# Automorphisms and derivations of finite-dimensional algebras

Matej Brešar  Faculty of Mathematics and Physics, University of Ljubljana, and
Faculty of Natural Sciences and Mathematics, University of Maribor, Slovenia
<EMAIL_ADDRESS>
###### Abstract.

Let $A$ be a finite-dimensional algebra over a field $F$ with char$(F)\neq 2$.
We show that a linear map $D:A\to A$ satisfying $xD(x)x\in[A,A]$ for every
$x\in A$ is the sum of an inner derivation and a linear map whose image lies
in the radical of $A$. Assuming additionally that $A$ is semisimple and
char$(F)\neq 3$, we show that a linear map $T:A\to A$ satisfies
$T(x)^{3}-x^{3}\in[A,A]$ for every $x\in A$ if and only if there exist a
Jordan automorphism $J$ of $A$ lying in the multiplication algebra of $A$ and
a central element $\alpha$ satisfying $\alpha^{3}=1$ such that $T(x)=\alpha
J(x)$ for all $x\in A$. These two results are applied to the study of local
derivations and local (Jordan) automorphisms. In particular, the second result
is used to prove that every local Jordan automorphism of a finite-dimensional
simple algebra $A$ (over a field $F$ with char$(F)\neq 2,3$) is a Jordan
automorphism.

###### Key words and phrases:

Derivation, automorphism, antiautomorphism, Jordan automorphism, local
derivation, local automorphism, finite-dimensional algebra, simple algebra,
semisimple algebra, radical, functional identity

###### 2020 Mathematics Subject Classification:

16W20, 16W25, 16R60

Supported by the Slovenian Research Agency (ARRS) Grant P1-0288.

##  1\. Introduction

The theory of functional identities deals with the description of functions on
rings and algebras that satisfy certain identities [5]. In this paper, we
consider a more general type of problems where expressions involving functions
on an algebra $A$ are, instead of being always 0 as is usually the case with
functional identities, contained in a relatively large subset of $A$, namely
in $[A,A]$, the linear span of all commutators in $A$. In the special case
where $A$ is the matrix algebra $M_{n}(F)$ this can be equivalently stated as
that the trace of these expressions is always zero. Therefore, the relations
that we will study are, in some sense, also more general than trace identities
(see, e.g., [1]). We will not, however, develop some general theory, but
consider only two special cases that merely indicate a possible new approach
to “generalized identities” in rings and algebras. The two new type theorems
will be shown to have applications to a well-studied research topic, i.e., to
the theory of local derivations and local automorphisms.

Let us be more specific. All our results consider a finite-dimensional algebra
$A$ over a field $F$ with char$(F)\neq 2$. In Section 3, we additionally
assume that char$(F)\neq 3$. Section 2 is centered around the condition that a
linear map $D:A\to A$ satisfies

(1.1) $xD(x)x\in[A,A]\quad\mbox{for all $x\in A$},$

and Section 3 is centered around the condition that a linear map $T:A\to A$
satisfies

(1.2) $T(x)^{3}-x^{3}\in[A,A]\quad\mbox{for all $x\in A$.}$

It is immediate that inner derivations satisfy (1.1) and inner automorphisms
satisfy (1.2).

Our first main result, Theorem 2.3, states that (1.1) implies that $D$ is the
sum of an inner derivation and a linear map having the image in rad$(A)$, the
radical of $A$. Although maps with the image in rad$(A)$ do not always satisfy
(1.1), it is still reasonable that they appear in the conclusion since
rad$(A)$ is sometimes contained in $[A,A]$.

Condition (1.2) can be studied similarly as condition (1.1), but the results
are more involved. Theorem 3.3, which is our second main result, states that
if $A$ is semisimple then condition (1.2) is equivalent to the condition that
there exist a central element $\alpha$ satisfying $\alpha^{3}=1$ and a Jordan
automorphism $J$ of $A$ belonging to the multiplication algebra of $A$ such
that $T(x)=\alpha J(x)$ for all $x\in A$. In Corollaries 3.7 and 3.8, we
consider the situation where $A$ is a general, not necessarily semisimple
finite-dimensional algebra. However, these results are not as definitive as
Theorem 2.3.

The proofs of Theorems 2.3 and 3.3 use the classical theory of finite-
dimensional algebras together with some results on Jordan maps. We also
provide several examples that justify the assumptions.

As already indicated, these two theorems are applicable to the study of local
derivations and local automorphisms. A local derivation of an algebra $A$ is a
linear map $D:A\to A$ with the property that for each $x\in A$, there is a
derivation $D_{x}:A\to A$ such that $D(x)=D_{x}(x)$. This notion was
introduced in 1990 by Kadison [11] and independently by Larson and Sourour
[12] who also introduced local automorphisms. These are defined analogously,
i.e., as linear maps $T:A\to A$ such that for each $x\in A$, there is an
automorphism $T_{x}:A\to A$ satisfying $T(x)=T_{x}(x)$. The definitions of
other types of local maps should now be self-explanatory. The standard
question is whether local derivations, local automorphisms, etc. are
derivations, automorphisms, etc. Over the last three decades, positive answers
were obtained in various algebras $A$ (occurring not only in algebra but also
if not primarily in functional analysis). We refer to a few recent
publications [4, 8, 9] which contain some historical remarks and further
references.

Theorem 2.3 immediately implies that every local inner derivation of a finite-
dimensional semisimple algebra $A$ (over a field $F$ with char$(F)\neq 2$) is
an inner derivation (Corollary 2.8). In Example 2.9 we show that a similar
conclusion for general derivations does not always hold. The automorphism case
is more complex and interesting. As will be explained in Section 3, local
Jordan automorphisms are more natural in finite dimensions than local
automorphisms. Using Theorem 3.3, we will show that every local Jordan
automorphism of a finite-dimensional simple algebra $A$ over a field $F$ with
char$(F)\neq 2,3$ is a Jordan automorphism (Theorem 3.11). This is the third
main result of this paper.

To the best of our knowledge, our results on local maps are new and cover a
basic class of algebras which is quite different from those treated by other
authors. Papers on local maps are often based on the existence of some special
elements like idempotents. The class of simple algebras, however, includes
division algebras which contain no such elements.

##  2\. Derivations

We start with a simple but important lemma. As usual, we write $[x,y]$ for the
commutator $xy-yx$, and $[A,A]$ for the linear span of all $[x,y]$, $x,y\in
A$.

######  Lemma 2.1.

Let $A$ be a finite-dimensional simple algebra. If $c\in A$ is such that
$cA\subseteq[A,A]$, then $c=0$.

