Derivations on rings; the basics (1)

All rings in this post are assumed to have identity and $Z(R)$ denotes the center of a ring $R.$

Those who have the patience to follow my posts have seen derivation on rings several times but they have not seen a post exclusively about the basics on derivations of rings because, strangely, that post didn’t exist until now.

In the polynomial ring $\mathbb{C}[x],$ we have the familiar concept of differentiation with respect to $x,$ i.e. $\delta:= \frac{d}{dx}.$ This is a map form $\mathbb{C}[x] \to \mathbb{C}[x]$ which is additive and satisfies the product rule. Also, $\delta(c)=0$ for all $c \in \mathbb{C}.$ We now extend this concept to rings in general.

Definition 1. Let $R$ be a ring. A derivation of $R$ is any additive map $\delta : R \to R$ that satisfies the product rule: $\delta(ab)=\delta(a)b+a\delta(b),$ for all $a,b \in R.$ We denote by $\text{Der}(R)$ the set of all derivations of $R.$

We can now extend the familiar concept of differentiating polynomials to differentiating polynomials over any ring $R.$ Note that the variable $x$ is assumed to be in the center of $R[x].$

Example 1. Let $R$ be a ring, and let $R[x]$ be the ring of polynomials over $R.$ Define the map $\delta : R[x] \to R[x]$ by

$\displaystyle \delta\left(\sum_{k=0}^na_kx^k\right)=\sum_{k=0}^nka_kx^{k-1}, \ \ \ \ \ n \in \mathbb{Z}_{\ge 0}, \ \ a_k \in R.$

Then $\delta \in \text{Der}(R[x]).$ The derivation $\delta$ is usually denoted by $\frac{d}{dx}.$ Note that, for the definition to make sense, the term $ka_kx^{k-1}$ for $k=0$ is defined to be $0$ and $x^0$ is defined to be $1.$

Proof. We need to show that $\delta$ is additive and also it satisfies the product rule. Let $p(x), q(x) \in R[x].$ So we can write $p(x)=\sum_{k=0}^na_kx^k, \ q(x)=\sum_{k=0}^nb_kx^k.$ Then

$\displaystyle \delta(p(x)+q(x))=\delta \left(\sum_{k=0}^n(a_k+b_k)x^k\right)=\sum_{k=0}^nk(a_k+b_k)x^{k-1}$

$\displaystyle =\sum_{k=0}^nka_kx^{k-1}+\sum_{k=0}^nkb_kx^{k-1}=\delta(p(x))+\delta(q(x)).$

To prove the product rule, we have

$\displaystyle \delta(p(x)q(x))=\delta \left(\sum_{k=0}^{2n}\sum_{j=0}^ka_jb_{k-j}x^k\right)=\sum_{k=0}^{2n}k\sum_{j=0}^ka_jb_{k-j}x^{k-1}, \ \ \ \ \ \ \ \ \ (*)$

and

$\displaystyle x(\delta(p(x))q(x)+p(x)\delta(q(x))=x\left(\sum_{k=0}^nka_kx^{k-1}\sum_{k=0}^nb_kx^k+\sum_{k=0}^na_kx^k\sum_{k=0}^nkb_kx^{k-1}\right)$

$\displaystyle \begin{aligned}=\sum_{k=0}^nka_kx^k\sum_{k=0}^nb_kx^k+\sum_{k=0}^na_kx^k\sum_{k=0}^nkb_kx^k=\sum_{k=0}^{2n}\sum_{j=0}^kja_jb_{k-j}x^k+\sum_{k=0}^{2n}\sum_{j=0}^k(k-j)a_jb_{k-j}x^k\end{aligned}$

$\displaystyle =\sum_{k=0}^{2n}k\sum_{j=0}^ka_jb_{k-j}x^k=x \delta(p(x)q(x)), \ \ \ \ \ \text{by} \ (*).$

Therefore $\delta(p(x)q(x))=\delta(p(x))q(x)+p(x)\delta(q(x)). \ \Box$

Example 2. Let $R$ be a ring, and let $r \in R.$ Define the map $\delta : R \to R$ by $\delta(a)=ra-ar,$ for all $a \in R.$ Then $\delta \in \text{Der}(R)$ and it’s called an inner derivation.

