All rings considered are associative with unity. Let $ R $ be a ring. The set of all units, the set of all idempotents and the Jacobson radical of $ R $ are denoted by $ U(R) $, $ idem(R) $ and $ J(R) $, respectively. The symbol $ M_{n}(R) $ stands for the $ n\times n $ matrix ring over $ R $ whose identity element we write as $ I_{n} $.

Rings whose elements are sums of certain special elements have been widely studied in ring theory. Recall that a ring $ R $ is called {clean} if every element of $ R $ is the sum of an idempotent and a unit. Clean rings were introduced by Nicholson [1] in relation to exchange rings. A ring $ R $ is called {strongly clean} [2] if every element of $ R $ is the sum of an idempotent and a unit that commutes. According to [3, 4], a ring $ R $ is called {J-clean} if for each $ a\in R $, $ a=e+j $ for some $ e^2=e\in R $ and $ j\in J(R) $ (also called a {semiboolean ring} in [5]). Recently, Danchev [6] and Kosan et al. [3] called a ring $ R $ UJ if every unit of $ R $ is the sum of an idempotent and an element from $ J(R) $, or equivalently, $ U(R) = 1 + J(R) $. It was shown in [3] that a ring $ R $ is J-clean if and only if $ R $ is a clean UJ-ring.

Due to Harte [7], an element $ a\in R $ is called quasinilpotent if $ 1-ax\in U(R) $ for every $ x\in {\rm comm}(a); $ the set of all quasinilpotents of $ R $ is denoted by $ R^{ qnil} $. Clearly, $ J(R)\subseteq R^{qnil} $. Motivated by the above, we say that a unit $ u $ of a ring $ R $ is $ UQ $ if $ u=1+q $ for some $ q\in R^{qnil} $; and a ring $ R $ is $ UQ $ if every unit of $ R $ is $ UQ $ (equivalently, $ U(R)=1+R^{qnil} $). Elementary properties of $ UQ $-elements are studied in section 2, and some characterizations of $ UQ $-rings are provided in section 3. In section 4, we investigate rings for which every unit is the sum of two idempotents and an element from the Jacobson radical that commute with each other.

Let $ R $ be a ring. We say that a unit $ u $ of $ R $ is $ UJ $ if $ u=1+j $ for some $ j\in{J(R)} $. Clearly, all $ UJ $-elements are $ UQ $. In this section, we study the properties of $ UQ $-elements (including $ UJ $-elements).

The following result can be obtained by a direct check.

Proposition 2.1   The product of $ UJ $-elements is $ UJ $.

Remark 2.2   The product of $ UQ $-elements needs not to be $ UQ $. For example, let $ \mathbb{Z}_{2} $ be the ring of integers modulo $ 2 $, and let

$ u= \left( \begin{array}{cc} 1 & 0 \\ 1 & 1 \\ \end{array} \right), v= \left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} \right)\in{U(M_{2}(\mathbb{Z}_{2}))}. $

Then $ u, v $ are clearly $ UQ $. But

$ uv=\left( \begin{array}{cc} 1 & 0 \\ 1 & 1 \\ \end{array} \right)\left(\begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} \right)=\left(\begin{array}{cc} 1 & 1 \\ 1 & 0 \\ \end{array} \right)=\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right)+\left(\begin{array}{cc} 0 & 1 \\ 1 & 1 \\ \end{array} \right) $

is not $ UQ $ since $ \left(\begin{array}{cc} 0 & 1 \\ 1 & 1 \\ \end{array} \right) \notin U(M_2(\mathbb{Z}_{2})) $.

For a ring $ R $, we denote $ 2\times2 $ upper triangular matrix ring over $ R $ by $ T_2(R) $.

Proposition 2.3  Let $ R $ be a ring, $ u, v \in U(R) $. Then

$ (1) $ $ u, v $ are $ UJ $ if and only if $ \left( \begin{array}{cc} u & x \\ 0 & v \\ \end{array} \right) $ is $ UJ $ in $ T_2(R) $ for any $ x\in{R} $.

$ (2) $ $ u, v $ are $ UQ $ if and only if, for any $ x\in R $, $ \left( \begin{array}{cc} u & x \\ 0 & v \\ \end{array} \right) $ is $ UQ $ in $ T_{2}(R) $.

