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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Vespri, Vincenzo / Marano, Salvatore Angelo

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Volume 15, Issue 1

# Integrations on rings

Iztok Banič
• Corresponding author
• Faculty of Natural Sciences and Mathematics, University of Maribor, Koroška 160, SI-2000 Maribor, Slovenia
• Institute of Mathematics, Physics and Mechanics, Jadranska 19, SI-1000 Ljubljana, Slovenia
• Andrej Marušič Insitute, University of Primorska, Muzejski trg 2, SI-6000 Koper, Slovenia
• Email
• Other articles by this author:
Published Online: 2017-04-01 | DOI: https://doi.org/10.1515/math-2017-0034

## Abstract

In calculus, an indefinite integral of a function f is a differentiable function F whose derivative is equal to f. The main goal of the paper is to generalize this notion of the indefinite integral from the ring of real functions to any ring. We also investigate basic properties of such generalized integrals and compare them to the well-known properties of indefinite integrals of real functions.

MSC 2010: 16W25; 17C50; 16W10; 16W99

## 1 Introduction

In calculus, an antiderivative, primitive function or indefinite integral of a function f is a differentiable function F whose derivative is equal to f, i.e., F′ = f. In present paper we generalize this notion of the indefinite integral from the ring of real functions to any ring (by a ring we always mean an associative ring not necessarily with an identity element unless explicitly mentioned otherwise). We will use the concept of well-studied derivations on rings (see Definition 1.1) to generalize this notion of an indefinite integral from calculus. Going through the literature, one may easily notice that the derivations on rings have been the subject of intensive research for decades, while the notion of integrations on rings has never been introduced.

#### Definition 1.1

Let R be a ring. An additive mapping d : RR is a derivation on R, if $d(x⋅y)=d(x)⋅y+x⋅d(y)$

for all x, yR.

This concept of derivations on rings has also been generalized (as well as restricted) in many ways, for instance, there are inner derivations, Jordan derivations, Lie derivations, (θ, ϕ)-derivations, generalized derivations, left derivations, and more; for examples see [111]. In present paper we also deal with two such functions that are related to a derivation, see Definition 1.2.

#### Definition 1.2

Let R be a ring. An additive mapping

1. d : RR is an inner derivation on R, if there is aR such that $d(x)=[x,a]=x⋅a−a⋅x,forallx∈R.$

2. δ : RR is a Jordan derivation on R, if $δ(x∘y)=δ(x)∘y+x∘δ(y)$

for all x, yR, where xy = xy + yx denotes the Jordan product on R.

It is a well-known fact that every inner derivation is a derivation, but there are derivations that are not inner derivations. Also, every derivation is a Jordan derivation. It is also a well-known fact that there are Jordan derivations that are not derivations. A classical result of Herstein asserts that any Jordan derivation on 2-torsion free prime ring (recall that a ring is n-torsion free, where n > 1 is an integer, if nx = 0 implies x = 0 for any xR) is a derivation (see also [12]). Cusak has generalized the Herstein’s theorem to 2-torsion free semiprime rings [13] (see also [14]).

Derivations and Jordan derivations on rings have become very popular since their introduction and have been studied by many authors and many papers appeared. Both have been studied intensively not only on rings but also on other algebraic structures; for examples see [1217], where more references can be found.

It is a well-known fact that the indefinite integral of a function always gives rise to a family of functions and therefore it can be interpreted as the transformation, which maps each function f to the family of functions {g|g′ = f}. Therefore, to define integrations on rings as such set-valued transformations, the following notation is needed. For a ring R, let 2R denote the family of all subsets of R. Then, an integration (a Jordan integration) i on the ring R will be defined as such (set-valued) transformation i : R → 2R. For any A, B ∈ 2R \ {∅}, we also define A + B and AB in a natural way as A + B = {a + b| aA, bB} and AB = {ab| aA, bB}. In a special case when A = {a}, we write a + B or aB instead of A + B or AB, respectively.

