to prove the Leibniz Law.

Given two elements $(r)$$( r)$ and $(r^{\prime })$$( r' )$, they are defined as follows: $r=\sum _{i=0}^n{r}_i$$r = \sum_{i=0}^{n} r_i$ and $r^{\prime }=\sum _{j=0}^m{r}_j^{\prime }$$r' = \sum_{j=0}^{m} r'_j$.

Their product $r{r}^{\prime }$$rr'$ is:

$$\begin{array}{}\text{(10)}& r{r}^{\prime }=\sum _{l=0}^{n+m}\sum _{i+j=l}{r}_i{r}_j^{\prime }\end{array}$$

Now, apply $d$$d$ :

$$\begin{array}{}\text{(11)}& d(r{r}^{\prime })=\sum _{l=0}^{n+m}l\sum _{i+j=l}{r}_i{r}_j^{\prime }\end{array}$$

We need to check if:

$$\begin{array}{}\text{(12)}& d(r{r}^{\prime })=d(r)r^{\prime }+rd(r^{\prime })\end{array}$$

Let's calculate $d(r)r^{\prime }$$d(r)r'$:

$$\begin{array}{}\text{(13)}& d(r)r^{\prime }=\left(\sum _{i=0}^{n}i{r}_i\right)\left(\sum _{j=0}^m{r}_j^{\prime }\right)=\sum _{i=0}^n\sum _{j=0}^{m}i{r}_i{r}_j^{\prime }\end{array}$$

Similarly, let's calculate$rd(r^{\prime })$$r d(r')$:

$$\begin{array}{}\text{(14)}& rd(r^{\prime })=\left(\sum _{i=0}^n{r}_i\right)\left(\sum _{j=0}^{m}j{r}_j^{\prime }\right)=\sum _{i=0}^n\sum _{j=0}^{m}j{r}_i{r}_j^{\prime }\end{array}$$

Add these two results:

$$\begin{array}{}\text{(15)}& d(r)r^{\prime }+rd(r^{\prime })=\sum _{i=0}^n\sum _{j=0}^m(i+j)r_i{r}_j^{\prime }\end{array}$$

Now, compare this result with $d(r{r}^{\prime })$$d(rr')$:

$$\begin{array}{}\text{(16)}& d(r{r}^{\prime })=\sum _{l=0}^{n+m}l\sum _{i+j=l}{r}_i{r}_j^{\prime }\end{array}$$

Easy to see that

$$\begin{array}{}\text{(17)}& \sum _{l=0}^{n+m}l\sum _{i+j=l}{r}_i{r}_j^{\prime }=\sum _{i=0}^n\sum _{j=0}^m(i+j)r_i{r}_j^{\prime }\end{array}$$

by consider the matrix $[(i+j)a_{i,j}]$$[(i+j)a_{i,j}]$. Some readers may remind this blog Math Essays: Using matrix to count the product of two polynomias (marco-yuze-zheng.blogspot.com).

Hence we prove that $d(r{r}^{\prime })=d(r)r^{\prime }+rd(r^{\prime }).$$d(rr')=d(r)r'+rd(r').$ $◻$$\square$​

In particular, we could define this derivation on tensor algebra.

$$\begin{array}{}\text{(18)}& d[(x\otimes y)\otimes (a\otimes b\otimes c)]=5(x\otimes y\otimes a\otimes b\otimes c)=2(x\otimes y)\otimes (a\otimes b\otimes c)+(x\otimes y)\otimes 3(a\otimes b\otimes c)\end{array}$$

Square matrix as a functor on Category of Differential Ring.

We already know that $M_{n,n}:\mathrm{Ring}\to \mathrm{Ring}$$M_{n,n}:\mathrm{Ring}\to \mathrm{Ring}$ is a functor (see this blog Math Essays: Why det is a natural transformation? (marco-yuze-zheng.blogspot.com))

Let $(R,d)$$(R,d)$ be a differential ring, we want to prove that $M_{n,n}$$M_{n,n}$ is also a functor from differential ring to differential ring.

Let $(a_{i,j})\in M_{n,n}(R)$$(a_{i,j})\in M_{n,n}(R)$

We only need to shows that for $d:R\to R$$d:R\to R$, define derivation on $M_{n,n}(R)$$M_{n,n}(R)$ as:

$$\begin{array}{}\text{(19)}& d(a_{i,j})=(d(a_{i,j}))\end{array}$$

It is clear that the derivation is $\mathbb{Z}-$$\mathbb Z-$linear. Now we need to check the Leibniz law.

$$\begin{array}{}\text{(20)}& d[(a_{i,j})(b_{i,j})]=d(c_{i,j})=d\left(\sum _{k=1}^n{a}_{i,k}{b}_{k,j}\right)=\sum _{k=1}^n(d(a_{i,k})b_{k,j}+a_{i,k}d(b_{k,j})=d(a_{i,j})(b_{i,j})+(a_{i,j})d(b_{i,j})\end{array}$$