In this section we introduce the new concept of integration on a ring and show many properties of such integrations. Our definition is motivated by the definition of an indefinite integral (a primitive function) in mathematical analysis, calculus.

#### Definition 2.1

*Let R be a ring, x* ∈ *R and d : R* → *R a derivation on R. Let*
$${i}_{d}(x)=\{y\in R\phantom{\rule{thinmathspace}{0ex}}|\phantom{\rule{thinmathspace}{0ex}}x\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}d(y)\}.$$

*We say that y* ∈ *R is a d-primitive element of the element x* ∈ *R, if*
$$y\phantom{\rule{thinmathspace}{0ex}}\in \phantom{\rule{thinmathspace}{0ex}}{i}_{d}\phantom{\rule{thinmathspace}{0ex}}(x).$$

*We call the function i*_{d} : *R* → 2^{R}, *x* ↦ *i*_{d} (*x*), *the d-integration on R* *and the set i*_{d} (*x*) *the d-integral of x*.

*The function i* : *R* → 2^{R} *is an integration on R*, *if there is a derivation d on R*, *such that i* = *i*_{d}. *For each x* ∈ *R we call i* (*x*) *an integral of x*.

#### Example 2.2

*Let R be a ring and d* : *R* → *R the trivial derivation* (*d*(*x*) = 0 *for all x* ∈ *R*). *Then i*_{d} (*x*) = *R if x* = 0 *and i*_{d} (*x*) = ∅ *if x* ≠ 0.

Many results that follow are easily obtained from the definition of an integration and the fact that any derivation *d* : *R* → *R* is a homomorphism of the additive structure on *R*. Since the proofs of the results are short, we give them anyway.

The following results follow directly from Definition 2.1

#### Proposition 2.3

*Let R be a ring and d* : *R* → *R a derivation on R. Then the following holds*.

0 ∈ *i*_{d} (0).

*For each x* ∈ *R*, *x* ∈ *i*_{d} (*d*(*x*)).

*If i*_{d} (*x*) ≠ ∅ *then d* (*i*_{d}(*x*)) = {*x*} *for all* *x* ∈ *R*.

#### Proof

Since *d*(0) = 0, it follows that 0 ∈ *i*_{d}(0), which proves (1). To prove (2), let *y* = *d*(*x*). Then it follows from Definition 2.1 that *x* ∈ *i*_{d}(*y*) = *i*_{d}(*d*(*x*)). Finally we prove (3). Let *z* ∈ *i*_{d}(*x*). Then *d*(*z*) = *x* and hence *d*(*i*_{d}(*x*)) = {*x*}. □

#### Proposition 2.4

*Let* *R be a ring and d* : *R* → *R* *a derivation on R*. *Then*

*d* *is surjective if and only if* *i*_{d}(*x*) ≠ ∅ *for all x* ∈ *R*.

*d is injective if and only if* |*i*_{d}(*x*)| = 1 *for all x* ∈ *R*.

#### Proof

First we prove (1). Suppose that *d* is surjective and let *x* ∈ *R*. Then there is *y* ∈ *R* such that *x* = *d*(*y*). It follows that *y* ∈ *i*_{d}(*x*) and *i*_{d}(*x*) is not empty. Suppose that *i*_{d}(*x*) ≠ ∅ for all *x* ∈ *R*. Let *x* ∈ *R* and *y* ∈ *i*_{d}(*x*). It follows that *x* = *d*(*y*).

Next we prove (2). Suppose that *d* is injective, *x* ∈ *R*, and *y*, *z* ∈ *i*_{d}(*x*). Then *d*(*y*) = *d*(*z*) = *x*. Since *d* is injective, it follows that *y* = *z* and therefore *i*_{d}(*x*) consists only of a single element. Suppose that |*i*_{d}(*x*)| = 1 for all *x* ∈ *R*. Let *x*, *y*, *z* ∈ *R* be such elements that *x* = *d*(*y*) = *d*(*z*). Then *y*, *z* ∈ *i*_{d}(*x*) and therefore *y* = *z* (since |*i*_{d}(*x*)| = 1). □

The theorems that follow describe basic properties of integrations, introduced in Definition 2.1 First we show some properties of the integral *i*_{d}(0) (Theorem 2.6). Note that *i*_{d}(0) = Ker(*d*) = {*x* ∈ *R*|*d*(*x*) = 0} for any ring *R* and any derivation *d* on *R*.