###### Proof.

From $xcy=[x,cy]+cyx\in[A,A]$ we see that the ideal of $A$ generated by $c$ is
contained in $[A,A]$. However, it is easy to see that $[A,A]$ is a proper
subspace of $A$ (in fact, it has codimension $1$ if viewed as a space over the
center $Z$ of $A$ [2, Exercise 4.12]). Hence, $c=0$. ∎

######  Remark 2.2.

We will also need the following technical variation of Lemma 2.1: If
char$(F)\neq 2$ and $A$ is as in the lemma, then $cx^{2}\in[A,A]$ for every
$x\in A$ implies $c=0$. This follows immediately from
$x=\frac{1}{2}((x+1)^{2}-x^{2}-1^{2})$ (here we used that a finite-dimensional
simple algebra always contains $1$). Note that we may replace the condition
$cx^{2}\in[A,A]$ by $xcx\in[A,A]$ since $xcx=cx^{2}+[x,cx]$.

Let $D$ be a linear map from an algebra $A$ to itself. A linear map
$\Delta:A\to A$ is called a Jordan $(D,D)$-derivation if it satisfies

$\Delta(x^{2})=D(x)x+xD(x)\quad\mbox{for all $x\in A$}.$

This notion was introduced in [3] (as a special case of more general Jordan
$(D,G)$-derivations) in order to study the classical Jordan derivations (the
case where $\Delta=D$) on tensor products. Somewhat to the author’s surprise,
Jordan $(D,D)$-derivations naturally occur in the proof of the next theorem,
and the result from [3] stating that they satisfy $\Delta(xy)=D(x)y+xD(y)$
provided that $A$ is a semiprime algebra (over a field of characteristic not
$2$) is applicable.

Let us also recall a few standard definitions and facts. The radical of a
finite-dimensional algebra $A$, denoted rad$(A)$, is the unique maximal
nilpotent ideal of $A$. An equivalent description is that rad$(A)$ is the
intersection of all maximal ideals of $A$. If rad$(A)=\\{0\\}$, then $A$ is a
semisimple algebra, i.e., $A$ is a direct sum of ideals each of which is a
simple algebra. The quotient algebra $A/{\rm rad}(A)$ is always semisimple.
Finally, the algebra of all linear maps from $A$ to $A$ of the form
$x\mapsto\sum_{i}a_{i}xb_{i}$ for some $a_{i},b_{i}\in A$ is called the
multiplication algebra of $A$. It will be denoted by $M(A)$.

We can now state our first main theorem.

######  Theorem 2.3.

Let $A$ be a finite-dimensional algebra over a field $F$ with char$(F)\neq 2$.
If a linear map $D:A\to A$ satisfies $xD(x)x\in[A,A]$ for every $x\in A$, then
$D$ is the sum of an inner derivation of $A$ and a linear map from $A$ to
rad$(A)$.

###### Proof.

We write $x\equiv y$ for $x-y\in[A,A]$. Our assumption is thus $xD(x)x\equiv
0$ for all $x\in A$. As char$(F)\neq 2$, replacing $x$ by $x\pm y$ implies

(2.1) $yD(x)x+xD(y)x+xD(x)y\equiv 0\quad\mbox{for all $x,y\in A$.}$

This will be our basic relation in the course of the proof.

Let $M$ be a maximal ideal of $A$. Taking $y\in M$ it follows from (2.1) that
$xD(y)x\in M+[A,A]$ for every $x\in A$. Hence, $c=D(y)+M\in A/M$ satisfies
$ucu\in[A/M,A/M]$ for every $u\in A/M$. As $A/M$ is simple, Remark 2.2 tells
us that $c=0$, i.e., $D(y)\in M$. We have thus proved that $D(M)\subseteq M$
for every maximal ideal $M$ of $A$.

As rad$(A)$ is the intersection of all maximal ideals of $A$, it follows that
$D({\rm rad}(A))\subseteq{\rm rad}(A)$. We can thus define
$\overline{D}:A/{\rm rad}(A)\to A/{\rm rad}(A)$ by

$\overline{D}(x+{\rm rad}(A))=D(x)+{\rm rad}(A).$

Note that $\overline{D}$ is a linear map satisfying

$v\overline{D}(v)v\in[A/{\rm rad}(A),A/{\rm rad}(A)]\quad\mbox{for all $v\in
A/{\rm rad}(A)$}.$

Assuming that the theorem is true for semisimple algebras, it follows from
this relation that $\overline{D}$ is an inner derivation of $A/{\rm rad}(A)$,
which further implies that $D$ is of the desired form. Therefore, we may
assume without loss of generality that $A$ is a semisimple algebra.

Thus, $A=A_{1}\oplus\dots\oplus A_{r}$ where each $A_{i}$ is a simple algebra.
Since $A_{i}$ is the intersection of the maximal ideals
$A_{1}\oplus\dots\oplus A_{j-1}\oplus A_{j+1}\dots\oplus A_{r}$ with $j\neq
i$, it is invariant under $D$ by what we proved above. Considering the
restriction of $D$ to each $A_{i}$, we see that there is no loss of generality
in assuming that $A$ is a simple algebra.

Let $Z$ denote the center of $A$. Take $z\in Z$. Substituting $zy$ for $y$ in
(2.1) we obtain

$zyD(x)x+xD(zy)x+zxD(x)y\equiv 0\quad\mbox{for all $x,y\in A$.}$

On the other hand, since $z[A,A]\subseteq[A,A]$, (2.1) shows that

$zyD(x)x+zxD(y)x+zxD(x)y\equiv 0\quad\mbox{for all $x,y\in A$.}$

Comparing both relations we obtain

$x\big{(}D(zy)-zD(y)\big{)}x\equiv 0\quad\mbox{ for all $x,y\in A$.}$

Since $A$ is simple, we see from Remark 2.2 that $D(zy)=zD(y)$, i.e., $D$ is
$Z$-linear. But then $D$ belongs to the multiplication algebra $M(A)$ [2,
Lemma 1.25].