Proof   $ (1) $ Assume that $ u, v $ are UJ. Let $ u=1+j_{1} $, $ v=1+j_{2} $ where $ j_1, j_2\in J(R) $. Then

$ \left( \begin{array}{cc} u & x \\ 0 & v \\ \end{array} \right)= \left( \begin{array}{cc} 1+j_{1} & x \\ 0 & 1+j_{2} \\ \end{array} \right)= \left( \begin{array}{cc} j_{1} & x \\ 0 & j_{2} \\ \end{array} \right)+\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right). $

Since $ \left( \begin{array}{cc} j_{1} & x \\ 0 & j_{2} \\ \end{array} \right)\in{J(T_{2}(R))} $ for any $ x\in R $, we get $ \left( \begin{array}{cc} u & x \\ 0 & v \\ \end{array} \right) $ is $ UJ $ in $ T_{2}(R) $.

For the converse, since $ \left( \begin{array}{cc} u&x \\ 0&v \\ \end{array} \right) $ is $ UJ $, we have $ \left( \begin{array}{cc} u&x \\ 0&v \\ \end{array} \right)=\left( \begin{array}{cc} 1&0 \\ 0&1 \\ \end{array} \right)+\left( \begin{array}{cc} u-1&x \\ 0 &v-1 \\ \end{array} \right) $ where $ u-1 $ and $ v-1 $ are in $ J(R) $. Hence $ u, v $ are $ UJ $.

$ (2) $ Suppose that $ u, v $ are UQ. Let $ u=1+q_{1} $, $ v=1+q_{2} $ where $ q_{1}, q_2\in{R^{qnil}} $. Then $ \left( \begin{array}{cc} u & x \\ 0 & v \\ \end{array} \right)-\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right)=\left( \begin{array}{cc} q_{1} & x \\ 0 & q_{2} \\ \end{array} \right)\in{T_{2}(R)}. $ Next we show that $ \left( \begin{array}{cc} q_{1} & x \\ 0 & q_{2} \\ \end{array} \right) \in (T_2(R))^{ qnil} $. Let $ \left( \begin{array}{cc} a & y \\ 0 & b \\ \end{array} \right)\in T_2(R) $ with $ \left( \begin{array}{cc} a & y \\ 0 & b \\ \end{array} \right)\left( \begin{array}{cc} q_{1} & x \\ 0 & q_{2} \\ \end{array} \right)=\left( \begin{array}{cc} q_{1} & x \\ 0 & q_{2} \\ \end{array} \right)\left( \begin{array}{cc} a & x \\ 0 & b \\ \end{array} \right) $. Then $ aq_{1}=q_{1}a $, $ bq_{2}=q_{2}b $. Since $ q_{1}, q_{2}\in{R^{qnil}} $, $ 1+aq_{1}\in{U(R)} $ and $ 1+bq_{2}\in{U(R)} $. So $ I_{2}+\left( { \begin{array}{cc} aq_{1} & ax+yq_2 \\ 0 & bq_{2} \\ \end{array}} \right)=\left({ \begin{array}{cc} 1+aq_{1} & ax+yq_2 \\ 0 & 1+bq_{2} \\ \end{array}} \right)\in U(T_2(R)) $. This proves $ \left( \begin{array}{cc} u & x \\ 0 & v \\ \end{array} \right) $ is $ UQ $ in $ T_{2}(R) $.

Conversely, it suffices to prove that both $ u-1 $ and $ v-1 $ are quasinilpotents in $ R $. We may let $ x=0 $. Since $ \left( \begin{array}{cc} u & 0 \\ 0 & v \\ \end{array} \right) $ is $ UQ $ in $ T_{2}(R) $, we have $ \left( \begin{array}{cc} u-1 & 0 \\ 0 & v-1 \\ \end{array} \right) \in (T_{2}(R))^{qnil} $. By an easy computation, one gets $ u-1\in R^{qnil} $ and $ v-1\in R^{qnil}, $ as required.

Proposition 2.4   Let $ R $ be a ring, and $ a, b\in R. $

$ (1) $ If $ ab $ is $ UJ $, then $ ba $ is $ UJ $ if and only if $ a, b\in{U(R)} $.

$ (2) $ If $ ab $ is $ UQ $, then $ ba $ is $ UQ $ if and only if $ a, b\in{U(R)} $.