#### Definition 2.5

*Let* *R be a ring with unity* 1 ∈ *R*. *For an integer* *n*, **n** *will always denote the element* **n** = *n* ⋅ 1 =
$\underset{n}{\underset{\u23df}{1+1+1+\dots +1}}$∈ *R*.

#### Theorem 2.6

*Let* *R* *be a ring with unity* 1 ∈ *R* *and* *d* : *R* → *a derivation on R*.

*Then*–**n**, **n** ∈ *i*_{d}(0) *for each positive integer n*.

*If* **n** *is invertible, then*
$${\mathbf{n}}^{-1}\cdot \mathbf{m},\mathbf{m}\cdot {\mathbf{n}}^{-1}\in {i}_{d}\phantom{\rule{thinmathspace}{0ex}}(0)$$
*for all integers m*, *n*.

#### Proof

First we prove (1). From *d*(1) = *d*(1 ⋅ 1) = *d*(1) ⋅ 1 + 1 ⋅ *d*(1) = 2*d*(1) it follows that *d*(1) = 0. Therefore 1 ∈ *i*_{d}(0). Since
$$d(\mathbf{n})=d\underset{n}{\underset{\u23df}{(1+1+1+\dots +1)}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\underset{n}{\underset{\u23df}{d(1)+d(1)+d(1)+\dots +d(1)}}=0,$$
it follows that **n** ∈ *i*_{d}(0). Also, from *d*(−**n**) = −*d*(**n**) = 0 it follows that −**n** ∈ *i*_{d}(0).

To prove (2), we first observe that *d*(**n**^{−1})= 0. This follows from the fact that 0 = *d*(1) = *d*(**n** ⋅ **n**^{−1}) = *d*(**n**) ⋅ **n**^{−1} + **n** ⋅ *d*(**n**^{−1}) = 0 + **n** ⋅ *d*(**n**^{−1}) = **n** ⋅ *d*(**n**^{−1}), and since **n** is invertible, *d*(**n**^{−1}) = 0. Next, it follows from *d*(**n**^{−1} ⋅ **m**) = *d*(**n**^{−1}) ⋅ **m** + **n**^{−1} ⋅ *d*(**m**) = 0 + 0 = 0, that **n**^{−1} ⋅ **m** ∈ *i*_{d}(0). A similar proof gives **m** ⋅ **n**^{−1} ∈ *i*_{d}(0). □

#### Corollary 2.7

*Let* *R be a ring with unity* 1 ∈ *R* *and* *d* : *R* → *R a derivation on R. If y* ∈ *i*_{d}(*x*) *and* **n** *is invertible, then*
$$y+{\mathbf{n}}^{-1}\cdot \mathbf{m},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}y\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}\mathbf{m}\cdot {\mathbf{n}}^{-1}\in {i}_{d}(x)\phantom{\rule{thinmathspace}{0ex}}$$
*for all integers m*, *n and all x*, *y* ∈ *R*.

#### Proof

The result follows from *d*(*y* + **n**^{−1}⋅ **m**) = *d*(*y*) + *d*(**n**^{−1} ⋅ **m**) = *x* + 0 = *x* by Theorem 2.6. □

Next we generalize the result from calculus, saying that if *g* is an indefinite integral of *f*, then also *g* + *c* is an indefinite integral of *f* for any constant function *c*.

#### Theorem 2.8

*Let* *R be a ring*, *x*, *z* ∈ *R*, *d* : *R* → *R a derivation on* *R* *and y* ∈ *i*_{d}(*x*). *The following statements are equivalent*.

*z* ∈ *i*_{d}(*x*).

*There is a unique element* *w*_{0} ∈ Ker(*d*), *such that z* = *y* + *w*_{0}.

