Throughout, $ R $ is an associative ring with identity. For a ring $ R $, the set of all nilpotents is denoted by $ Nil(R) $. We write $ \mathbb{Z}_{n} $ for the ring of integral modulo $ n $, $ M_{n}(R) $ for the $ n\times n $ matrix ring with identity matrix $ I_n $.

Rings whose elements are sums of certain special elements have been widely studied in ring theory. Recall that an element $ a $ of a ring is a tripotent if $ a^{3}=a $. Zhou [1] investigated that, for a ring $ R $, every element of $ R $ is a sum of a nilpotent, an idempotent and a tripotent that commute with one another, if and only if every element of $ R $ is a sum of a nilpotent and two idempotents that commute with one another. In [2], the authors determined the rings for which every element is a sum of two commuting idempotents, and characterized the rings for which every element is a sum of an idempotent and a tripotent that commute with one another. In a recent paper [3], Cui and Xia determined the rings for which every element is a sum of a nilpotent and three tripotents that commute with one another.

In [4], the authors investigated rings in which every element is a sum of two idempotents, and proved that for every $ n>1 $ and any ring $ R $, there exists a matrix $ A\in M_n(R) $ such that $ A $ cannot be written as a sum of two idempotents. In [5], Tang, Zhou and Su proved that, for a field $ F $ and any integer $ n\geq 1 $, every matrix over $ F $ is a sum of three idempotents if and only if $ F\cong\mathbb{Z}_{2} $ or $ F\cong\mathbb{Z}_{3} $. Xia, Tang and Zhou [6] showed that, for an integral domain $ R $ and any integer $ n\geq1 $, every $ n\times n $ matrix over $ R $ is a sum of two tripotents if and only if $ R\cong\mathbb{Z}_{p} $ for $ p=2, 3 $ or $ 5 $. In [7], Abyzov and Tapkin determined the rings for which every matrix is a sum of two tripotents, and proved that every $ n\times n $ matrix over a field $ F $ is a sum of two tripotents if and only if $ F $ is a prime field with $ Char{\it{(}}F{\it{)}}\leq 5 $, where Char(F) is the characteristic of $ F $.

In this paper, we go one step further investigation of this subject. In Section 2, as an extension of the work in [3], the main objective is to present the structure of reduced rings for which every element is a sum of three tripotents that commute with one another. In Section 3, we determine the integral domain $ R $ over which every $ n\times n $ matrix is a sum of three tripotents, and present equivalent conditions for an $ n\times n $ matrix over an integral domain $ R $ to be a sum of three tripotents.

In this section, we consider reduced rings for which every element is a sum of three tripotents. We call a ring $ R $ has property $ \mathcal{P} $ if every element of $ R $ is a sum of three tripotents that commute. For an integer $ n $, write $ n=n\cdot 1_R\in R $ for short.

Lemma 2.1  Let $ R $ be a ring. If $ 4=e+f+g $ with $ e^{3}=e, f^{3}=f, g^{3}=g $ and $ e, f, g $ all commute, then $ 2\cdot 3\cdot 5\cdot 7 \in Nil(R) $.

Proof  The proof is similar to that of [3, Lemma 2], we give a proof for a convenience. By assumption, $ 4^{3}-4=15(e+f+g)=(e+f+g)^{3}-(e+f+g) $ implies

$ \begin{equation} 60=15(e+f+g)=3e^{2}f+3e^{2}g+3ef^{2}+3f^{2}g+3eg^{2}+3fg^{2}+6efg\nonumber. \end{equation}$

(1.1)

Multiplying both sides of (1.1) by $ efg $ gives $ 6e^{2}f^{2}g^{2}=36efg $, which implies $ 210efg=0 $. In view of (1.1), we have

$ \begin{align} 4^{3} &=3e^{2}f+3e^{2}g+3ef^{2}+3f^{2}g+3eg^{2}+3fg^{2}+6efg+(e+f+g)\nonumber\\ &=3(e^{2}+f^{2}+g^{2})e+3(e^{2}+f^{2}+g^{2})f+3(e^{2}+f^{2}+g^{2})g+6efg-2(e+f+g)\nonumber\\ &=3(e^{2}+f^{2}+g^{2})(e+f+g)-2(e+f+g)+6efg\nonumber\\ &=12(e^{2}+f^{2}+g^{2})-8+6efg. \end{align}$

(1.2)