Proof. We need to show that $\delta$ is additive and also it satisfies the product rule. Let $a,b \in R.$ Then

$\delta(a+b)=r(a+b)-(a+b)r=ra-ar+rb-br=\delta(a)+\delta(b),$

and

$\begin{aligned} \delta(a)b+a\delta(b)=(ra-ar)b+a(rb-br)=rab-arb+arb-abr=rab-abr=\delta(ab). \ \Box \end{aligned}$

Remarks. Let $R$ be a ring, and let $\delta \in \text{Der}(R).$ Let $K:=\ker \delta=\{a \in R: \ \delta(a)=0\}.$

i) $\{0,1\} \subseteq K,$

ii) $K$ is a subring of $R$ called the ring of constants of $\delta,$

iii) if $a \in K$ is invertible in $R,$ then $a^{-1} \in K;$ so if $R$ is a division ring, $K$ is a division ring too,

iv) if $r \in R,$ then $\delta(ra)=r\delta(a)$ for all $a \in R$ if and only if $r \in K,$

v) $\delta(Z(R)) \subseteq Z(R),$

vi) if $\delta(a) \in Z(R),$ then $\delta(a^n)=na^{n-1}\delta(a)$ for all integers $n \ge 1,$ where, for $n=1,$ we define $a^0=1,$

vii) The identity given in vi) may not hold true if $\delta(a) \notin Z(R).$

Proof. i) Since $\delta$ is additive, $\delta(0)=\delta(0+0)=\delta(0)+\delta(0)=2\delta(0),$ and so $\delta(0)=0,$ hence $0 \in K.$ Also, by product rule, $\delta(1)=\delta(1 \cdot 1)=\delta(1) \cdot 1 + 1 \cdot \delta(1)=2\delta(1),$ and so $\delta(1)=0$ hence $1 \in K.$

ii) If $a,b \in \ker \delta,$ then since $\delta$ is additive, $\delta(a+b)=\delta(a)+\delta(b)=0+0=0$ and so $a+b \in \ker \delta.$ Also, by product rule, $\delta(ab)=\delta(a)b+a\delta(b)=0+0=0$ and so $ab \in \ker \delta.$

iii) By i) and product rule, $0=\delta(1)=\delta(aa^{-1})=\delta(a)a^{-1}+a\delta(a^{-1})=a\delta(a^{-1})$ and so $\delta(a^{-1}) =0.$

iv) If $\delta(ra)=r\delta(a),$ for some $r \in R$ and all $a \in R,$ then $a=1$ gives $\delta(r)=r\delta(1)=0$ and so $r \in K.$ Conversely, if $r \in K, \ a \in R,$ then, by product rule, $\delta(ra)=\delta(r)a+r\delta(a)=0+r\delta(a)=r\delta(a).$

v) Let $a \in Z(R), \ b \in R.$ Then

$\delta(a)b+a\delta(b)=\delta(ab)=\delta(ba)=\delta(b)a+b\delta(a)=a\delta(b)+b\delta(a),$

and so $\delta(a)b=b\delta(a),$ proving that $\delta(a) \in Z(R).$

vi) First notice that, by v), the condition $\delta(a) \in Z(R)$ is stronger than $a \in Z(R).$ Now, the proof is by induction over $n.$ It is clear for $n=1.$ Assuming the identity holds for $n,$ we have

$\begin{aligned} \delta(a^{n+1})=\delta(a^na)=\delta(a^n)a+a^n\delta(a)=na^{n-1}\delta(a)a+a^n\delta(a)=na^n\delta(a)+a^n\delta(a)=(n+1)a^n\delta(a).\end{aligned}$

vii) Consider $R:=M_2(\mathbb{C}),$ the ring of $2 \times 2$ matrices with complex entries. Let $\delta$ be the inner derivation (see Example 2) of $R$ corresponding to $r=\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}.$ Now, choose $a=\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}.$ Then $a^2$ is the identity matrix and so $\delta(a^2)=0.$ Also,

$\delta(a)=ra-ar=\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix} \notin Z(R), \ \ \ 2a\delta(a) =\begin{pmatrix}-2 & 0 \\ 0 & 2\end{pmatrix} \ne 0=\delta(a^2). \ \Box$

Definition 2. Let $R$ be a ring, $\delta \in \text{Der}(R),$ and $C$ a subring of $R$ contained in the center of $R.$ If $C \subseteq \ker \delta,$ then, by Remark iv), $\delta(ca+b)=c\delta(a)+\delta(b)$ for all $c \in C,a,b \in R,$ i.e. $\delta$ is $C$ -linear. The set of all $C$ -linear derivations of $R$ is denoted by $\text{Der}_C(R).$

Exercise. Consider the commutative polynomial ring $C]x]$ and let $R:=C[x]/I,$ where $I:=\langle x^2\rangle.$ Define the map $\delta: R \to R$ by $\delta(\alpha x+ \beta+I)=\alpha x + I,$ for all $\alpha, \beta \in C.$ Show that $\delta \in \text{Der}_C(R).$