#### Proof

Suppose first that *z* ∈ *i*_{d}(*x*). Let *w*_{0} = *z* − *y*. Obviously, *z* = *y* + *w*_{0} and *d*(*w*_{0}) = *d*(*z* − *y*) = 0. If *z* = *y* + *w*_{0} = *y* + *w*_{1} for *w*_{0}, *w*_{1} ∈ Ker(*d*), then 0 = *z* − *z* = *w*_{0} − *w*_{1} and therefore *w*_{0} = *w*_{1}. For the converse, let *z* = *y* + *w*_{0} for any *w*_{0}∈ Ker(*d*). Then *d*(*z*) = *d*(*y* + *w*_{0}) = *d*(*y*) + *d*(*w*_{0}) = *x* + 0 = *x* and therefore *z* ∈ *i*_{d}(*x*). □

#### Corollary 2.9

*Let R be a ring, x* ∈ *R*, *d* : *R* → *R* *a derivation on R and y* ∈ *i*_{d}(*x*). *Then*
$${i}_{d}(x)=\{y+z\phantom{\rule{thinmathspace}{0ex}}|\phantom{\rule{thinmathspace}{0ex}}z\phantom{\rule{thinmathspace}{0ex}}\in \mathrm{K}\mathrm{e}\mathrm{r}\phantom{\rule{thinmathspace}{0ex}}(d)\}.$$

#### Example 2.10

*Let*
$D=\mathbb{R}\phantom{\rule{thinmathspace}{0ex}}\mathrm{\setminus}\phantom{\rule{thinmathspace}{0ex}}\{{\displaystyle \frac{(2k+1)\pi}{2}|\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}k\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{is an integer}\}}$
*and let R be the ring of all differentiable functions on D with the standard addition and multiplication of functions, i.e*. (*f* + *g*)(*x*) = *f*(*x*) + *g*(*x*) *and* (*f* ⋅ *g*)(*x*) = *f*(*x*)*g*(*x*) *for each x* ∈ *D*. *Let d be the standard derivation on D, i.e*. *d*(*f*) = *f*′ *for each f* ∈ *R*. *Next, let h* ∈ *R be defined by*

$$h(x)={\displaystyle \frac{(2k+1)\pi}{2},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{if}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x\in \left(\frac{(2k-1)\pi}{2},\frac{(2k+1)\pi}{2}\right).}$$

*Then clearly h* ∈ Ker(*d*). *Finally, let f, g* ∈ *R be defined by*
$f(x)={\displaystyle \frac{1}{{\mathrm{cos}}^{{2}_{X}}}}$
*and g*(*x*) = tan *x* *for all* *x*∈ *D*. *Since g* ∈ *i*_{d}(*f*), *it follows from* Theorem 2.8, *that also g* + *h* ∈ *i*_{d}(*f*).

Next we realize the well-known formula
$$\int c\cdot f(x)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x=c\phantom{\rule{thinmathspace}{0ex}}\cdot \int f\phantom{\rule{thinmathspace}{0ex}}(x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x$$
from calculus for arbitrary rings.

#### Theorem 2.11

*Let R be a ring, d* : *R* → *R* *a derivation on R and w* ∈ Ker(*d*). *Then the following holds*.

*w* ⋅ *i*_{d}(*x*) ⊆ *i*_{d} (*w* ⋅ *x*) *and* *i*_{d}(*x*)⋅ *w* ⊆ *i*_{d}(*x* ⋅ *w*).

*If* *i*_{d}(*x*) ≠ ∅, *then* *i*_{d} (*w* ⋅ *x*), *i*_{d} (*x* ⋅ *w*) ≠ ∅.

*If R is a ring with unity* 1 ∈ *R*, *w* *an invertible element and* *i*_{d}(*x*) ≠ ∅, *then* *i*_{d} (*w* ⋅ *x*) = *w* ⋅ *i*_{d}(*x*) *and* *i*_{d} (*x* ⋅ *w*) = *i*_{d}(*x*)⋅ *w*.