Multiplying both sides of (1.2) by 35 gives $ 35\cdot 4^{3}=35\cdot12(e^{2}+f^{2}+g^{2})-35\cdot8 $, so we have $ 35\cdot12(e^{2}+f^{2}+g^{2})=2^{3}\cdot3^{2}\cdot5\cdot7 $. Note that $ e^{2}+f^{2}+g^{2}=(e+f+g)^2-2(ef+eg+fg)=16-2(ef+eg+fg) $. Thus, $ 35\cdot12(2ef+2eg+2fg)=2^{3}\cdot3\cdot5^{2}\cdot7 $. Now multiplying both sides of $ (1.1) $ by $ e, f $ and $ g $ respectively, we obtain

$ \begin{align} 15e+12ef+12eg&=3e^{2}f^{2}+3e^{2}g^{2}+3ef^{2}g+3efg^{2}+6e^{2}fg; \end{align} $

(1.3)

$ \begin{align*} 15f+12ef+12fg&=3e^{2}f^{2}+3g^{2}f^{2}+3e^{2}fg+3efg^{2}+6ef^{2}g; \end{align*} $

(1.4)

$ \begin{align} 15g+12eg+12fg&=3e^{2}g^{2}+3f^{2}g^{2}+3e^{2}fg+3ef^{2}g+6efg^{2}. \end{align}$

(1.5)

By (1.3)+(1.4)+(1.5), we have

$ \begin{align} 15(e+f+g)&=12efg(e+f+g)+6(e^{2}f^{2}+e^{2}g^{2}+f^{2}g^{2})-24(ef+eg+fg)\nonumber\\ &=12efg(e+f+g)+6(ef+eg+fg)^{2}-12efg(e+f+g)-24(ef+eg+fg)\nonumber\\ &=6(ef+eg+fg)^{2}-24(ef+eg+fg)\nonumber. \end{align} $

Hence $ 2\cdot60=3(2ef+2eg+2fg)^{2}-24(2ef+2eg+2fg) $, multiplying both sides by $ 35^{2}\cdot48 $. It follows that $ 2^{6}\cdot 3^{2}\cdot 5^{3}\cdot 7^{2}\in Nil(R) $, so $ 2\cdot 3\cdot 5\cdot 7\in Nil(R) $.

Recall that a ring $ R $ is reduced if it contains no nonzero nilpotents (i.e., $ Nil(R)=\{0\} $).

Lemma 2.2  Let $ R $ be a reduced ring. Then $ R $ has property $ \mathcal{P} $ if and only if $ R\cong R_{1}\times R_{2}\times R_{3}\times R_{4} $ where each $ R_{i} $ has property $ \mathcal{P} $ with $ 2=0 $ in $ R_{1} $, $ 3=0 $ in $ R_{2} $, $ 5=0 $ in $ R_{3} $ and $ 7=0 $ in $ R_{4} $.

Proof  It is clear that the direct product of rings with property $ \mathcal{P} $ is also with property $ \mathcal{P} $, so the sufficiency follows. We now prove the necessity. In view of Lemma 2.1, we get $ 2\cdot 3\cdot 5\cdot 7=0\in R $ since $ R $ is reduced. Thus, $ 2R\cap 3R\cap 5R\cap 7R=0 $. By the Chinese Remainder Theorem,

$ R\cong R/2R\times R/3R\times R/5R\times R/7R. $

Write $ R_{1}=R/2R, \; R_{2}=R/3R, \; R_{3}=R/5R $ and $ R_{4}=R/7R $. Then every $ R_{1}, R_{2}, R_{3}, R_{4} $ has property $ \mathcal{P} $ for $ i=1, 2, 3, 4. $ Clearly, $ 2=0 $ in $ R_{1} $, $ 3=0 $ in $ R_{2} $, $ 5=0 $ in $ R_{3} $ and $ 7=0 $ in $ R_{4} $, and $ R\cong R_{1}\times R_{2}\times R_{3}\times R_{4} $.

We use $ J(R) $ to denote the Jacobson radical of a ring $ R $.

Lemma 2.3  Let $ R $ be a reduced ring, and $ p\in \{2, 3, 5, 7\} $. The following are equivalent:

$ (1) $ $ R $ has property $ \mathcal{P} $ with $ p\in J(R) $.

$ (2) $ $ R $ satisfies the identity $ x^{p}=x $ with $ p=0 $ in $ R $.

$ (3) $ $ R $ is a subdirect product of $ \mathbb{Z}_{p} $'s.