Let $a_{i},b_{i}\in A$ be such that

$D(x)=\sum_{i}a_{i}xb_{i}\quad\mbox{for all $x\in A$}.$

We have

$xD(y)x=\sum_{i}xa_{i}yb_{i}x=\sum_{i}b_{i}x^{2}a_{i}y+\sum_{i}[xa_{i}y,b_{i}x]$

and so $xD(y)x\equiv\sum_{i}b_{i}x^{2}a_{i}y$. Using also $yD(x)x\equiv
D(x)xy$ we now see that (2.1) can be rewritten as

$\big{(}D(x)x+\sum_{i}b_{i}x^{2}a_{i}+xD(x)\big{)}y\equiv 0\quad\mbox{for all
$x,y\in A$.}$

Therefore, by Lemma 2.1,

$D(x)x+\sum_{i}b_{i}x^{2}a_{i}+xD(x)=0\quad\mbox{for all $x\in A$.}$

This means that the linear map $\Delta:A\to A$ defined by

$\Delta(x)=-\sum_{i}b_{i}xa_{i}$

is a Jordan $(D,D)$-derivation. By [3, Theorem 4.3], $\Delta$ satisfies

(2.2) $\Delta(xy)=D(x)y+xD(y)\quad\mbox{for all $x,y\in A$.}$

Writing first $1$ for $x$ and then $1$ for $y$ we see that $\beta=D(1)$ lies
in $Z$. From $\Delta(x)=D(x)+\beta x$ and (2.2) it follows that $d:A\to A$
defined by

$d(x)=D(x)-\beta x$

is a derivation. Moreover, $d$ is $Z$-linear. It is a standard fact that such
a derivation is inner (see, e.g., [2, Exercise 4.20]). Thus, there is an $a\in
A$ such that $D(x)=\beta x+[a,x]$ for all $x\in A$.

The proof will be complete by showing that $\beta=0$. Since

(2.3) $x[a,x]x=[xax,x]\equiv 0$

it follows from $xD(x)x\equiv 0$ that $\beta x^{3}\equiv 0$. If $\beta$ was
not $0$, every $x\in A$ would satisfy

(2.4) $x^{3}\equiv 0.$

To show that this is not true, first note that a complete linearization of
(2.4) gives

(2.5)
$x_{1}x_{2}x_{3}+x_{1}x_{3}x_{2}+x_{2}x_{1}x_{3}+x_{2}x_{3}x_{1}+x_{3}x_{1}x_{2}+x_{3}x_{2}x_{1}\equiv
0$

for all $x_{1},x_{2},x_{3}\in A$. Since char$(F)\neq 2$, we also have

(2.6) $x_{1}^{2}x_{2}+x_{1}x_{2}x_{1}+x_{2}x_{1}^{2}\equiv 0$

for all $x_{1},x_{2}\in A$. Let $K$ be the algebraic closure of $Z$ and let
$A_{K}=K\otimes A$ be the scalar extension of $A$ to $K$. Take
$y=\sum_{j}k_{j}\otimes x_{j}\in A_{K}$. Observe that $y^{3}$ is a sum of
terms of the form

$k\otimes x_{i}^{3},$
$k^{\prime}\otimes(x_{i}^{2}x_{j}+x_{i}x_{j}x_{i}+x_{j}x_{i}^{2}),$

and

$k^{\prime\prime}\otimes(x_{i}x_{j}x_{k}+x_{i}x_{k}x_{j}+x_{j}x_{i}x_{k}+x_{j}x_{k}x_{i}+x_{k}x_{i}x_{j}+x_{k}x_{j}x_{i}).$

Using (2.4), (2.5), and (2.6) it follows that $y^{3}\in[A_{K},A_{K}]$.
However, $A_{K}\cong M_{n}(K)$ and so this obviously cannot hold for every
$y\in A_{K}$ (e.g., for an idempotent of rank $1$). This contradiction proves
that $\beta=0$. ∎

######  Remark 2.4.

The last paragraph of the proof could be shortened if char$(F)$ was different
from $3$. Indeed, since $x_{1}^{2}x_{2}\equiv x_{1}x_{2}x_{1}\equiv
x_{2}x_{1}^{2}$, in this case it is enough to apply Lemma 2.1 to (2.6). If,
however, char$(F)=3$, then (2.6) holds for any algebra over $F$ as
$x_{1}^{2}x_{2}+x_{1}x_{2}x_{1}+x_{2}x_{1}^{2}$ is then equal to
$[x_{1},[x_{1},x_{2}]]$.

As noticed in (2.3), every inner derivation $D$ satisfies $xD(x)x\in[A,A]$.
Thus, if the algebra $A$ is such that rad$(A)\subseteq[A,A]$, then Theorem 2.3
turns into an “if and only if” theorem. A simple concrete example is the
algebra $A$ of all upper triangular matrices over $F$ (which actually
satisfies rad$(A)=[A,A]$). Another example is of course any semisimple
algebra. We record this as a corollary.

######  Corollary 2.5.

Let $A$ be a finite-dimensional semisimple algebra over a field $F$ with
char$(F)\neq 2$. The following conditions are equivalent for a linear map
$D:A\to A$:

  1. (i)

$xD(x)x\in[A,A]$ for every $x\in A$.

  2. (ii)

$D$ is an inner derivation.

Inner derivations $D$ also satisfy a simpler condition $xD(x)\in[A,A]$ for all
$x\in A$. However, so do many other maps, as the next example shows. There are
thus good reasons for considering the condition $xD(x)x\in[A,A]$.

######  Example 2.6.

Every map of the form $D(x)=\sum_{i}a_{i}xb_{i}-b_{i}xa_{i}$ satisfies
$xD(x)\in[A,A]$ for all $x\in A$. This follows from
$x(a_{i}xb_{i}-b_{i}xa_{i})=[xa_{i},xb_{i}]$.

The following example shows that the assumption that char$(F)\neq 2$ is
necessary in Corollary 2.5.

######  Example 2.7.

Let $F=\mathbb{F}_{2}$ be the field with 2 elements and let $A=M_{2}(F)$.
Define $D:A\to A$ by

$D\left(\left[\begin{matrix}x_{11}&x_{12}\cr
x_{21}&x_{22}\cr\end{matrix}\right]\right)=\left[\begin{matrix}x_{22}&x_{12}\cr
0&x_{11}\cr\end{matrix}\right].$

Using $xy(x+y)=0$ for all $x,y\in F$ one can check that the trace of the
matrix $xD(x)x$ is $0$ for every $x\in A$. Therefore, $xD(x)x$ lies in
$[A,A]$. However, $D$ is not a derivation.

In the rest of this section we consider local (inner) derivations.

######  Corollary 2.8.

Let $A$ be a finite-dimensional semisimple algebra over a field $F$ with
char$(F)\neq 2$. Then every local inner derivation $D:A\to A$ is an inner
derivation.

###### Proof.

The condition that $D$ is a local inner derivations means that for every $x\in
A$, there exists an $a_{x}\in A$ such that $D(x)=[a_{x},x]$. This obviously
implies $xD(x)x=[xa_{x}x,x]\in[A,A]$. Therefore, $D$ is an inner derivation by
Corollary 2.5. ∎

The next example shows that Corollary 2.8 cannot be extended to general, not
necessarily inner derivations. In fact, this fails to hold even when $A$ is a
field. Of course, this can occur only if $A$ is an inseparable field extension
of $F$ in order to have nontrivial derivations. We remark that our example is
similar to the one from [11] which concerns the algebra $\mathbb{C}(X)$
(which, however, is infinite-dimensional over $\mathbb{C}$).