#### Proof

If *i*_{d}(*x*) = ∅, then obviously (1) holds. Suppose that *i*_{d}(*x*) ≠ ∅. We only prove that *w* ⋅ *i*_{d}(*x*) ⊆ *i*_{d} (*w* ⋅ *x*) (*i*_{d}(*x*) ⋅ *w* ⊆ *i*_{d} (*x* ⋅ *w*) can be proved similarly). Let *z* ∈ *i*_{d}(*x*). Then *d*(*w* ⋅ *z*) = *d*(*w*) ⋅ *z* + *w* ⋅ *d*(*z*) = 0 + *w* ⋅ *x* = *w* ⋅ *x* and therefore *w* ⋅ *z* ∈ *i*_{d} (*w* ⋅ *x*). This proves (1) and (2). To prove (3), suppose that (*w*) is invertible and *i*_{d}(*x*) ≠ ∅. First observe, that from 0 = *d*(1) = *d*(*w* ⋅ *w*^{−1}) = *d*(*w*)⋅ *w*^{−1} + *w*⋅ *d*(*w*^{−1}) = *w* ⋅ *d*(*w*^{−1}) we get *d*(*w*^{−1}) = 0. It also follows from (2) that *i*_{d} (*w* ⋅ *x*) ≠ ∅. Take any *y* ∈ *i*_{d} (*w* ⋅ *x*) and let *z* = *w*^{−1}⋅ *y*. Then
$$d(z)=d({w}^{-1}\cdot y)=d({w}^{-1})\cdot y+{w}^{-1}\cdot d(y)=0+{w}^{-1}\cdot w\cdot x\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}x.$$
This means that *z* ∈ *i*_{d}(*x*) and therefore *y* = *w* ⋅ *z* ∈ *w* ⋅ *i*_{d}(*x*). A similar proof gives *i*_{d} (*x* ⋅ *w*) = *i*_{d}(*x*)⋅ *w*. □

#### Example 2.12

*Applying* Theorem 2.11 *to* Example 2.10, *one gets* *i*_{d}(*h* ⋅ *f*) = *h* ⋅ *i*_{d}(*f*). *Presenting this with the standard integral notation, one gets*:
$$\int h(x)\phantom{\rule{thinmathspace}{0ex}}f(x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}h(x)\phantom{\rule{thinmathspace}{0ex}}\int f(x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x.$$
*Note that* *h* *is not a constant function (but it is constant on every connected component of the domain* *D*).

Next, for *y* ∈ *i*(*x*), *y*_{1} ∈ *i*(*x*_{1}), *y*_{2}∈ *i*(*x*_{2}), we describe in Theorem 2.14 the integrals that contain *y*_{1} + *y*_{2}, *y*_{1} ⋅ *y*_{2}, and *y*^{−1}, respectively.

#### Theorem 2.14

*Let* *R* *be a ring and* *d* : *R* → *R* *a derivation on* *R*.

*If* *y*_{1} ∈ *i*_{d}(*x*_{1}) *and* *y*_{2} ∈ *i*_{d}(*x*_{2}), *then* *y*_{1} + *y*_{2} ∈ *i*_{d}(*x*_{1}+*x*_{2}) *for all* *x*_{1}, *x*_{2}, *y*_{1}, *y*_{2} ∈ *R*.

*If* *y*_{1} ∈ *i*_{d}(*x*_{1}) *and* *y*_{2} ∈ *i*_{d}(*x*_{2}), *then* *y*_{1} ⋅ *y*_{2} ∈ *i*_{d}(*x*_{1} ⋅ *y*_{2} + *y*_{1} ⋅ *x*_{2}) *for all* *x*_{1}, *x*_{2}, *y*_{1}, *y*_{2} ∈ *R*.

*If* *R* *is a ring with unity* 1 ∈ *R* *and* *y* ∈ *i*_{d}(*x*), *then* *y*^{−1} ∈ *i*_{d}(−*y*^{−1} ⋅ *x* ⋅ *y*^{−1}) *for all* *x* ∈ *R* *and all invertible* *y* ∈ *R*.

#### Proof

(1) follows from *d*(*y*_{1}+*y*_{2}) = *d*(*y*_{1}) + *d*(*y*_{2}) = *x*_{1} + *x*_{2}, (2) follows from *d*(*y*_{1} ⋅ *y*_{2}) = *d*(*y*_{1}) ⋅ *y*_{2} + *y*_{1}. *d*(*y*_{2}) = *x*_{1} ⋅ *y*_{2} + *y*_{1}⋅ *x*_{2}. Finally we prove (3). It follows from 0 = *d*(1) = *d*(*y* ⋅ *y*^{−1}) = *d*(*y*) ⋅ *y*^{−1} + *y* ⋅ *d*(*y*^{−1}) that *d*(*y*^{−1}) = −*y*^{−1}⋅ *x* ⋅ *y*^{−1}. Therefore *y*^{−1} ∈ *i*_{d}(−*y*^{−1} ⋅ *x* ⋅ *y*^{−1}). □

#### Corollary 2.15

*Let* *R* *be a ring and* *d* : *R* → *R* *a derivation on R. If y, z* ∈ *i*_{d}(*x*), *then*

*y* + *z* ∈ *i*_{d}(2*x*) *and*

*y* ⋅ *z* ∈ *i*_{d}(*x* ⋅ *z* + *y* ⋅ *x*)

*for all x*,

*y*,

*z* ∈

*R*.