Proof  $ (1)\Rightarrow (2) $ By Lemma 2.1, $ 2\cdot 3\cdot 5\cdot 7=0\in R. $ Since $ p\in \{2, 3, 5, 7\} $ and $ p\in J(R) $, the rest three elements of $ \{2, 3, 5, 7\} $ are all units. So $ p=0\in R $.

We first show that for any $ h^3=h\in R, $ $ h^p-h=0\in R $. Indeed, if $ p\in \{3, 5, 7\} $ then $ h^p=h $; if $ p=2 $ then $ (e^2-e)^2=0 $ since $ 2=0\in R $, which yields $ e^2-e=0 $. Now, let $ x\in R. $ Then $ x=e+f+g $ with $ e^{3}=e, f^{3}=f, g^{3}=g $ and $ e, f, g $ all commute. From $ p=0\in R, $ we get $ x^p-x=(e+f+g)^p-(e+f+g)=(e^p-e)+(f^p-f)+(g^p-g)=0 $, and therefore, $ x^{p}=x $.

$ (2)\Leftrightarrow(3) $ follows from [8, Ex. 12.11].

$ (3)\Rightarrow (1) $ By [3, Theorem 10], the result follows.

Combining Lemma 2.2 and Lemma 2.3, we have the following result.

Theorem 2.4  Let $ R $ be a reduced ring. The following are equivalent:

$ (1) $ A ring $ R $ has property $ \mathcal{P} $.

$ (2) $ $ R\cong R_{1}\times R_{2}\times R_{3}\times R_{4} $, where $ R_{1} $ is zero or a subdirect product of $ \mathbb{Z}_{2} $'s, $ R_{2} $ is zero or a subdirect product of $ \mathbb{Z}_{3} $'s, $ R_{3} $ is zero or a subdirect product of $ \mathbb{Z}_{5} $'s and $ R_{4} $ is zero or a subdirect product of $ \mathbb{Z}_{7} $'s.

Recall that if $ K $ is a field, then every $ n\times n $ matrix over $ K $ with $ n\geq2 $ is similar to its rational canonical form. Let $ R $ be a ring and $ a_{0}, a_{1}, ..., a_{n-1} $ be elements in $ R $. The matrix

$ C=\left(\begin{matrix} 0&0&\cdots&0&a_{0}\\ 1&0&\cdots&0&a_{1}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&\cdots&0&a_{n-2}\\ 0&0&\cdots&1&a_{n-1}\\ \end{matrix} \right) $

is called the companion matrix associated with $ a_{0}, a_{1}, ..., a_{n-1} $. We will set $ tr(C)=a_{n-1} $.

Lemma 3.1 [6, Lemma 4.3]. Let $ R $ be a ring and $ n\geq 2 $. For any idempotent $ (n-1)\times(n-1) $ block $ E $ and for any $ (n-1)\times1 $ block $ X $ over $ R $, the $ n\times n $ matrices

$ \begin{equation*} \left(\begin{matrix} E&X\\ 0&-1\\ \end{matrix} \right) \ and \ \left(\begin{matrix} -E&X\\ 0&1\\ \end{matrix} \right) \end{equation*} $

are tripotent.

Lemma 3.2  Let $ R $ be a ring and $ C $ be an $ n\times n $ companion matrix associated with $ a_{0}, a_{1}, ..., a_{n-1} $$ \in R $. If $ tr(C)=a_{n-1}\in \{3, 2, 1, 0\} $, then $ C $ is a sum of three tripotents in $ M_{n}(R) $.

Proof  It is clear if $ n=1 $, so we are assuming $ n\geq2 $. If $ tr(C)=a_{n-1}\in \{2, 1, 0\} $, then by [7, Proposition 3], $ C $ is a sum of an idempotent and a tripotent, and so the result follows. Consequently, we can assume $ tr(C)=3 $, and decompose $ C=E+F+G $ as follows, where $ E^{3}=E $ is a consequence of Lemma 3.1, and $ F, G $ are tripotents.