Proof  The proof of $ (2) $ follows from [8, Theorem 2.11]. We give a new proof of $ (1). $

$ (1) $ Suppose that $ a, b\in U(R) $. As $ ab $ is $ UJ $, one has $ 1-ab\in{J(R)}=\cap{M_{i}} $ where $ M_{i} $ is all maximal right ideal of $ R $. If $ ba $ is not $ UJ $, then $ 1-ba\notin{J(R)} $. So, there exists a maximal right ideal $ M_{i_0} $ such that $ 1-ba\notin{M_{i_{0}}} $. It follows that $ R=M_{i_{0}}+(1-ba)R=M_{i_{0}}+a^{-1}a(1-ba)R=M_{i_{0}}+a^{-1}(1-ab)aR $. Since $ 1-ab\in J(R), $ $ a^{-1}(1-ab)aR \subseteq J(R) $. So $ M_{i_{0}}+J(R)=R $, which is a contradiction. Therefore, $ ba $ is $ UJ. $ The converse is trivial.

Jacobson's Lemma states that for any $ a, b\in R, $ $ 1-ab\in U(R) $ if and only if $ 1-ba\in U(R). $ Recall that a ring $ R $ is {reversible} if $ ab=0 $ implies $ ba=0 $ for any $ a, b\in R. $

Proposition 2.5   Let $ R $ be a ring.

$ (1) $ For any $ a, b\in R $ with $ 1-ab $ is $ UJ $, then $ 1-ba $ is $ UJ $ if and only if $ R/J(R) $ is reversible.

$ (2) $ $ 1-ab $ is $ UQ $ if and only if $ 1-ba $ is $ UQ $.

Proof   By a direct computation, we can prove $ (1) $.

For $ (2), $ we can deduce from [9, Lemma 2.1]. Here, we give a simple proof for a convenience. Note that $ (2) $ is equivalent to the comment "$ ab $ is quasinilpotent if and only if so is $ ba $". Now we assume that $ ab\in R^{qnil} $ but $ ba\notin R^{qnil}. $ Then there exists $ y\in R $ such that $ (ba)y=y(ba) $ and $ 1+bay \notin U(R). $ From $ bay=yba, $ we obtain $ ab(ay^2b)=(ay^2b)ab $. Since $ ab\in R^{qnil}, $ $ 1-ab(ay^2b)\in U(R). $ By Jacobson's Lemma, we have $ 1-babay^2=1-(bay)^2\in U(R), $ which implies $ 1+bay\in U(R) $, a contradiction. So $ ba\in R^{qnil} $.

This section is devoted to the study of $ UQ $-rings.

Proposition 3.1   Let $ R $ be a $ UQ $-ring.

$ (1) $ For any $ q_{1}, q_{2}\in{R^{qnil}}, q_{1}+q_{2}+q_{1}q_{2}\in{R^{qnil}} $.

$ (2) $ If $ q_{1}, q_{2}\in{R^{qnil}} $ and $ q_{1}q_{2}={q_{2}q_{1}} $, then $ q_{1}+q_{2}\in{R^{qnil}}. $

Proof   $ (1) $ Note that $ (1+q_{1})(1+q_{2})\in{U(R)} $. Since $ R $ is a $ UQ $-ring, $ 1+q_{1}+q_{2}+q_{1}q_{2}\in{1+R^{qnil}}, $ and so $ q_{1}+q_{2}+q_{1}q_{2}\in{R^{qnil}}. $

$ (2) $ Let $ q_{1}, q_{2}\in{R^{qnil}} $ and $ q_{1}q_{2}={q_{2}q_{1}} $. Then

$ \begin{align*} 1+q_{1}+q_{2}&=(1+q_{1})(1+(1+q_{1})^{-1}q_{2})\in{U(R)}=1+R^{qnil}. \end{align*} $

So $ q_{1}+q_{2}\in{R^{qnil}} $.

Proposition 3.2  Let $ R $ be a $ UQ $-ring. Then$ : $

$ (1) $ $ 2\in{J(R)} $.

$ (2) $ $ eRe $ is $ UQ $-ring.

Proof  $ (1) $ Let $ -1=1+x $ with $ x\in{R^{qnil}} $. Then $ x=-2 $. Note that $ x $ is central. Hence, $ 2\in{J(R)} $.

$ (2) $ Let $ x\in{U(eRe)} $. Then there exists $ y\in eRe $ such that $ xy=e=yx $. So we have $ [x+(1-e)][y+(1-e)]=e+0+0+1-e=1=[y+(1-e)][x+(1-e)] $, which implies $ [x+(1-e)] \in{U(R)} $. Since $ R $ is a $ UQ $-ring, $ x+(1-e)=1+q $ for some $ q\in{R^{qnil}} $. Then $ x-e=q\in(R^{qnil})\cap{eRe}=(eRe)^{qnil} $ by [10, Lemma 3.5(2)]. So $ eRe $ is $ UQ $-ring.