#### Corollary 2.16

*If* *R* *be a commutative ring with unity* 1 ∈ *R* *and* *d* : *R* → *R* *a derivation on R. If y* ∈ *i*_{d}(*x*) *then* *y*^{−1} ∈ *i*_{d} (−*y*^{−2} ⋅ *x*) *for all* *x* ∈ *R* *and all invertible* *y* ∈ *R*.

#### Proof

By Theorem 2.14, *y*^{−1} ∈ *i*_{d}(−*y*^{−1}⋅ *x* ⋅ *y*^{−1}). Since *R* is commutative, it follows that −*y*^{−1}⋅ *x* ⋅ *y*^{−1} = − *y*^{−2} ⋅ *x*. Therefore *y*^{−1}∈ *i*_{d}(−*y*^{−2} ⋅ *x*). □

We use the following lemma to prove Theorem 2.18, where the integration by parts is introduced – it generalizes the integration by parts of functions
$\int u(x)\phantom{\rule{thinmathspace}{0ex}}{v}^{\prime}(x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}u(x)\phantom{\rule{thinmathspace}{0ex}}v(x)-\int {u}^{\prime}(x)\phantom{\rule{thinmathspace}{0ex}}v(x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x,$
or more compactly
$$uv=\int u\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}v+\int v\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}u.$$

#### Lemma 2.17

*Let* *R* *be a ring and* *d* : *R* → *R* *a derivation on R. If* *i*_{d}(*x*), *i*_{d}(*y*) ≠ ∅, *then* *i*_{d} (*x* + *y*) = *i*_{d}(*x*) + *i*_{d}(*y*) *for all* *x*, *y* ∈ *R*.

#### Proof

Let *z* ∈ *i*_{d}(*x*) + *i*_{d}(*y*). Then there are *a*, *b* ∈ *R*, *a* ∈ *i*_{d}(*x*), *b* ∈ *i*_{d}(*y*), such that *z* = *a* + *b*. Then *d*(*z*) = *d*(*a*+ *b*) = *d*(*a*) + *d*(*b*) = *x* + *y* and therefore *z* ∈ *i*_{d}(*x* + *y*) (this also proves that from *i*_{d}(*x*), *i*_{d}(*y*) ≠ ∅ it follows that *i*_{d}(*x* + *y*) ≠ ∅).

Let *z* ∈ *i*_{d}(*x* + *y*) . Then *d*(*z*) = *x* + *y*. Since *i*_{d}(*x*) ≠ ∅, let *a*∈ *i*_{d}(*x*). Then *d*(*z*−*a*) = *d*(*z*)−*d*(*a*) = *d*(*z*)−*x* = *y* and therefore *z*−*a*∈ *i*_{d}(*y*). We have proved that *z* = *a* + (*z* − *a*) ∈ *i*_{d}(*x*)+*i*_{d}(*y*). □

#### Theorem 2.18

(Integration by parts). *Let* *R* *be a ring and* *d* : *R* → *R* *a derivation on R. If* *i*_{d}(*d*(*x*)⋅ *y*), *i*_{d} (*x* ⋅ *d*(*y*)) ≠ ∅, *then x* ⋅ *y* ∈ *i*_{d}(*d*(*x*)⋅ *y*) + *i*_{d}( *x* ⋅ *d*(*y*)) *for all* *x*, *y* ∈ *R*.