######  Example 2.9.

Let $p$ be an odd prime and let $F=\mathbb{F}_{p}(t)$ be the rational function
field over the field with $p$ elements $\mathbb{F}_{p}$. If $\alpha$ is a root
of the (irreducible and inseparable) polynomial $X^{p}-t$, then $A=F(\alpha)$
has degree $p$ over $F$. For any $f=\sum_{k=0}^{p-1}a_{k}\alpha^{k}\in A$,
$a_{k}\in F$, define $f^{\prime}=\sum_{k=0}^{p-1}ka_{k}\alpha^{k-1}$. Observe
that $f\mapsto f^{\prime}$ is an $F$-linear derivation of $A$ whose kernel is
$F$. For any $g\in A$, $f\mapsto gf^{\prime}$ is also an $F$-linear derivation
of $A$. This readily implies that every $F$-linear map of $A$ that sends $1$
to $0$ is a local derivation. However, such a map is not necessarily a
derivation. For example, if it sends $\alpha$ to $1$ and $\alpha^{2}$ to $0$,
then it certainly is not.

Corollary 2.8 also does not hold without the assumption of finite
dimensionality.

######  Example 2.10.

There exist (infinite-dimensional) division algebras $D$ in which every
nonzero inner derivation is surjective [7]. Every linear map from $D$ to $D$
that vanishes at central elements is then a local inner derivation. However,
such a map does not to be an inner derivation.

##  3\. Automorphisms, antiautomorphisms, and Jordan automorphisms

Let $A$ be an algebra over a field $F$. Recall that a Jordan automorphism of
$A$ is a bijective linear map $J:A\to A$ satisfying

$J(xy+yx)=J(x)J(y)+J(y)J(x)\quad\mbox{for all $x,y\in A$.}$

If char$(F)\neq 2$, this condition is equivalent to $J(x^{2})=J(x)^{2}$ for
all $x\in A$. Obvious examples of Jordan automorphisms are automorphisms and
antiautomorphisms. These obvious examples are also the only examples if $A$ is
a simple algebra over a field $F$ with char$(F)\neq 2$. This is a special case
of the classical theorem of Herstein [10] (the assumption from [10] that the
characteristic is not $3$ was later removed).

Our first lemma in this section essentially concerns central simple algebras,
but for notational consistency we state it in a slightly different form.

######  Lemma 3.1.

Let $A$ be a finite-dimensional simple algebra with center $Z$. If $J$ is a
$Z$-linear Jordan automorphism of $A$, then $J(x)-x\in[A,A]$ for all $x\in A$.

###### Proof.

Let $K$ be the algebraic closure of $Z$ and let $A_{K}=K\otimes A$ be the
scalar extension of $A$ to $K$. Then $\overline{J}={\rm id}_{K}\otimes J$ is a
Jordan automorphism, and hence an automorphism or an antiautomorphism of
$A_{K}\cong M_{n}(K)$. If $\overline{J}(u)-u\in[A_{K},A_{K}]$ for every $u\in
A_{K}$, then for $u=1\otimes x$ we obtain that $1\otimes(J(x)-x)$ is equal to
an element of the form $\sum_{i}k_{i}\otimes[x_{i},y_{i}]$, which implies that
$J(x)-x\in[A,A]$. Therefore, it is enough to consider the case where $A$ is
the $K$-algebra $M_{n}(K)$.

If $J$ is an automorphism, then $J$ is inner by the Skolem-Noether Theorem, so
there is an invertible $a\in A$ such that

$J(x)-x=axa^{-1}-x=[a,xa^{-1}]\in[A,A]$

for every $x\in A$. Assume that $J$ is an antiautomorphism. Then $x\mapsto
J(x^{t})$, where $x^{t}$ is the transpose of $x$, is an automorphism of $A$.
Hence, $J(x^{t})=axa^{-1}$ for all $x\in A$, or written equivalently,
$J(x)=ax^{t}a^{-1}$. Since $ax^{t}a^{-1}$ and $x$ have the same trace it
follows that $J(x)-x\in[A,A]$. ∎

We now begin our study of condition $T(x)^{3}-x^{3}\in[A,A]$.

######  Lemma 3.2.

Let $A$ be an algebra over a field $F$ with char$(F)\neq 2,3$. If a linear map
$T:A\to A$ satisfies $T(x)^{3}-x^{3}\in[A,A]$ for every $x\in A$, then

(3.1) $T(x)^{2}T(y)-x^{2}y\in[A,A]\quad\mbox{for all $x,y\in A$.}$

Moreover, if $A$ is finite-dimensional and $\ker T\cap{\rm rad}(A)=\\{0\\}$,
then $T$ is bijective and leaves every maximal ideal of $A$ invariant.

###### Proof.

As in the preceding section, we write $x\equiv y$ for $x-y\in[A,A]$.

Since char$(F)\neq 2$, replacing $x$ by $x\pm y$ in $T(x)^{3}\equiv x^{3}$
gives

$T(x)^{2}T(y)+T(x)T(y)T(x)+T(y)T(x)^{2}\equiv x^{2}y+xyx+yx^{2}.$

As $T(x)^{2}T(y)\equiv T(x)T(y)T(x)\equiv T(y)T(x)^{2}$, $x^{2}y\equiv
xyx\equiv yx^{2}$, and char$(F)\neq 3$, (3.1) follows.

Now assume that $A$ is finite-dimensional and $\ker T\cap{\rm
rad}(A)=\\{0\\}$. Take $y\in\ker T$. From (3.1) we see that $x^{2}y\in[A,A]$
for all $x\in A$. Hence, for any ideal $M$ of $A$ we have
$u^{2}(y+M)\in[A/M,A/M]$ for all $u\in A/M$. Assuming that $M$ is maximal it
follows from Remark 2.2 that $y\in M$. This implies that $y\in{\rm rad}(A)$
and so $y=0$ by our assumption. Thus, $T$ is bijective.

Finally, from (3.1) it now follows that $w^{2}T(y)\in M+[A,A]$ for every ideal
$M$ of $A$, $y\in M$, and $w\in A$. As in the preceding paragraph we see that
this implies $T(y)\in M$ if $M$ is maximal. ∎

We are ready to prove our second main result.

######  Theorem 3.3.

Let $A$ be a finite-dimensional semisimple algebra over a field $F$ with
char$(F)\neq 2,3$. The following conditions are equivalent for a linear map
$T:A\to A$:

  1. (i)

$T(x)^{3}-x^{3}\in[A,A]$ for every $x\in A$.