We proceed as follows. In the next section we introduce the notion of integration on a ring and show some of its basic properties. In Section 3 we define a Jordan integration on a ring and show that every integration is also a Jordan integration. We also show that there is a Jordan integration which is not an integration.

## 2 Integrations on rings

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, xR and d : RR a derivation on R. Let $id(x)={y∈R|x=d(y)}.$

We say that yR is a d-primitive element of the element xR, if $y∈id(x).$

We call the function id : R → 2R, xid (x), the d-integration on R and the set id (x) the d-integral of x.

The function i : R → 2R is an integration on R, if there is a derivation d on R, such that i = id. For each xR we call i (x) an integral of x.

#### Example 2.2

Let R be a ring and d : RR the trivial derivation (d(x) = 0 for all xR). Then id (x) = R if x = 0 and id (x) = ∅ if x ≠ 0.

Many results that follow are easily obtained from the definition of an integration and the fact that any derivation d : RR 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 : RR a derivation on R. Then the following holds.

1. 0 ∈ id (0).

2. For each xR, xid (d(x)).

3. If id (x) ≠ ∅ then d (id(x)) = {x} for all xR.

#### Proof

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

#### Proposition 2.4

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

1. d is surjective if and only if id(x) ≠ ∅ for all xR.

2. d is injective if and only if |id(x)| = 1 for all xR.

#### Proof

First we prove (1). Suppose that d is surjective and let xR. Then there is yR such that x = d(y). It follows that yid(x) and id(x) is not empty. Suppose that id(x) ≠ ∅ for all xR. Let xR and yid(x). It follows that x = d(y).

Next we prove (2). Suppose that d is injective, xR, and y, zid(x). Then d(y) = d(z) = x. Since d is injective, it follows that y = z and therefore id(x) consists only of a single element. Suppose that |id(x)| = 1 for all xR. Let x, y, zR be such elements that x = d(y) = d(z). Then y, zid(x) and therefore y = z (since |id(x)| = 1). □

The theorems that follow describe basic properties of integrations, introduced in Definition 2.1 First we show some properties of the integral id(0) (Theorem 2.6). Note that id(0) = Ker(d) = {xR|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 = $\underset{n}{\underset{⏟}{1+1+1+\dots +1}}$R.

#### Theorem 2.6

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

1. Thenn, nid(0) for each positive integer n.

2. If n is invertible, then $n−1⋅m,m⋅n−1∈id(0)$ for all integers m, n.

#### Proof

First we prove (1). From d(1) = d(1 ⋅ 1) = d(1) ⋅ 1 + 1 ⋅ d(1) = 2d(1) it follows that d(1) = 0. Therefore 1 ∈ id(0). Since $d(n)=d(1+1+1+…+1)⏟n=d(1)+d(1)+d(1)+…+d(1)⏟n=0,$ it follows that nid(0). Also, from d(−n) = −d(n) = 0 it follows that −nid(0).

To prove (2), we first observe that d(n−1)= 0. This follows from the fact that 0 = d(1) = d(nn−1) = d(n) ⋅ n−1 + nd(n−1) = 0 + nd(n−1) = nd(n−1), and since n is invertible, d(n−1) = 0. Next, it follows from d(n−1m) = d(n−1) ⋅ m + n−1d(m) = 0 + 0 = 0, that n−1mid(0). A similar proof gives mn−1id(0). □

#### Corollary 2.7

Let R be a ring with unity 1 ∈ R and d : RR a derivation on R. If yid(x) and n is invertible, then $y+n−1⋅m,y+m⋅n−1∈id(x)$ for all integers m, n and all x, yR.

#### Proof

The result follows from d(y + n−1m) = d(y) + d(n−1m) = 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, zR, d : RR a derivation on R and yid(x). The following statements are equivalent.

1. zid(x).