#### Proof

It follows from Proposition 2.3 and the definition of a derivation that *x* ⋅ *y* ∈ *i*_{d}(*d*(*x* ⋅ *y*)) = *i*_{d}(*d*(*x*) ⋅ *y* + *x* ⋅ *d*(*y*)) and therefore *x* ⋅ *y*∈ *i*_{d}(*d*(*x*) ⋅ *y* + *x* ⋅ *d*(*y*)). By Lemma 2.17, *i*_{d}(*d*(*x*)⋅ *y* + *x* ⋅ *d*(*y*)) = *i*_{d}(*d*(*x*) ⋅ *y*) + *i*_{d}(*x* ⋅ *d*(*y*)). □

As seen in Theorem 2.18, the integration by parts only works through if
$${i}_{d}\phantom{\rule{thinmathspace}{0ex}}(d(x)\phantom{\rule{thinmathspace}{0ex}}\cdot \phantom{\rule{thinmathspace}{0ex}}y),\phantom{\rule{thinmathspace}{0ex}}{i}_{d}\phantom{\rule{thinmathspace}{0ex}}(x\cdot \phantom{\rule{thinmathspace}{0ex}}d(y))\phantom{\rule{thinmathspace}{0ex}}\ne \phantom{\rule{thinmathspace}{0ex}}\mathrm{\varnothing}.$$

On the other hand, the set *i*_{d}(*d*(*x*)⋅ *y* + *x* ⋅ *d*(*y*)) is never empty (we can always find *x* ⋅ *y* in it). This gives rise to the following question. Is it possible that *i*_{d}(*d*(*x*) ⋅ *y*), *i*_{d}(*x* ⋅ *d*(*y*)) = ∅? The following example demonstrates that the answer to the question is affirmative.

#### Example 2.19

*Let R be the ring of real* 2 × 2 − *matrices*
$\begin{array}{}{M}_{2}(\mathbb{R}),\phantom{\rule{thinmathspace}{0ex}}A=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\end{array}$
*and let d* : *R* → *R* *be the inner derivation on R, defined by* *d*(*X*) = *X* ⋅ *A* − *A*. *X*. *Next, let*
$\begin{array}{}X=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]and\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}Y=\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]\end{array}$.
*Then* *d*(*X*) = −*X*, *d*(*Y*) = *Y*. *It follows from*
$$\begin{array}{}d\left(\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\left[\begin{array}{cc}0& -b\\ c& 0\end{array}\right],\end{array}$$
*d*(*X*)⋅ *Y* = − *A* *and X* ⋅ *d*(*Y*) = *A*, *that* *i*_{d}(*d*(*X*) ⋅ *Y*) = ∅ *and* *i*_{d}(*X* ⋅ *d*(*Y*)) = ∅.

We conclude the section with theorems that present results generalizing the formula
$$\int {x}^{n}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\frac{{x}^{n+1}}{n+1}+c$$
from calculus. First we state and prove the following lemma.

#### Lemma 2.20

*Let R be a ring with unity* 1 ∈ *R* *and* *d* : *R* → *R* *a derivation on R*.

*If x* ∈ *i*_{d}(**n** ⋅ *y*) *and* **n** *is invertible, then* **n**^{ −1} ⋅ *x* ∈ *i*_{d}(*y*) *for all x*, *y* ∈ *R* *and all integers n*.

*If* **n** ⋅ *x* ∈ *i*_{d}(*y*) *and* **n** *is invertible, then x* ∈ *i*_{d}(**n**^{−1}⋅ *y*) *for all x*, *y* ∈ *R* *and all integers n*.

#### Proof

It follows from *d*(**n**^{−1} ⋅ *x*) = *d*(**n**^{−1})⋅ *x* + **n**^{−1}⋅ *d*(*x*) = 0 + **n**^{−1} ⋅ *d*(*x*) = **n**^{−1} ⋅ *d*(*x*) = **n**^{−1} ⋅ **n** ⋅ *y* = *y* that **n**^{−1} ⋅ *x* ∈ *i*_{d}(*y*). This proves (1). Next, from *d*(*x*) = *d*(**n**^{−1}⋅ **n** ⋅ *x*) = *d*(**n**^{−1})⋅(**n** ⋅ *x*) + **n**^{−1} ⋅ *d*(**n** ⋅ *x*) = 0 + **n**^{−1} ⋅ *d*(**n** ⋅ *x*) = **n**^{−1} ⋅ *y* it follows that *x* ∈ *i*_{d}(**n**^{−1} ⋅ *y*). This proves (2) and we are done. □

#### Theorem 2.21

*Let R be a commutative ring with unity* 1 ∈ *R* *and* *d* : *R* → *R* *a derivation on R. Then*
$${x}^{n}\phantom{\rule{thinmathspace}{0ex}}\in {i}_{d}\phantom{\rule{thinmathspace}{0ex}}(\mathbf{n}\cdot {x}^{n-1}\cdot d(x))$$

*for all x* ∈ *R* *and all positive integers n*.