If $ n $ is even, then

$ C=\left(\begin{matrix} -1&&&&&&a_{0}\\ &-1&0&&&&a_{1}\\ &1&0&&&&a_{2}\\ &&&\ddots&&&\vdots\\ &&&&-1&0&a_{n-3}\\ &&&&1&0&a_{n-2}\\ &&&&&&1\\ \end{matrix} \right)+ \left(\begin{matrix} 0&0&&&\\ 1&1&&&\\ &&\ddots&&\\ &&&0&0\\ &&&1&1\\ \end{matrix} \right)+ \left(\begin{matrix} 1&&&&\\ &0&&&\\ &&\ddots&&\\ &&&0&\\ &&&&1\\ \end{matrix} \right). $

If $ n $ is odd, then

$ C=\left(\begin{matrix} -1&0&&&&a_{0}\\ 1&0&&&&a_{1}\\ &&\ddots&&&\vdots\\ &&&-1&0&a_{n-3}\\ &&&1&0&a_{n-2}\\ &&&&&1\\ \end{matrix} \right)+ \left(\begin{matrix} 1&&&&&\\ &0&0&&&\\ &1&1&&&\\ &&&\ddots&&\\ &&&&0&0\\ &&&&1&1\\ \end{matrix} \right)+ \left(\begin{matrix} 0&&&&\\ &0&&&\\ &&\ddots&&\\ &&&0&\\ &&&&1\\ \end{matrix} \right). $

According to Lemma 3.2, we come to the following result. We use $ U(R) $ to denote the set of all units of a ring $ R $.

Theorem 3.3  Let $ R $ be an integral domain. The following are equivalent:

$ (1) $ Every matrix in $ M_{n}(R) $ is a sum of three tripotents for any integer $ n\geq 1 $.

$ (2) $ Every matrix in $ M_{2}(R) $ is a sum of three tripotents.

$ (3) $ $ R\cong \mathbb{Z}_{p} $ for $ p=2, 3, 5 $ or $ 7 $.

Proof  $ (1)\Rightarrow (2) $ This is clear.

$ (2)\Rightarrow (3) $ Note that the integral domain $ R $ can be embed into its field of fraction. For any $ E^3=E\in M_2(R) $. In view of [6, Lemma 3.1], $ tr(E)\in \{-2, -1, 0, 1, 2\} \subseteq \mathbb{Z}\cdot 1_{R} $. Let $ a\in R $, and let $ A=aE_{11} $, where $ E_{11} $ is a matrix with $ (1, 1) $-entry $ 1 $ and zeros elsewhere. By hypothesis, there exist tripotents $ A_{1}, A_{2}, A_{3}\in M_{2}(R) $ such that

$ A=aE_{11}=A_{1}+A_{2}+A_{3}. $

Then $ a=tr(A)=tr(A_{1})+tr(A_{2})+tr(A_{3}) \in \mathbb{Z}\cdot 1_{R} $. So $ R $ is a finite ring since each $ tr(A_{i}) $ is of the form $ k\cdot1_{R} $ and $ k\in \{-2, -1, 0, 1, 2\} $, $ i=1, 2, 3 $. Hence, $ R $ is a field.

Now, let $ 4I_{2}=E+F+G $ where $ E^{3}=E $, $ F^{3}=F $ and $ G^{3}=G $. Then

$ 8=tr(4I_{2})=tr(E)+tr(F)+tr(G)\in 2\mathbb{Z}\cdot 1_R. $

Note that $ tr(E)+tr(F)+tr(G)\in \{-6, -4, -2, 0, 2, 4, 6\} $. So we have $ 2m=0 $ and $ m\in \{1, 2, \ldots, 7\} $. If $ 2\in U(R) $, then $ R $ is isomorphic to one of the fields $ \mathbb{Z}_{3}, \mathbb{Z}_{5} $ or $ \mathbb{Z}_{7} $. If $ 2\notin U(R) $, then $ R\cong \mathbb{Z}_{2} $. It is clear that every matrix over $ \mathbb{Z}_{2} $ is a sum of three tripotents. Therefore, $ R \cong \mathbb{Z}_{p} $ where $ p=2, 3, 5 $ or $ 7 $.

$ (3)\Rightarrow (1) $ By Lemma 3.2, if $ tr(A)\in \{3, 2, 1, 0\} $ then $ A $ is a sum of three tripotents. Notice that a matrix $ A $ is a sum of three tripotents if and only if $ -A $ is a sum of three tripotents. So the case for $ tr(A)\in \{-3, -2, -1\} $ follows immediately. Thus, every matrix in $ M_{n}(R) $ is a sum of three tripotents.

Now, combining Theorem 2.4 and Theorem 3.3, we have the following result.

Proposition 3.4  Let $ R $ be a reduced ring with property $ \mathcal{P} $. Then every $ n\times n $ matrix over $ R $ is a sum of three tripotents.