2. There is a unique element w0 ∈ Ker(d), such that z = y + w0.

#### Proof

Suppose first that zid(x). Let w0 = zy. Obviously, z = y + w0 and d(w0) = d(zy) = 0. If z = y + w0 = y + w1 for w0, w1 ∈ Ker(d), then 0 = zz = w0w1 and therefore w0 = w1. For the converse, let z = y + w0 for any w0∈ Ker(d). Then d(z) = d(y + w0) = d(y) + d(w0) = x + 0 = x and therefore zid(x).   □

#### Corollary 2.9

Let R be a ring, xR, d : RR a derivation on R and yid(x). Then $id(x)={y+z|z∈Ker(d)}.$

#### Proof

The result follows directly from Theorem 2.8   □

#### Example 2.10

Let $D=\mathbb{R}\phantom{\rule{thinmathspace}{0ex}}\mathrm{\setminus }\phantom{\rule{thinmathspace}{0ex}}\left\{\frac{\left(2k+1\right)\pi }{2}|\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}k\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{is an integer}\right\}$ 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 (fg)(x) = f(x)g(x) for each xD. Let d be the standard derivation on D, i.e. d(f) = ffor each fR. Next, let hR be defined by

$h(x)=(2k+1)π2,ifx∈(2k−1)π2,(2k+1)π2.$

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

Next we realize the well-known formula $∫c⋅f(x)dx=c⋅∫f(x)dx$ from calculus for arbitrary rings.

#### Theorem 2.11

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

1. wid(x) ⊆ id (wx) and id(x)⋅ wid(xw).

2. If id(x) ≠ ∅, then id (wx), id (xw) ≠ ∅.

3. If R is a ring with unity 1 ∈ R, w an invertible element and id(x) ≠ ∅, then id (wx) = wid(x) and id (xw) = id(x)⋅ w.

#### Proof

If id(x) = ∅, then obviously (1) holds. Suppose that id(x) ≠ ∅. We only prove that wid(x) ⊆ id (wx) (id(x) ⋅ wid (xw) can be proved similarly). Let zid(x). Then d(wz) = d(w) ⋅ z + wd(z) = 0 + wx = wx and therefore wzid (wx). This proves (1) and (2). To prove (3), suppose that (w) is invertible and id(x) ≠ ∅. First observe, that from 0 = d(1) = d(ww−1) = d(w)⋅ w−1 + wd(w−1) = wd(w−1) we get d(w−1) = 0. It also follows from (2) that id (wx) ≠ ∅. Take any yid (wx) and let z = w−1y. Then $d(z)=d(w−1⋅y)=d(w−1)⋅y+w−1⋅d(y)=0+w−1⋅w⋅x=x.$ This means that zid(x) and therefore y = wzwid(x). A similar proof gives id (xw) = id(x)⋅ w. □

#### Example 2.12

Applying Theorem 2.11 to Example 2.10, one gets id(hf) = hid(f). Presenting this with the standard integral notation, one gets: $∫h(x)f(x)dx=h(x)∫f(x)dx.$ Note that h is not a constant function (but it is constant on every connected component of the domain D).

#### Remark 2.13

Let R be a ring with unity 1 ∈ R. If w is not an invertible element, then id(x) ≠ ∅ does not necessarily imply id (wx) = wid(x) or id (xw) = id(x) ⋅ w. This follows immediately from the fact that $\int w\left(x\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\phantom{\rule{thinmathspace}{0ex}}\ne \phantom{\rule{thinmathspace}{0ex}}w\left(x\right)\cdot \int \mathrm{d}x,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}for\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}w\left(x\right)=0$ for each x ∈ ℝ.

Next, for yi(x), y1i(x1), y2i(x2), we describe in Theorem 2.14 the integrals that contain y1 + y2, y1y2, and y−1, respectively.

#### Theorem 2.14

Let R be a ring and d : RR a derivation on R.