#### Proof

We prove the theorem by induction on *n*.

–

*n* = 1. Obviously *x* ∈ *i*_{d}(*d*(*x*)) *by* Proposition 2.3.

–

*n* = 2. We prove that *x*^{2} ∈ *i*_{d}(*x* ⋅ *d*(*x*)). Since *d*(*x*^{2}) = *d*(*x*) ⋅ *x* + *x* ⋅ *d*(*x*) = **2** ⋅ *x* ⋅ *d*(*x*), it follows that *x*^{2} ∈ *i*_{d}(**2** ⋅ *x* ⋅ *d*(*x*)).

–

Suppose that *x*^{n} ∈ *i*_{d}(**n** ⋅ *x*^{n−1} ⋅ *d*(*x*)). We prove that *x*^{n+1} ∈ *i*_{d}(**n**+1) ⋅ *x*^{n} ⋅ *d*(*x*)). Observe that *d*(*x*^{n+1})= *d*(*x* ⋅ *x*^{n}) = *d*(*x*) ⋅ *x*^{n} + *x* ⋅ *d*(*x*^{n}) = *d*(*x*) ⋅ *x*^{n} + *x* ⋅(**n** ⋅ *x*^{n−1} ⋅ *d*(*x*)) = *x*^{n} ⋅ *d*(*x*)+**n** ⋅ *x*^{n} ⋅ *d*(*x*) = (**n**+1) ⋅ *x*^{n} ⋅ *d*(*x*). Therefore *x*^{n+1}∈ *i*_{d}((**n**+1) ⋅ *x*^{n} ⋅ *d*(*x*)). □

#### Corollary 2.22

*Let R* *be a commutative ring with unity* 1 ∈ *R* *and* *d* : *R* → *R* *a derivation on R. If* **n** *is invertible, then*
$${\mathbf{n}}^{-1}\cdot {x}^{n}\in {i}_{d}({x}^{n-1}\cdot d(x))$$
*for all* *x* ∈ *R* *and all positive integers* *n*.

#### Theorem 2.23

*Let R* *be a commutative ring with unity* 1 ∈ *R* *and d* : *R* → *R* *a derivation on R. Then*
$${x}^{-n}\in {i}_{d}(-\mathbf{n}\cdot {x}^{-n-1}\cdot d(x))$$
*for all invertible* *x* ∈ *R* *and all positive integers n*.

#### Proof

We prove the theorem by induction on *n*.

–

*n* = 1. It follows from 0 = *d*(1) = *d*(*x* ⋅ *x*^{−1}) = *d*(*x*) ⋅ *x*^{−1} + *x* ⋅ *d*(*x*^{−1}) that *d*(*x*^{−1}) = − *x*^{−2} ⋅ *d*(*x*). Therefore *x*^{−1}∈ *i*_{d}(−*x*^{−2} ⋅ *d*(*x*)).

–

Suppose that *x*^{−n}∈ *i*_{d}(−**n** ⋅ *x*^{−n−1} ⋅ *d*(*x*)). We prove that *x*^{−n−1} ∈ *i*_{d}((−**n**−1) ⋅ *x*^{−n−2} ⋅ *d*(*x*)). Observe that *d*(*x*^{−n−1}) = *d*(*x*^{−n} ⋅ *x*^{−1}) = *d*(*x*^{−n}) ⋅ *x*^{−1} + *x*^{−n} ⋅ *d*(*x*^{−1}) = (−**n** ⋅ *x*^{−n−1} ⋅ *d*(*x*)) ⋅ *x*^{−1} + *x*^{−n} ⋅(−*x*^{−2} ⋅ *d*(*x*)) = −**n** ⋅ *x*^{−n−2} ⋅ *d*(*x*)−*x*^{−n−2} ⋅ *d*(*x*) = (−**n**−1) ⋅ *x*^{−n−2} ⋅ *d*(*x*). Therefore *x*^{−n−1} ∈ *i*_{d}((−**n**−1) ⋅ *x*^{−n−2} ⋅ *d*(*x*)). □

#### Corollary 2.24

*Let R* *be a commutative ring with unity* 1 ∈ *R* *and d* : *R* → *R* *a derivation on R. Then*
$${x}^{n}\in \phantom{\rule{thinmathspace}{0ex}}{i}_{d}\phantom{\rule{thinmathspace}{0ex}}(\mathbf{n}\cdot {x}^{n-1}\cdot d(x))$$
*for all invertible* *x* ∈ *R* *and all integers n*.