  2. (ii)

There exist a Jordan automorphism $J$ of $A$ and an element $\alpha$ from the
center $Z$ of $A$ such that $T(x)=\alpha J(x)$ for all $x\in A$. Moreover, $J$
belongs to the multiplication algebra $M(A)$ and $\alpha$ satisfies
$\alpha^{3}=1$.

###### Proof.

(i)$\implies$(ii). Lemma 3.2 tells us that $T$ is bijective and leaves maximal
ideals of $A$ invariant. The latter implies that every simple component of $A$
is also invariant under $T$. We may therefore assume without loss of
generality that $A$ is a simple algebra.

Take $z$ from the center $Z$ of $A$. Replacing $y$ by $zy$ in (3.1) we obtain

$T(x)^{2}T(zy)\equiv zx^{2}y\equiv T(x)^{2}zT(y).$

Since $T$ is surjective, we thus have $u^{2}\big{(}T(zy)-zT(y)\big{)}\equiv 0$
for every $u\in A$. Hence, $T(zy)=zT(y)$ by Remark 2.2. This implies that
$T\in M(A)$ [2, Lemma 1.25], that is, there exist $a_{i},b_{i}\in A$ such that

$T(x)=\sum_{i}a_{i}xb_{i}\quad\mbox{for all $x\in A$}.$

Hence,

$T(x)^{2}T(y)=\sum_{i}T(x)^{2}a_{i}yb_{i}\equiv\sum_{i}b_{i}T(x)^{2}a_{i}y,$

which along with (3.1) gives

$\Big{(}\sum_{i}b_{i}T(x)^{2}a_{i}-x^{2}\Big{)}y\equiv 0\quad\mbox{for all
$x,y\in A$.}$

Therefore, by Lemma 2.1,

(3.2) $W(T(x)^{2})=x^{2}\quad\mbox{for all $x\in A$,}$

where

$W(x)=\sum_{i}b_{i}xa_{i}.$

Substituting $x+1$ for $x$ in (3.2) it follows that

$W(T(x)T(1)+T(1)T(x))=2x.$

This shows that $W$ is invertible and that $S=W^{-1}$ satisfies

(3.3) $2S(x)=T(x)T(1)+T(1)T(x)\quad\mbox{for all $x\in A$.}$

By (3.2), $S(x^{2})=T(x)^{2}$ for all $x\in A$, and hence

(3.4) $S(xy+yx)=T(x)T(y)+T(y)T(x)\quad\mbox{for all $x,y\in A$.}$

As above, we denote by $K$ the algebraic closure of $Z$ and by $A_{K}=K\otimes
A$ the scalar extension of $A$ to $K$. Observe that $\overline{S}={\rm
id}_{K}\otimes S$ and $\overline{T}={\rm id}_{K}\otimes T$ are $K$-linear maps
of $A_{K}\cong M_{n}(K)$ satisfying

(3.5)
$\overline{S}(xy+yx)=\overline{T}(x)\overline{T}(y)+\overline{T}(y)\overline{T}(x)\quad\mbox{for
all $x,y\in A_{K}$.}$

Take an idempotent $e\in A_{K}$. From (3.5) we see that
$\overline{S}(e)=\overline{T}(e)^{2}$ and
$2\overline{S}(e)=\overline{T}(e)u+u\overline{T}(e)$ where
$u=\overline{T}(1\otimes 1)$. Hence,
$2\overline{T}(e)^{2}=\overline{T}(e)u+u\overline{T}(e)$, and so
$\overline{T}(e)u+u\overline{T}(e)$ commutes with $\overline{T}(e)$. Observe
that this can be equivalently stated as that $\overline{T}(e)^{2}$ commutes
with $u$. That is, $\overline{S}(e)$ commutes with $u$ for every idempotent
$e\in A_{K}$. The algebra $A_{K}\cong M_{n}(K)$ is linearly spanned by its
idempotents (indeed, observe that the matrix unit $e_{ij}$, $i\neq j$, is a
difference of two idempotents: $e_{ij}=(e_{ii}+e_{ij})-e_{ii}$). As
$\overline{S}$ is surjective it follows that $u$ is a scalar multiple of
$1\otimes 1$. Since $u=\overline{T}(1\otimes 1)=1\otimes T(1)$, this shows
that $\alpha=T(1)\in Z$. Of course, $\alpha\neq 0$. From (3.3) we see that
$\alpha^{-1}S(x)=T(x)$ for all $x\in A$. Hence, (3.4) implies that
$J(x)=\alpha^{-1}T(x)$ is a Jordan automorphism.

It remains to show that $\alpha^{3}=1$. By Lemma 3.1, $J(x)^{3}=J(x^{3})\equiv
x^{3}$, and by our assumption, $T(x)^{3}\equiv x^{3}$. Hence, $J(x)^{3}\equiv
T(x)^{3}$. Since $T(x)=\alpha J(x)$ it follows that
$(\alpha^{3}-1)J(x)^{3}\equiv 0$. As we saw at the end of the proof of Theorem
2.3, there exist elements in $A$ whose cube does not lies in $[A,A]$. Since
$J$ is surjective it follows that $\alpha^{3}-1=0$.

(ii)$\implies$(i). As $T$ lies in $M(A)$, it leaves every ideal of $A$
invariant. We may therefore assume without loss of generality that $A$ is
simple. Now, $J\in M(A)$ implies that $J$ is $Z$-linear, and so Lemma 3.1
shows that $J(x^{3})\equiv x^{3}$. Since

$J(x^{3})=J(x)^{3}=\alpha^{-3}T(x)^{3}=T(x)^{3},$

this proves (i). ∎

The author is thankful to Misha Chebotar for suggesting the trick with
idempotents after equation (3.5).

The purpose of the following example is to show that the simpler condition
$T(x)^{2}-x^{2}\in[A,A]$ is not characteristic for Jordan automorphisms
(compare Example 2.6).

######  Example 3.4.

If $a,b\in A$ are such that $a^{2}=ab=ba=b^{2}=0$, then $T(x)=x+axb-bxa$
satisfies

$T(x)^{2}-x^{2}=[xa,xb]+[ax,bx]\in[A,A]$

and $T(1)=1$. However, $T$ is not always a Jordan automorphism.

The next two examples show that Theorem 3.3 does not hold if char$(F)$ is $2$
or $3$.

######  Example 3.5.

Let $F$ be a field with char$(F)=2$, let $A=M_{2}(F)$, and let $T:A\to A$ be
given by $T(x)=x+{\rm tr}(x)1$. It is easy to see that $T$ is neither an
automorphism nor an antiautomorphism. However,

$\displaystyle T(x)^{3}-x^{3}$ $\displaystyle={\rm tr}(x)\big{(}x^{2}+{\rm
tr}(x)x\big{)}+{\rm tr}(x)^{3}1$ $\displaystyle={\rm tr}(x){\det}(x)1+{\rm
tr}(x)^{3}1$ $\displaystyle=\big{(}{\rm tr}(x){\det}(x)+{\rm
tr}(x)^{3}\big{)}[e_{12},e_{21}]\in[A,A].$

Moreover, $T(1)=1$. Thus, $T$ satisfies condition (i), but does not satisfy
condition (ii).