1. If y1id(x1) and y2id(x2), then y1 + y2id(x1+x2) for all x1, x2, y1, y2R.

2. If y1id(x1) and y2id(x2), then y1y2id(x1y2 + y1x2) for all x1, x2, y1, y2R.

3. If R is a ring with unity 1 ∈ R and yid(x), then y−1id(−y−1xy−1) for all xR and all invertible yR.

#### Proof

(1) follows from d(y1+y2) = d(y1) + d(y2) = x1 + x2, (2) follows from d(y1y2) = d(y1) ⋅ y2 + y1. d(y2) = x1y2 + y1x2. Finally we prove (3). It follows from 0 = d(1) = d(yy−1) = d(y) ⋅ y−1 + yd(y−1) that d(y−1) = −y−1xy−1. Therefore y−1id(−y−1xy−1). □

#### Corollary 2.15

Let R be a ring and d : RR a derivation on R. If y, zid(x), then

1. y + zid(2x) and

2. yzid(xz + yx)

for all x, y, zR.

#### Proof

The results follow directly from Theorem 2.14. □

#### Corollary 2.16

If R be a commutative ring with unity 1 ∈ R and d : RR a derivation on R. If yid(x) then y−1id (−y−2x) for all xR and all invertible yR.

#### Proof

By Theorem 2.14, y−1id(−y−1xy−1). Since R is commutative, it follows that −y−1xy−1 = − y−2x. Therefore y−1id(−y−2x). □

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 $\int u\left(x\right)\phantom{\rule{thinmathspace}{0ex}}{v}^{\prime }\left(x\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}u\left(x\right)\phantom{\rule{thinmathspace}{0ex}}v\left(x\right)-\int {u}^{\prime }\left(x\right)\phantom{\rule{thinmathspace}{0ex}}v\left(x\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x,$ or more compactly $uv=∫udv+∫vdu.$

#### Lemma 2.17

Let R be a ring and d : RR a derivation on R. If id(x), id(y) ≠ ∅, then id (x + y) = id(x) + id(y) for all x, yR.

#### Proof

Let zid(x) + id(y). Then there are a, bR, aid(x), bid(y), such that z = a + b. Then d(z) = d(a+ b) = d(a) + d(b) = x + y and therefore zid(x + y) (this also proves that from id(x), id(y) ≠ ∅ it follows that id(x + y) ≠ ∅).

Let zid(x + y) . Then d(z) = x + y. Since id(x) ≠ ∅, let aid(x). Then d(za) = d(z)−d(a) = d(z)−x = y and therefore zaid(y). We have proved that z = a + (za) ∈ id(x)+id(y). □

#### Theorem 2.18

(Integration by parts). Let R be a ring and d : RR a derivation on R. If id(d(x)⋅ y), id (xd(y)) ≠ ∅, then xyid(d(x)⋅ y) + id( xd(y)) for all x, yR.

#### Proof

It follows from Proposition 2.3 and the definition of a derivation that xyid(d(xy)) = id(d(x) ⋅ y + xd(y)) and therefore xyid(d(x) ⋅ y + xd(y)). By Lemma 2.17, id(d(x)⋅ y + xd(y)) = id(d(x) ⋅ y) + id(xd(y)). □

As seen in Theorem 2.18, the integration by parts only works through if $id(d(x)⋅y),id(x⋅d(y))≠∅.$

On the other hand, the set id(d(x)⋅ y + xd(y)) is never empty (we can always find xy in it). This gives rise to the following question. Is it possible that id(d(x) ⋅ y), id(xd(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 $\begin{array}{}{M}_{2}\left(\mathbb{R}\right),\phantom{\rule{thinmathspace}{0ex}}A=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\end{array}$ and let d : RR be the inner derivation on R, defined by d(X) = XAA. 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 $dabcd=0−bc0,$ d(X)⋅ Y = − A and Xd(Y) = A, that id(d(X) ⋅ Y) = ∅ and id(Xd(Y)) = ∅.

We conclude the section with theorems that present results generalizing the formula $∫xndx=xn+1n+1+c$ from calculus. First we state and prove the following lemma.