Proposition 3.3  Let $ R $ be a $ UQ $-ring. Then $ M_{n}(R) $ is not $ UQ $ for any $ n\geq{2}. $

Proof  By Proposition 3.2(2), it suffices to prove $ M_{2}(R) $ is not $ UQ $. Let $ A_{1}=\left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array} \right) , $ $ A_{2}=\left( \begin{array}{cc} 0 & 0 \\ 1 & 0 \\ \end{array} \right) $. Then $ A_{1}, A_{2}\in{M_{2}(R)^{qnil}} $. But $ A_{1}+A_{2}+A_{1}A_{2}=\left({ \begin{array}{cc} 1 & 1 \\ 1 & 0 \\ \end{array}} \right) \notin{M_{2}(R)^{qnil}} $. By Proposition 3.1(1), $ M_2(R) $ is not $ UQ $.

We next determine when a single matrix over a field is $ UQ $. For a matrix $ A $ over a filed, we write $ {\rm tr}(A) $ and $ {\rm det}A $ for the trace and the determinate of $ A, $ respectively.

Example 3.4  Let $ F $ be a field, $ U\in U({M_{2}(F)}) $. Then $ U $ is $ UQ $ if and only if $ U=I_{2} $ or there exists an invertible matrix $ P\in M_2(F) $ such that $ P^{-1}UP=\left({ \begin{array}{cc} 0 & -1 \\ 1 & 2 \\ \end{array}} \right) $.

Proof  By [8, Example 2.2], $ (M_2(F))^{qnil} $ equals the set of all nilpotents in $ M_2(F) $.

Suppose that $ U $ is $ UQ $. By Hamilton-Cayley Theorem, we have $ U^{2}={\rm tr}(U)\cdot{U}-{\rm det}U\cdot I_2 $. Since $ U $ is $ UQ $, we have $ U-I_2 $ is nilpotent, and so $ (U-I_2)^{2}=0 $. It follows that $ {\rm tr}(U)\cdot{U}-{\rm det}U\cdot I_2=2U-I_2 $, and then $ ({\rm tr}(U)-2)U=({\rm det}U-1)I_2 $. Notice that $ 0={\rm det}(U-I_2)={\rm det}U-{\rm tr}(U)+1 $. Thus, $ {\rm det}U={\rm tr}(U)-1 $. Combing this equation with $ ({\rm tr}(U)-2)U=({\rm det}U-1)I_2 $, we have $ ({\rm tr}(U)-2)U=({\rm tr}U-2)I_2 $. If $ {\rm tr}(U)\neq 2 $, then $ U=I_2 $ and $ {\rm det}U={\rm tr}(U)-1 \neq 1 $, which is a contradiction. So $ {\rm tr}(U)=2 $, and $ {\rm det}U=1 $. Assume that $ U\neq I_2. $ Then there exists an invertible matrix $ P\in M_2(F) $ such that $ P^{-1}UP=\left({ \begin{array}{cc} 0 & a_{1} \\ 1 & a_{2} \\ \end{array}} \right) $ where $ a_1, a_2\in F $. Since $ {\rm det}(P^{-1}UP)={\rm det}U $ and $ {\rm tr}(P^{-1}UP)={\rm tr}(U) $, one gets $ a_1=-1 $ and $ a_2=2. $ Therefore, $ P^{-1}UP=\left({ \begin{array}{cc} 0 & -1 \\ 1 & 2 \\ \end{array}} \right) $. The converse is clear.

Proposition 3.5  Let $ R $ be a ring.

$ (1) $ $ R $ is $ UJ $-ring if and only if $ J(R)=\{{x\in R:1+x\in{U(R)}}\} $.

$ (2) $ $ R $ is $ UQ $-ring if and only if $ R^{qnil}=\{{q\in R:1+q\in{U(R)}}\} $.

Proof   We only need to prove $ (1) $. The proof of $ (2) $ is similar to that of $ (1) $.