######  Example 3.6.

If $A$ is an algebra over a field $F$ with char$(F)=3$, then for all $x,y\in
A$ we have

$(x+y)^{3}-x^{3}-y^{3}=[x,[x,y]]+[y,[y,x]]\in[A,A]$

(compare Remark 2.4). Taking any $a\in A$ such that $a^{3}\in[A,A]$ (say,
$a^{3}=0$) and any linear functional $\varphi$ on $A$, we thus see that the
map $T:A\to A$ given by $T(x)=x+\varphi(x)a$ satisfies condition (i). However,
$T$ does not necessarily satisfy condition (ii).

The next corollary considers general finite-dimensional algebras.

######  Corollary 3.7.

Let $A$ be a finite-dimensional algebra over a field $F$ with char$(F)\neq
2,3$. If a linear map $T:A\to A$ satisfies $T(x)^{3}-x^{3}\in[A,A]$ for every
$x\in A$, then $T(x^{4})-T(x)^{4}\in{\rm rad}(A)$ for every $x\in A$.
Moreover, if $A$ is unital and $T(1)=1$, then $T(x^{2})-T(x)^{2}\in{\rm
rad}(A)$ for every $x\in A$.

###### Proof.

Lemma 3.2 implies that $T({\rm rad}(A))\subseteq{\rm rad}(A)$. Therefore, we
can define $\overline{T}:A/{\rm rad}(A)\to A/{\rm rad}(A)$ by

$\overline{T}(x+{\rm rad}(A))=T(x)+{\rm rad}(A).$

Since

$\overline{T}(v)^{3}-v^{3}\in[A/{\rm rad}(A),A/{\rm rad}(A)]\quad\mbox{for all
$v\in A/{\rm rad}(A)$}$

it follows from Theorem 3.3 that there exist a Jordan automorphism
$\overline{J}$ of $A/{\rm rad}(A)$ and an element $\overline{\alpha}$ from the
center of $A/{\rm rad}(A)$ such that $\overline{\alpha}^{3}=1$ and
$\overline{T}(v)=\overline{\alpha}\overline{J}(v)$ for all $v\in A/{\rm
rad}(A)$. Accordingly,

$\overline{T}(v^{4})=\overline{\alpha}\overline{J}(v^{4})=\overline{\alpha}\overline{J}(v)^{4}=\overline{\alpha}^{4}\overline{J}(v)^{4}=\overline{T}(v)^{4}$

for all $v\in A/{\rm rad}(A)$, which shows that $T(x^{4})-T(x)^{4}\in{\rm
rad}(A)$ for all $x\in A$. Finally, if $A$ is unital and $T(1)=1$, then also
$\overline{T}(1)=1$ and hence $\overline{\alpha}=1$. Thus, $\overline{T}$ is a
Jordan automorphism and so $T(x^{2})-T(x)^{2}\in{\rm rad}(A)$ for every $x\in
A$. ∎

Assuming that the field $F$ is perfect, which makes it possible for us to use
the Wedderburn Principal Theorem, we obtain a nicer result that is more
similar to Theorem 2.3. We will assume for simplicity that $A$ is unital and
$T(1)=1$.

######  Corollary 3.8.

Let $A$ be a unital finite-dimensional algebra over a perfect field $F$ with
char$(F)\neq 2,3$. If a linear map $T:A\to A$ satisfies $T(1)=1$ and
$T(x)^{3}-x^{3}\in[A,A]$ for every $x\in A$, then $T$ is the sum of a Jordan
endomorphism of $A$ and a linear map from $A$ to rad$(A)$.

###### Proof.

By the Wedderburn Principal Theorem, $A$ contains a subalgebra $S$ (isomorphic
to $A/{\rm rad}(A)$) such that $A$ is the vector space direct sum of $S$ and
rad$(A)$. Let $\pi$ be the projection on $S$ along rad$(A)$. As $T-\pi T$ has
image in rad$(A)$ we must only prove that $\pi T$ is a Jordan endomorphism.
Now, Corollary 3.7 tells us that $\pi(T(x^{2})-T(x)^{2})=0$, and since $\pi$
is an endomorphism this can be written as $(\pi T)(x^{2})=(\pi T)(x)^{2}$. ∎

We continue with an analog of Corollary 2.8.

######  Corollary 3.9.

Let $A$ be a finite-dimensional semisimple algebra over a field $F$ with
char$(F)\neq 2,3$. Then every local inner automorphism $T:A\to A$ is a Jordan
automorphism belonging to $M(A)$.

###### Proof.

Our assumption can be read as that for each $x\in A$, there is an invertible
$a_{x}\in A$ such that $T(x)=a_{x}xa_{x}^{-1}$. Hence,

$T(x)^{3}-x^{3}=[a_{x}x^{3},a_{x}^{-1}]\in[A,A].$

As $T(1)=1$, Theorem 3.3 gives the desired conclusion. ∎

There are many algebras in which every local automorphism is an automorphism.
However, the matrix algebra $M_{n}(F)$ is not one of them. Indeed, any matrix
$x\in M_{n}(F)$ is similar to its transpose $x^{t}$, so $x\mapsto x^{t}$ is an
example of a local inner automorphism which is not an automorphism but an
antiautomorphism. This explains why Jordan automorphisms appear in the
conclusion of Corollary 3.9. Moreover, it indicates that in just about any
reasonable class of finite-dimensional algebras, the question whether local
Jordan automorphisms are Jordan automorphisms is more natural than the usual
question whether local automorphisms are automorphisms.

Our last theorem gives an answer to the question just raised. In its proof we
will use the following elementary lemma. Actually, we will need only its
special case where each $m_{i}=3$. The general form, however, may be useful
elsewhere.

######  Lemma 3.10.

Let $F$ be an infinite field, let $V$and $W$ be vector spaces over $F$, and
let $n$ and $m_{1},\dots,m_{n}$ be positive integers. Suppose that
$m_{i}$-linear maps $f_{i}:V^{m_{i}}\to W$ are such that for each $x\in V$, at
least one of the elements $f_{1}(x,\dots,x),\dots,f_{n}(x,\dots,x)$ is $0$.
Then there exists an $i\in\\{1,\dots,n\\}$ such that $f_{i}(x,\dots,x)=0$ for
every $x\in V$.