#### Lemma 2.20

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

1. If xid(ny) and n is invertible, then n −1xid(y) for all x, yR and all integers n.

2. If nxid(y) and n is invertible, then xid(n−1y) for all x, yR and all integers n.

#### Proof

It follows from d(n−1x) = d(n−1)⋅ x + n−1d(x) = 0 + n−1d(x) = n−1d(x) = n−1ny = y that n−1xid(y). This proves (1). Next, from d(x) = d(n−1nx) = d(n−1)⋅(nx) + n−1d(nx) = 0 + n−1d(nx) = n−1y it follows that xid(n−1y). This proves (2) and we are done.   □

#### Theorem 2.21

Let R be a commutative ring with unity 1 ∈ R and d : RR a derivation on R. Then $xn∈id(n⋅xn−1⋅d(x))$

for all xR and all positive integers n.

#### Proof

We prove the theorem by induction on n.

• n = 1. Obviously xid(d(x)) by Proposition 2.3.

• n = 2. We prove that x2id(xd(x)). Since d(x2) = d(x) ⋅ x + xd(x) = 2xd(x), it follows that x2id(2xd(x)).

• Suppose that xnid(nxn−1d(x)). We prove that xn+1id(n+1) ⋅ xnd(x)). Observe that d(xn+1)= d(xxn) = d(x) ⋅ xn + xd(xn) = d(x) ⋅ xn + x ⋅(nxn−1d(x)) = xnd(x)+nxnd(x) = (n+1) ⋅ xnd(x). Therefore xn+1id((n+1) ⋅ xnd(x)).   □

#### Corollary 2.22

Let R be a commutative ring with unity 1 ∈ R and d : RR a derivation on R. If n is invertible, then $n−1⋅xn∈id(xn−1⋅d(x))$ for all xR and all positive integers n.

#### Proof

The result follows directly from Theorem 2.21 and Lemma 2.20.   □

#### Theorem 2.23

Let R be a commutative ring with unity 1 ∈ R and d : RR a derivation on R. Then $x−n∈id(−n⋅x−n−1⋅d(x))$ for all invertible xR and all positive integers n.

#### Proof

We prove the theorem by induction on n.

• n = 1. It follows from 0 = d(1) = d(xx−1) = d(x) ⋅ x−1 + xd(x−1) that d(x−1) = − x−2d(x). Therefore x−1id(−x−2d(x)).

• Suppose that xnid(−nxn−1d(x)). We prove that xn−1id((−n−1) ⋅ xn−2d(x)). Observe that d(xn−1) = d(xnx−1) = d(xn) ⋅ x−1 + xnd(x−1) = (−nxn−1d(x)) ⋅ x−1 + xn ⋅(−x−2d(x)) = −nxn−2d(x)−xn−2d(x) = (−n−1) ⋅ xn−2d(x). Therefore xn−1id((−n−1) ⋅ xn−2d(x)).   □

#### Corollary 2.24

Let R be a commutative ring with unity 1 ∈ R and d : RR a derivation on R. Then $xn∈id(n⋅xn−1⋅d(x))$ for all invertible xR and all integers n.

#### Proof

If n ≠ 0, the result follows from Theorem 2.21 and Theorem 2.23. If n = 0, then x0 = 1 ∈ id(0) by Theorem 2.6.   □

#### Corollary 2.25

Let R be a commutative ring with unity 1 ∈ R and d : RR a derivation on R. If n is invertible, then $n−1⋅xn∈id(xn−1⋅d(x))$ for all invertible xR and all integers n.

#### Proof

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

Let ℛd = {id(x)|id(x) ≠ ∅, xR}. Since ℛd = R/Kerd, it follows from the first isomorphism theorem that ℛd is, as an additive group, isomorphic to d(R). Therefore the ring structure on ℛd can be easily obtained from d(R). However, ℛd 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 : RR the trivial derivation on R. Obviously d is a proper derivation.

#### Example 2.28

Let R, d, and A, X, YR be such as in Example 2.19. Then XY = 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 id(XY) = ∅ although id(x), id(Y) ≠ ∅. Hence, d is not a proper derivation.