Proof  In view of Theorem 2.4, $ R $ is isomorphic to the product of $ R_{1}, R_{2}, R_{3}, R_{4} $, where $ R_{1} $ is a subdirect product of $ \mathbb{Z}_{2} $'s, $ R_{2} $ is a subdirect product of $ \mathbb{Z}_{3} $'s, $ R_{3} $ is a subdirect product of $ \mathbb{Z}_{5} $'s and $ R_{4} $ is a subdirect product of $ \mathbb{Z}_{7} $'s.

Take $ M\in M_{n}(R_{1}) $. Let $ S $ be a subring of $ R_{1} $ generated by the elements of the matrix $ M $. So the ring $ S $ is finite, and whence $ S $ is a finite direct product of $ \mathbb{Z}_{2} $'s. By Theorem 3.3, every matrix over $ S $ is a sum of three tripotents. Hence, $ M\in M_{n}(R_{1}) $ is a sum of three tripotents. A similar argument as the above reveals that if $ M\in M_{n}(R_{i}) $, then $ M $ is also a sum of three tripotents for $ i=2, 3, 4 $.

Let $ A\in M_{n}(R) $. Write $ A=(A_{1}, A_{2}, A_{3}, A_{4}) \in M_{n}(R_{1})\times M_{n}(R_{2})\times M_{n}(R_{3})\times M_{n}(R_{4}) $, where $ A_{i}\in M_{n}(R_{i}), i=1, 2, 3, 4 $. Clearly, $ A $ is a sum of three tripotents in $ M_n(R) $.

The following example is useful to reflect conclusions.

Example 3.5 Every $ n\times n $ matrix over $ \mathbb{Z}_{210} $ is a sum of three tripotents.

Proof  By the Chinese Reminder Theorem, $ \mathbb{Z}_{210}\cong \mathbb{Z}_{2}\times \mathbb{Z}_{3}\times \mathbb{Z}_{5}\times \mathbb{Z}_{7} $. Then $ \mathbb{Z}_{210} $ is a reduced ring. By Theorem 2.4, $ \mathbb{Z}_{210} $ has property $ \mathcal{P} $. In view of Proposition 3.4, every $ n\times n $ matrix over $ \mathbb{Z}_{210} $ is a sum of three tripotents.

Proposition 3.6  Let $ R $ be a reduced ring. The following are equivalent:

$ (1) $ Every matrix in $ M_{n}(R) $ is a sum of three tripotents.

$ (2) $ $ R\cong S\times T $ where $ x^{5}=x $ for all $ x\in S $ and $ x^{7}=x $ for all $ x\in T $ with $ 7=0 $.

Proof  $ (1)\Rightarrow (2) $ Suppose that every matrix in $ M_{n}(R) $ is a sum of three tripotents. As $ R $ is a reduced ring, $ R $ is a subdirect product of integral domain {$ R_{\alpha} $}. Since $ M_{n}(R_{\alpha}) $ is a homomorphic image of $ M_{n}(R) $, it follows that every matrix in $ M_{n}(R_{\alpha}) $ is a sum of three tripotents. Thus, by Theorem 3.3 $ R_{\alpha}\cong \mathbb{Z}_{p} $ for $ p\in \{2, 3, 5, 7\} $. Let $ S=R_{1}\times R_{2}\times R_{3} $ where $ R_{1} $ is a subdirect product of $ \mathbb{Z}_{2} $'s, $ R_{2} $ is a subdirect product of $ \mathbb{Z}_{3} $'s, $ R_{3} $ is a subdirect product of $ \mathbb{Z}_{5} $'s and $ T $ is a subdirect product of $ \mathbb{Z}_{7} $'s. So $ R\cong S\times T $ where $ x^{5}=x $ for all $ x\in S $ and $ x^{7}=x $ for all $ x\in T $ with $ 7=0 $.

$ (2)\Rightarrow (1) $ Notice that $ M_n(R)\cong M_n(S)\times M_n(T) $. By [6, Theorem 4.7], every matrix in $ M_n(S) $ is a sum of two tripotents. In view of [8, Ex. 12.11], $ T $ is a subdirect product of $ \mathbb{Z}_{7} $'s. So by the proof of Proposition 3.4, every matrix in $ M_n(T) $ is a sum of three tripotents. Therefore every matrix in $ M_{n}(R) $ is a sum of three tripotents.