###### Proof.

Suppose the lemma is not true. Then for each $i\in\\{1,\dots,n\\}$ there
exists an $x_{i}\in V$ such that $f_{i}(x_{i},\dots,x_{i})\neq 0.$ Choose a
linear functional $\tau_{i}$ on $W$ such that
$\tau_{i}(f_{i}(x_{i},\dots,x_{i}))\neq 0$. For any $z_{1},\dots,z_{n}\in F$,
define

$p_{i}(z_{1},\dots,z_{n})=\tau_{i}(f_{i}(z_{1}x_{1}+\dots+z_{n}x_{n},\dots,z_{1}x_{1}+\dots+z_{n}x_{n})).$

Note that the assumption of the lemma implies that for every
$(z_{1},\dots,z_{n})\in F^{n}$, there is an $i\in\\{1,\dots,n\\}$ such that
$p_{i}(z_{1},\dots,z_{n})=0$. Consequently,

(3.6) $p_{1}(z_{1},\dots,z_{n})\cdots p_{n}(z_{1},\dots,z_{n})=0$

for all $(z_{1},\dots,z_{n})\in F^{n}$.

Since $f_{i}$ is $m_{i}$-linear, we may consider $p_{i}$ as a polynomial in
$z_{1},\dots,z_{n}$. Its coefficient at $z_{i}^{m_{i}}$ is
$\tau_{i}(f_{i}(x_{i},\dots,x_{i}))$, so $p_{i}\neq 0$. The product
$p_{1}\cdots p_{n}$ is therefore a nonzero polynomial too. However, (3.6)
shows that this polynomial vanishes at every $(z_{1},\dots,z_{n})\in F^{n}$,
which contradicts the assumption that $F$ is infinite. ∎

Lemma 3.10 will make it possible for us to use Theorem 3.3 in the proof of the
following theorem, provided of course that the field $F$ is infinite. The
theorem also holds for finite fields, but for them we will have to use a
different method which is more similar to standard methods for tackling local
automorphisms.

######  Theorem 3.11.

Let $A$ be a finite-dimensional simple algebra over a field $F$ with
char$(F)\neq 2,3$. Then every local Jordan automorphism $T:A\to A$ is a Jordan
automorphism.

###### Proof.

We consider separately two cases.

Case 1: $F$ is infinite. The center $Z$ of $A$ is a finite extension of $F$,
so there are only finitely many $F$-linear automorphisms of $Z$. Denote them
by $\sigma_{1},\dots,\sigma_{n}$. The restriction of every Jordan automorphism
of $A$ to $Z$ is therefore one of the $\sigma_{i}$’s.

For each $x\in A$, there is a Jordan automorphism $T_{x}$ of $A$ such that
$T(x)=T_{x}(x)$. Therefore, $A$ is the union of its subsets

$A_{i}=\\{x\in A\,|\,\left.T_{x}\right|_{Z}=\sigma_{i}\\},\,\,\,i=1,\dots,n.$

For every $i$ such that $A_{i}\neq\emptyset$ choose an $x_{i}\in A_{i}$ and
set $T_{i}=T_{x_{i}}$.

Take $x\in A_{i}$. Then $T_{i}^{-1}T_{x}$ fixes elements from $Z$ and is
therefore a $Z$-linear Jordan automorphism. Hence,

$(T_{i}^{-1}T)(x)^{3}-x^{3}=(T_{i}^{-1}T_{x})(x)^{3}-x^{3}=(T_{i}^{-1}T_{x})(x^{3})-x^{3}\in[A,A]$

by Lemma 3.1. This shows that $A_{i}$ is a subset of the set

$B_{i}=\\{x\in A\,|\,(T_{i}^{-1}T)(x)^{3}-x^{3}\in[A,A]\\}.$

Therefore, $A=\bigcup_{i=1}^{n}B_{i}$.

Define $f_{i}:A^{3}\to A/[A,A]$ by

$f_{i}(x,y,z)=(T_{i}^{-1}T)(x)(T_{i}^{-1}T)(y)(T_{i}^{-1}T)(z)-xyz+[A,A].$

Note that $x\in B_{i}$ if and only if $f_{i}(x,x,x)=0$. The conditions of
Lemma 3.10 are therefore satisfied, and so there exists an
$i\in\\{1,\dots,n\\}$ such that $f_{i}(x,x,x)=0$ for every $x\in A$. That is,
$(T_{i}^{-1}T)(x)^{3}-x^{3}\in[A,A]$ for every $x\in A$. Theorem 3.3 shows
that $T_{i}^{-1}T$ is a Jordan automorphism ($\alpha=1$ since
$(T_{i}^{-1}T)(1)=1$). But then the same holds for $T=T_{i}(T_{i}^{-1}T)$.
This completes the proof for this case.

Case 2: $F$ is finite. Without loss of generality we may assume that $F$ is
equal to its prime subfield $\mathbb{F}_{p}$. The center $Z$ of $A$ is then
the finite field $\mathbb{F}_{p^{n}}$, and, by Wedderburn’s theorems, $A$ can
be identified with the matrix algebra $M_{s}(Z)$ for some $s\geq 1$.

Let $\sigma$ denote the Frobenius automorphism of $Z$. It is well known that
the only automorphisms of $Z$ are $\sigma^{i}$, $i=0,1,\dots,n-1$. Also, it is
well known that $Z$ contains a normal basis, i.e., there exists an $a\in Z$
such that the elements $\sigma^{i}(a)=a^{p^{i}}$, $i=0,1,\dots,n-1$, form a
basis of $Z$ over $F$.

The restriction of every Jordan automorphism of $A$ to $Z$ is an automorphism
of $Z$. Therefore, for each $z\in Z$ there exists an $i$ such that
$T(z)=\sigma^{i}(z)$. Let $k$ be such that $T(a)=\sigma^{k}(a)$. Take $j\geq
1$. Then there exist $l,m$ such that

$T(\sigma^{j}(a))=\sigma^{l}(a)$

and

$T(a-\sigma^{j}(a))=\sigma^{m}(a-\sigma^{j}(a))=\sigma^{m}(a)-\sigma^{j+m}(a).$

On the other hand,

$T(a-\sigma^{j}(a))=T(a)-T(\sigma^{j}(a))=\sigma^{k}(a)-\sigma^{l}(a).$

Hence,

$\sigma^{m}(a)-\sigma^{j+m}(a)=\sigma^{k}(a)-\sigma^{l}(a).$

Since $\sigma^{i}(a)$, $i=0,1,\dots,n-1$, are linearly independent and
char$(F)\neq 2$, it follows that $m=k$ and $j+m\equiv l$ (mod $n$). This means
that

$T(\sigma^{j}(a))=\sigma^{k}(\sigma^{j}(a))\quad\mbox{for every $j\geq 1$.}$

Consequently, $\left.T\right|_{Z}=\sigma^{k}$. We can extend $\sigma^{k}$ to
an automorphism $S$ of $A$ in the obvious way, that is,
$S((z_{ij}))=(\sigma^{k}(z_{ij}))$ for every matrix $(z_{ij})\in A$. Note that
$S^{-1}T$ is a local Jordan automorphism of $A$ that acts as the identity on
$Z$. Therefore, there is no loss of generality in assuming that $T$ itself is
the identity on $Z$. Also, we may now assume that $s\geq 2$.