$ (1) $ Assume that $ R $ is a $ UJ $-ring. Clearly, $ J(R)\subseteq \{{x\in R:1+x\in{U(R)}}\} $. Let $ x\in R $ be such that $ 1+x\in U(R) $. Since $ R $ is UJ, $ 1+x=1+j $ for some $ j\in J(R). $ So $ x=j\in J(R). $ Thus, $ J(R) \supseteq \{{x\in R:1+x\in{U(R)}}\} $, and then $ J(R)=\{{x\in R:1+x\in{U(R)}}\} $.

Conversely, let $ u\in U(R) $. Then $ u=1+(u-1) $. As $ J(R)=\{{x:1+x\in{U(R)}}\} $, we get $ u-1\in J(R), $ which implies that $ u $ is $ UJ $, and hence $ R $ is a $ UJ $-ring.

Recall that a ring $ R $ is {abelian} if every idempotent of $ R $ is central.

Theorem 3.6   Let $ R $ be a ring. Then the following are equivalent$ : $

$ (1) $ $ R $ is a regular $ UQ $-ring.

$ (2) $ $ R $ is an abelian regular $ UQ $-ring.

$ (3) $ $ R $ has the identity $ x^2=x $ (i.e., $ R $ is Boolean).

Proof   $ (1)\Rightarrow (2). $ Since $ R $ is regular, $ J(R)=0 $ and every nonzero right ideal contains a nonzero idempotent. Let $ a\in R $, if $ a\neq{0}, a^{2}=0 $. By [11, Theorem 2.1], there is an idempotent $ e\in RaR $ such that $ eRe \cong M_2(T) $ for some non-trivial ring $ T. $ As $ R $ is a $ UQ $-ring, $ eRe $ is $ UQ $. So $ M_2(T) $ is a $ UQ $-ring, which is a contradiction. So $ a=0, $ and this proves that $ R $ is reduced. As is well known, reduced rings are abelian.

$ (2)\Rightarrow (3) $. Let $ q\in Q(R). $ Since $ R $ is strongly regular, there exist $ e^2=e $ and $ u\in U(R) $ such that $ q=eu=ue. $ So $ 1-e=1-qu^{-1}\in U(R), $ and whence $ e=0 $, which implies $ q=0. $ Thus $ Q(R)=0 $, and so $ U(R)=1+Q(R)=1. $ Clearly, a strongly regular ring $ R $ with $ U(R)=1 $ is Boolean.

$ (3)\Rightarrow (1) $. Clearly, $ R $ is regular. Let $ u\in U(R). $ Then $ u^2=u, $ which implies that $ u=1 $, and thus, $ R $ is a $ UQ $-ring.

Corollary 3.7   Let $ R $ be a ring with $ J(R)=0 $. Then the followings are equivalent$ : $

$ (1) $ $ R $ is an exchange $ UQ $-ring.

$ (2) $ $ R $ is a clean $ UQ $-ring.

Proof $ (2)\Rightarrow {(1)} $ is clear.

$ (1)\Rightarrow {(2)} $. Since $ R $ is an exchange ring with $ J(R)=0 $, by Theorem 3.6, every nonzero right ideal contains a nonzero idempotent. In view of the proof $ (1)\Rightarrow (2) $ of Theorem 3.6, we have $ R $ is abelian. By [1, Propsition 1.8(2)], abelian exchange rings are clean.

Theorem 3.8   Let $ R $ be a ring. Then the followings are equivalent$ : $

$ (1) $ $ R $ is a strongly clean $ UQ $-ring.

$ (2) $ For any $ {a\in{R}} $, there exist $ e^2=e\in R $ and $ q\in R^{qnil} $ such that $ a=e+q $ and $ ae=ea $.

Proof    $ (2)\Rightarrow {(1)} $. By $ (2) $, $ a=e+q=(1-e)+(2e-1+q) $ where $ e^2=e\in R $ and $ q\in{R^{qnil}} $. Note that $ 2e-1-q=(2e-1)-q=(2e-1)[1-(2e-1)q]\in{U(R)} $ as $ eq=qe $. Hence, $ R $ is a strongly clean ring. Further, for any $ v\in{U(R)} $, we have $ v=e+q $. Then $ e=v-q=v(1-v^{-1})q\in{U(R)} $, which yields $ e=1 $ and $ v=1+q $. Thus $ R $ is also a $ UQ $-ring.