#### Example 2.29

Let R be the polynomial ring ℝ[X] and d : RR the standard derivation d(p) = p′. Then obviously, for any p1, p2R, p1p2R and there always exists a polynomial pR such that d (p) = p1p2. 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 M3(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 : RR be the inner derivation on R, defined by d(x) = XAA. X. It is easily seen that $d(R)=0ab00c000|a,b,c∈R$ is a proper subring of R. Therefore d is a proper derivation on R.

## 3 Jordan integrations on rings

In this section we introduce the new concept of Jordan integrations on rings and show basic properties of such Jordan integrations.

#### Definition 3.1

Let R be a ring, xR and δ : RR a Jordan derivation on R. Let $jδ(x)={y∈R|x=δ(y)}.$

We say that yR is a Jordan δprimitive element of the element xR, if $y∈jδ(x).$

We call the function jδ : R, → 2R, xjδ (x), the Jordan δintegration on R and the set jδ(x) the Jordan δintegral of x.

The function j : R → 2R is a Jordan integration on R, if there is a Jordan derivation δ on R, such that j = jδ. For each xR we call j (x) a Jordan integral of x.

Since every derivation d is a Jordan derivation, it follows that any d-integration is a Jordan d-integration. Since there are rings R and Jordan derivations δ on R that are not derivations, it follows that for such δ’s, the corresponding Jordan δ-integrations jδ are never δ-integrations. But, is it true that for any such δ there is a derivation d, such that jδ = id? The following theorem answers the question in negative.

#### Theorem 3.2

There is a ring R and a Jordan derivation δ on R such thatfor each derivation d on R, idjδ.

#### Proof

Let R be a ring and δ a Jordan derivation on R that is not a derivation. Assume that there is a derivation d on R such that id(x) = jδ(x) for all xR. Let yR be any element. Then there is xR such that yid(x) = jδ(x). It follows that d(y) = x = δ(y). Therefore d = δ, which is a contradiction. □

Following the same line of thoughts as in Section 2, one can obtain similar results for Jordan integrations – they are listed in the theorem below.

#### Theorem 3.3

Let R be a ring and δ : RR a Jordan derivation on R. Then the following holds.

1. 0 ∈ jδ(0).

2. If y, zjδ(x), then yzjδ(0) for all x, y, zR.

3. For all xR, xjδ(δ(x)).

4. If xR and yjδ(x), then $jδ(x)={y+z|z∈Ker(δ)}.$

5. If jδ(x) ≠ ∅, then δ (jδ(x)) = {x} for all xR.

6. If jδ(x), jδ(y) ≠∅, then jδ(x + y) = jδ(x) + jδ(y) for all x, yR.

7. (Jordan integration by parts)

If jδ(δ(x) ⋅ y), jδ(xδ(y)), jδ(δ(y) ⋅ x), jδ(yδ(x)) ≠∅, then $x∘y∈jδ(δ(x)⋅y)+jδ(x⋅δ(y))+jδ(δ(y)⋅x)+jδ(y⋅δ(x))$ for all x, yR.

8. If y1jδ(x1) and y2jδ(x2), then y1 + y2jδ(x1 + x2) for all x1, x2, y1, y2R.

9. If y1jδ(x1) and y2jδ (x2), then y1y2jδ (x1y2 + y1x2 + x2y1 + y2x1) for all x1, x2, y1, y2R.

10. The Jordan derivation δ is surjective if and only if jδ(x) ≠∅for all xR.

11. The Jordan derivation δ is injective if and only if|jδ(x)| = 1 for all xR.

#### Proof

One can easily prove the theorem using similar arguments to those in Section 2. We leave the details to a reader. □

## Acknowledgement

The author acknowledges the financial support from the Slovenian Research Agency (research core funding No. P1-0285).

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Accepted: 2016-01-11

Published Online: 2017-04-01

Citation Information: Open Mathematics, Volume 15, Issue 1, Pages 365–373, ISSN (Online) 2391-5455,

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