Let $A_{0}=M_{s}(F)$. It is obvious that $\left.T\right|_{A_{0}}$, the
restriction of $T$ to $A_{0}$, is an $F$-linear local Jordan homomorphism from
$A_{0}$ to $A$. By [6, Theorem 2.1], $\left.T\right|_{A_{0}}$ is the sum of a
homomorphism $\Phi$ and an antihomomorphism $\Psi$. Suppose $\Phi\neq 0$. From
$\Phi(1)=\Phi(e_{11})+\dots+\Phi(e_{ss})$ (where as above $e_{ii}$ are matrix
units) we see that $\Phi(1)$ is an idempotent that can be written as a sum of
$s$ mutually orthogonal nonzero idempotents in $A$. This implies that
$\Phi(1)=1$. Similarly, $\Psi\neq 0$ yields $\Psi(1)=1$. However, since
$1=T(1)=\Phi(1)+\Psi(1)$, one of $\Phi$ and $\Psi$ is $0$. That is,
$\left.T\right|_{A_{0}}$ is either a homomorphism or an antihomomorphism.
Hence, $R:A\to A$ given by

$R\big{(}\sum z_{ij}e_{ij}\big{)}=\sum z_{ij}T(e_{ij})\quad\mbox{for all
$z_{ij}\in Z$}$

is a ($Z$-linear) automorphism or antiautomorphism of $A$. Note that $R^{-1}T$
is a local Jordan automorphism of $A$ that satisfies $(R^{-1}T)(a_{0})=a_{0}$
for every $a_{0}\in A_{0}$. Therefore, without loss of generality we may
assume that $T$ itself is the identity on $A_{0}$.

The proof will be completed by showing that $T$ is the identity on $A$.

We start the proof with a general remark. Observe that a nonzero matrix $a\in
A$ has rank one if and only if $aAa\subseteq Za$. This shows that every Jordan
automorphism of $A$ maps the set of rank one matrices onto itself. The same is
then true for every local Jordan automorphism. Thus, $a$ has rank one if and
only if $T(a)$ has rank one.

Take and fix an arbitrary rank-one idempotent $e$ belonging to $A_{0}$. For
any $z\in Z$, we have

$T(ze)=T_{ze}(ze)=T_{ze}(z)T_{ze}(e).$

Note that $T_{ze}(z)\in Z$ and $T_{ze}(e)$ is a rank-one idempotent. Thus, for
every $z\in Z$ there exist a $z^{\prime}\in Z$ and a rank-one idempotent
$e_{z}\in A$ such that $T(ze)=z^{\prime}e_{z}$. Similarly,
$T(z(1-e))=z^{\prime\prime}f_{z}$ where $z^{\prime\prime}\in Z$ and $f_{z}$ is
an idempotent (of rank $s-1$, but we will not need this). Hence,

$z=T(z)=T(ze)+T(z(1-e))=z^{\prime}e_{z}+z^{\prime\prime}f_{z}.$

This implies that $z^{\prime\prime}(e_{z}f_{z}-f_{z}e_{z})=0$. Since
$z^{\prime\prime}\neq 0$ whenever $z\neq 0$ it follows that $e_{z}$ and
$f_{z}$ commute for every $z\in Z$. Further, squaring
$z=z^{\prime}e_{z}+z^{\prime\prime}f_{z}$ we obtain

$z^{2}=z^{\prime 2}e_{z}+z^{\prime\prime
2}f_{z}+2z^{\prime}z^{\prime\prime}f_{z}e_{z}=z^{\prime}(z-z^{\prime\prime}f_{z})+f_{z}(z^{\prime\prime
2}+2z^{\prime}z^{\prime\prime}e_{z}),$

which gives

$(z-z^{\prime})z=f_{z}(z^{\prime\prime
2}+2z^{\prime}z^{\prime\prime}e_{z}-z^{\prime}z^{\prime\prime}).$

Since $f_{z}$ is a nontrivial idempotent it follows that $z=z^{\prime}$ or
$z=0$. We have thereby proved that for every $z\in Z$, there is a rank-one
idempotent $e_{z}\in A$ such that $T(ze)=ze_{z}$. We may assume that
$e_{0}=e$.

Since $e\in A_{0}$, $T(e)=e$. For every $z\in Z$ we thus have

$(z+1)e_{z+1}=T((z+1)e)=T(ze)+T(e)=ze_{z}+e.$

By squaring we obtain

$(z+1)^{2}e_{z+1}=z^{2}e_{z}+z(e_{z}e+ee_{z})+e.$

On the other hand,

$(z+1)^{2}e_{z+1}=(z+1)(ze_{z}+e).$

Comparing the last two identities we see that $z(e_{z}-e)^{2}=0$. This shows
that $a_{z}=e_{z}-e$ has square $0$ for every $z\in Z$. Since $e_{z}=e+a_{z}$
is an idempotent it follows that $a_{z}=ea_{z}+a_{z}e$. Consequently,
$ea_{z}e=0$ and hence

$a_{z}=(1-e)a_{z}e+ea_{z}(1-e).$

Suppose $a_{z}\neq 0$. Then at least one of $(1-e)a_{z}e$ and $ea_{z}(1-e)$,
let us say the latter one, is nonzero. Take any $u\in(1-e)A_{0}e$ such that
$(1-e)za_{z}e\neq u$. Then $ze-u\in Ae$ has rank one, but

$T(ze-u)=z(e+a_{z})-u=ze+\big{(}(1-e)za_{z}e-u\big{)}+eza_{z}(1-e)$

has rank at least $2$ since it is a sum of an element from $Ze$, a nonzero
element from $(1-e)Ae$, and a nonzero element from $eA(1-e)$. This
contradiction shows that $a_{z}=0$. Thus, $T(ze)=ze$ for every $z\in Z$ and
every rank one idempotent $e\in A_{0}$.

Since $e_{ii}$ and $e_{ii}+e_{ij}$, where $i\neq j$, are rank-one idempotents
belonging to $A_{0}$, it follows that $T(ze_{ii})=ze_{ii}$ and
$T(ze_{ij})=ze_{ij}$ for all $z\in Z$ and all $1\leq i,j\leq s$. But then
$T(a)=a$ for every $a\in A$. ∎

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