$ (1)\Rightarrow {(2)} $. Let $ a\in{R} $. As $ R $ is a strongly clean ring, we have $ 1+a=e+u $ where $ e^{2}=e\in R $, $ u\in U(R) $ and $ ae=ea. $ Since $ R $ is a $ UQ $-ring, $ u=1+q $ for some $ q\in R^{qnil} $. Thus, $ 1+a=e+u=e+1+q $, and so $ a=e+q $ and $ ae=ea $.

Recall that a ring $ R $ is Dedekind finite if $ ab=1 $ implies $ ba=1 $ for any $ a, b\in R. $

Theorem 3.9  Every $ UQ $-ring is Dedekind finite.

Proof   We first show the following claim:

Claim    If a ring $ R $ contains a set of matrix units $ \{e_{ij}\}_{i, j\leq{2}} $ such that $ e_{ii}\neq{0} $ for $ i=1, 2, $ then $ R $ is not a $ UQ $-ring.

Proof of Claim. Let $ f=e_{11}+e_{22} $ and $ u=e_{12}+e_{21}+e_{22} $. Then it is easy to know that $ f $ is an idempotent, $ u(u-f)=(u-f)u=f $ and $ fuf=u $. Hence $ u $, $ u-f\in{U(fRf).} $ As $ (fRf)^{qnil} \cap{U(fRf)}=\varnothing $, we obtain $ u-f\notin{(fRf)^{qnil}} $, that is to say $ fRf $ is not $ UQ $-ring. By Proposition 3.2(2), $ R $ is not a $ UQ $-ring. So the Claim follows.

Let $ R $ be a $ UQ $-ring, and assume that $ R $ is not Dedekind finite. Let $ a, b\in{R} $ be such that $ ab=1 $ and $ e=ba\neq{1} $. Clearly, $ e $ is a nontrivial idempotent. In view of [12, Example 21.26], there exists a set of nonzero matrix units of the form $ e_{ij}=b^{i}(1-e)a^{j} $. Then applying the above Claim, we get that $ R $ is not a $ UQ $-ring, a contradiction.

We call a ring $ R $ a strong $ UJII $-ring if for each $ u\in{U(R)} $, $ u=j+e+f $ where $ j\in{J(R)} $, $ e, f\in{idem(R)} $ and $ j, e, f $ commute with one another. $ UJ $-rings are strong $ UJII $-rings.

Proposition 4.1 Let $ R $ be a strong $ UJII $-ring.

$ (1) $ $ R^{qnil}=J(R) $.

$ (2) $ $ J(R)=\sqrt{J(R)} $ where $ \sqrt{J(R)}=\{{x\in{R}\; |\; x^{k}}\in{J(R)}\; for\; some \; integer \; k\geq1\} $.

Proof   $ (1) $ Let $ b\in R^{qnil} $. Then $ 1+b\in{U(R)} $. Write $ 1+b=j+e+f $ where $ j\in{J(R)} $, $ e, f\in{idem(R)} $ and $ b, j, e, f $ all commute. So $ (1-e)-f=j-b $. Note that $ (1-e)-f=[(1-e)-f]^3 $. Then $ (b-j)^3=b-j $. It follows that $ b-b^3=b(1-b^2)\in J(R) $. Since $ 1-b^2\in U(R) $, we obtain $ b\in J(R). $ Thus, $ R^{qnil}\subseteq{J(R)} $, and whence $ R^{qnil}=J(R) $.

$ (2) $ It suffices to show that $ \sqrt{J(R)} \subseteq {J(R)} $. Let $ x\in{\sqrt{J(R)}} $. Then $ x^k\in J(R) $ with $ k\geq 1 $ and $ 1-x\in U(R). $ By hypothesis, $ 1-x=j+e+f $ for some $ j\in{J(R)} $, $ e, f\in{idem(R)} $, and $ j, e, f $ all commute. Clearly, $ (1-e)-f=j+x $. Note that $ (1-e)-f=((1-e)-f)^{2m+1} $ for any positive integer $ m. $ Since $ x^k\in J(R) $ and $ jx=xj $, we obtain

$ (1-e)-f=((1-e)-f)^{2k+1}=(j+x)^{2k+1}\in J(R)+x^{2k+1}=J(R). $

So $ 1-[(1-e)-f]^{2k} \in U(R) $. It follows from $ [(1-e)-f][1-((1-e)-f)^{2k}]=0 $ that $ (1-e)-f=0 $. Thus $ 1-e=f $, and whence $ x=-j\in{J(R)} $. Hence $ \sqrt{J(R)}=J(R) $.