#### Corollary 2.25

*Let R* *be a commutative ring with unity* 1 ∈ *R* *and d* : *R* → *R* *a derivation on R. If* **n** *is invertible, then*
$${\mathbf{n}}^{-1}\cdot {x}^{n}\in {i}_{d}\phantom{\rule{thinmathspace}{0ex}}({x}^{n-1}\cdot d(x))$$
*for all invertible* *x* ∈ *R* *and all integers n*.

#### Proof

The result follows directly from Corollary 2.24 and Lemma 2.20. □

Let ℛ_{d} = {*i*_{d}(*x*)|*i*_{d}(*x*) ≠ ∅, *x* ∈ *R*}. Since ℛ_{d} = *R*/_{Kerd}, it follows from the first isomorphism theorem that ℛ_{d} is, as an additive group, isomorphic to *d*(*R*). Therefore the ring structure on ℛ_{d} can be easily obtained from *d*(*R*). However, ℛ_{d} can be interpreted as a ring if *d*(*R*) is a subring of *R*. As seen in Example 2.28, *d*(*R*) is not necessarily a subring of *R*.

#### Definition 2.26

*Let d* *be a derivation on a ring R. The derivation* *d* *is a proper derivation, if* *d*(*R*) *is a subring of R*.

We conclude the section with several interesting examples of proper as well as non-proper derivations.

#### Example 2.27

*Let* *R* *be a ring and d* : *R* → *R* *the trivial derivation on R. Obviously d* *is a proper derivation*.

#### Example 2.28

*Let R, d, and* *A, X, Y* ∈ *R* *be such as in* Example 2.19. *Then* *X* ⋅ *Y* = *A*, *d*(−*X*) = *X* *and* *d*(*Y*) = *Y*, *and therefore* (*by a similar argument as the one in* Example 2.19) *one can easily see that* *i*_{d}(*X* ⋅ *Y*) = ∅ *although* *i*_{d}(*x*), *i*_{d}(*Y*) ≠ ∅. *Hence*, *d* *is not a proper derivation*.

#### Example 2.29

*Let* *R* *be the polynomial ring* ℝ[*X*] *and d* : *R* → *R* *the standard derivation* *d*(*p*) = *p*′. *Then obviously, for any* *p*_{1}, *p*_{2} ∈ *R*, *p*_{1} ⋅ *p*_{2} ∈ *R* *and there always exists a polynomial* *p* ∈ *R* *such that* *d* (*p*) = *p*_{1} ⋅ *p*_{2}. *Therefore d* *is a proper derivation*.

For derivations *d* in Examples 2.29, *d*(*R*) a subring of *R* but it is not a proper subring of *R*. The following example presents a ring *R* and a non-trivial inner derivation *d* on *R* such that *d*(*R*) is a proper subring of *R*.

#### Example 2.30

*Let*
$\begin{array}{}R=\left\{\left[\begin{array}{ccc}a& b& c\\ 0& 0& d\\ 0& 0& e\end{array}\right]|a,b,c,d,e\in \mathbb{R}\right\}\end{array}$
*be the subring of real* 3×3−*matrices* *M*_{3}(ℝ), *and let*
$\begin{array}{}A=\left[\begin{array}{ccc}1& 1& 1\\ 0& 0& 1\\ 0& 0& 1\end{array}\right].\end{array}$
*Then let* *d* : *R* → *R* *be the inner derivation on* *R*, *defined by* *d*(*x*) = *X* ⋅ *A* − *A*. *X*. *It is easily seen that*
$$\begin{array}{}d(R)=\left\{\left[\begin{array}{ccc}0& a& b\\ 0& 0& c\\ 0& 0& 0\end{array}\right]|a,b,c\in \mathbb{R}\right\}\end{array}$$
*is a proper subring of R. Therefore* *d* *is a proper derivation on R*.

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