Corollary 4.2   Let $ R $ be a ring. Then $ M_{n}(R) $ is not strong $ UJII $ for any $ n\geq 2 $.

Proof   Note that any matrix unit $ E_{ij} $ is quasinilpotent when $ i\neq j $. However, $ E_{ij}\notin{J(M_{n}(R))} $. By Proposition 4.1(1), $ M_{n}(R) $ is not strong $ UJII $.

Proposition 4.3   If $ R $ is a strong $ UJII $-ring, then $ 6\in{J(R)} $.

Proof   Write $ -1=j+e+f $, where $ j\in{J(R)} $, $ e, f\in{\rm{idem}(R)} $ and $ j, e, f $ all commute. Then

$ 1+2j+j^{2}=(-1-j)^{2}=(e+f)^{2}=e+f+2ef=(-1-j)+2e(-1-j-e)=-1-j-4e-2ej. $

We get $ 2+4e=-3j-j^{2}-2ej $, and then $ 6e=(2+4e)e=(-3j-j^{2}-2ej)e=-(5e+je)j\in{J(R)} $. Similarly, $ 6f\in{J(R)} $. Then $ -6=6j+6e+6f\in{J(R)} $, so $ 6\in{J(R)} $.

Clearly, a direct product of rings is strong $ UJII $ if and only if each ring is strong $ UJII $.

Corollary 4.4   Let $ R $ be a ring with $ J(R)=0 $. Then $ R $ is a strong $ UJII $-ring if and only if $ R=A\oplus{B} $ where $ A, B $ are strong $ UJII $-rings, and $ 2=0 $ in $ {A} $ and $ 3=0 $ in $ {B} $.

Proof   Suppose that $ R $ is a strong $ UJII $-ring. Since $ J(R)=0 $, by Proposition 4.3, $ 6=0 $. By Chinese Remainder Theorem, $ R\cong{R/2R\oplus{R/3R}} $. Let $ A=R/2R $, $ B=R/3R $. So $ A, B $ are strong $ UJII $-rings, and $ 2=0 $ in $ {A} $, $ 3=0 $ in $ {B} $. The other direction is obvious.

A ring $ R $ is {$ 2 $-UJ} [13] if for any $ u\in{U(R)} $, $ u^{2}=1+j $ for some $ j\in{J(R)} $. Recall that a ring $ R $ is called {reduced} if $ R $ contains no nonzero nilpotent elements (equivalently, $ a^2=0 $ implies $ a=0 $ for any $ a\in R $).

Proposition 4.5   If $ R $ is a strong $ UJII $-ring, then $ R $ is $ 2 $-UJ and $ R/J(R) $ is reduced.

Proof   Let $ u\in{U(R)} $. Write $ u=j+e+f $ where $ j\in{J(R)} $, $ e, f\in{idem(R)} $ and $ j, e, f $ all commute. Then $ u-j=e+f\in{U(R)} $. Set $ u^{'}=u-j=e-f+2f $. Then $ u^{'}-2f=e-f $, and we get $ (u^{'}-2f)^{3}={u^{'}}^{3}-6{u^{'}}^{2}f+12u^{'}f-8f={u^{'}}^{3}-2f+6({u^{'}}^{2}f+2u^{'}f)=u^{'}-2f $. By Proposition 4.3, $ {u^{'}}^{3}-u^{'}=6({u^{'}}^{2}f+2u^{'}f)\in{J(R)} $, and thus $ {u^{'}}^{2}-1\in{J(R)} $. It follows that $ u^{2}-1=({u^{'}}^{2}-1)+2u^{'}j+j^2 \in J(R) $. Hence $ R $ is a $ 2 $-$ UJ $ ring. Next we show that $ R/J(R) $ is reduced. Let $ x\in R $ with $ x^2 \in J(R). $ It is easy to see that $ x\in R^{qnil} $. In view of Proposition 4.1(2), we get $ x\in J(R) $, which implies that $ R/J(R) $ is reduced.

Theorem 4.6   Let $ R $ be a strong $ UJII $-ring. Then the followings are equivalent$ : $

$ (1) $ For any $ u\in{U(R)} $, there exists a unique $ j\in{J(R)} $ such that $ u=e+f+j $ where $ e, f\in{idem(R)} $.

$ (2) $ $ R $ is abelian.

Proof   $ (1)\Rightarrow {(2)} $. For any $ e^{2}=e\in{R} $ and $ r\in{R} $, we have $ er(1-e)\in N(R). $ By Proposition 4.1(1), $ er(1-e) \in J(R) $. So $ 1+er(1-e)\in{U(R)} $. Write $ t=1+er(1-e) $. Then $ t=1+0+er(1-e) =[e+er(1-e)]+(1-e)+0 $ are two $ UJII $ expressions of $ t $. By assumption, $ er(1-e)=0 $, and thus, $ er=ere $. Similarly, we can deduce that $ re=ere $, and so $ er=re $. Therefore, $ R $ is abelian.

$ (2)\Rightarrow {(1)} $. Let $ u\in U(R). $ We may assume that

$ \begin{align} u=e_{1}+f_{1}+j_{1} \end{align} $

(1)

and

$ \begin{align} u=e_{2}+f_{2}+j_{2}, \end{align} $

(2)

where $ e_i, f_i\in $ idem$ (R) $, $ j_i\in J(R) $ and $ i=1, 2. $ Multiplying the above two equations by $ 1-e_{1}, $ we get $ u(1-e_{1})=f_{1}(1-e_{1})+j_{1}(1-e_{1}) $ and $ u(1-e_{1})=e_{2}(1-e_{1})+f_{2}(1-e_{1})+j_{2}(1-e_{1}). $ Note that $ f_{1}(1-e_{1})\in U((1-e_1)R(1-e_1))\cap $idem$ ((1-e_1)R(1-e_1)) $. Thus, $ f_1(1-e_1)=1-e_1 $. It follows that

$ \begin{align*} 0&=[f_{1}(1-e_{1})+j_{1}(1-e_{1})]-[e_{2}(1-e_{1})+f_{2}(1-e_{1})+j_{2}(1-e_{1})]\\ &=[(1-e_{1})+j_{1}(1-e_{1})]-[e_{2}(1-e_{1})+f_{2}(1-e_{1})+j_{2}(1-e_{1})]\\ &=(1-e_{1})(1-e_{2}-f_{2})+j_{1}(1-e_{1})-j_{2}(1-e_{1}). \end{align*} $

So one has $ (1-e_{1})(1-e_{2}-f_{2})\in J(R). $ Since $ 1-e_{1}\in idem(R) $, $ (1-e_{2}-f_{2})^3=1-e_{2}-f_{2} $ and $ (1-e_{1})(1-e_{2}-f_{2}) $ are a tripotent. A direct check will reveal that $ (1-e_{1})(1-e_{2}-f_{2})=0. $ Similarly, multiplying equations $ (1) $ and $ (2) $ by $ 1-e_2 $, we obtain $ (1-e_{2})(1-e_{1}-f_{1})=0. $ Thus, $ (1-e_1)f_2=(1-e_2)f_1 $, and whence $ f_1-f_2=e_2f_1-e_1f_2 $. One may note that $ e_1, e_2 $ and $ f_1, f_2 $ are parallel. So we can also get $ e_1-e_2=f_2e_1-f_1e_2 $. Therefore, $ e_1+f_1=e_2+f_2 $. This shows that $ j_{1}=j_{2} $, and we are done.

Theorem 4.7   Let $ R $ be a ring. Then the followings are equivalent:

$ (1) $ $ R $ is a strong $ UJII $-ring and $ 2\in{J(R)} $.

$ (2) $ $ R $ is a $ UJ $ ring.

Proof   $ (1)\Rightarrow (2) $. Let $ u\in{U(R)} $. Then $ u=j+e+f $ where $ j\in{J(R)} $, $ e, f\in{idem(R)} $ and $ j, e, f $ all commute. Write $ g:=e+f-2ef $ and $ c:=j+2ef $. Then $ g^{2}=g $ and $ c \in{J(R)} $ as $ 2\in J(R) $. So $ u=g+c $ and $ ug=gu $. It follows that $ g=u-c\in{U(R)} $. Thus $ g=1 $, and therefore, $ u=1+c $, which implies that $ R $ is a $ UJ $ ring.

$ (2)\Rightarrow (1) $. By hypothesis, $ -1=1+j $. So $ 2\in{J(R)} $. It is clear that $ R $ is a strong $ UJII $-ring.

We finish this short paper with following problems:

Problem  $ (1) $ If $ R $ is a strong $ UJII $-ring, is $ eRe $ strong $ UJII $ for any $ e^2=e\in R $?$ (2) $ Characterize when a group ring is strong $ UJII $.