Matrix l-algebras over l-fields

PURE MATHEMATICS RESEARCH ARTICLE Matrix l-algebras over l-fields Jingjing M * Received: 05 January 2015 Accepted: 11 May 2015 Published: 15 June 2015 *Corresponding author: Jingjing Ma, Department of Mathematics, University of Houston-Clear Lake, 2700 Bay Area Boulevard, Houston, TX 77059, USA E-mail: ma@uhcledu Reviewing editor: Shaoyong Lai, Southwestern University of Finance and Economics, China Additional information is available at the end of the article Abstract: It is shown that if a matrix l-algebra M n (K) over certain l-fields K contains a positive n-cycle e such that I + e + + e n 1 is a d-element on K then it is isomorphic to the l-algebra M n (K) over K with the entrywise lattice order Subjects: Advanced Mathematics; Algebra; Fields & Rings; Mathematics & Statistics; Science Keywords: d-element; lattice-ordered algebra; lattice-ordered field; matrix algebra; positive cycle AMS subject classification: 06F25 1 Introduction Let R be a lattice-ordered ring (l-ring) with the positive identity element and M n (R) (n 2) be the n n matrix ring over R M n (R) may be made into an l-ring by defining a matrix in M n (R) positive if each entry of the matrix is positive in R This lattice order on M n (R) is called the entrywise lattice order Since R has the positive identity element, the identity matrix of M n (R) is positive with respect to the entrywise lattice order In 1966, Weinberg first proved that if M 2 (Q) is an l-ring in which the identity matrix is positive, where Q is the field of rational numbers, then it is isomorphic to the l-ring M 2 (Q) with the entrywise lattice order (Weinberg, 1966) Then in Ma and Wojciechowski (2002) it was proven that this fact is true for any totally ordered subfield F of the field R of real numbers and any n n matrix algebra over F Whether or not the above fact is true for an n n (n > 2) matrix algebra over an arbitrary totally ordered field is still an open question Under some stronger conditions, though, the result is true; for instance, if an l-algebra M n (T) (n 2) contains a positive n-cycle, where T is an arbitrary totally ordered field, then it (T) with the ABOUT THE AUTHOR Jingjing Ma s research is in the area of Latticeordered Rings and Algebras that was first systematically studied by G Birkhoff and RS Pierce about 60 years ago His interests involve different topics on lattice-ordered rings with the focus on trying to better understand the algebraic structures of such systems The algebraic structure of lattice-ordered matrix algebras has been one of his research topics In 2000, working together with Wojciechowski (University of Texas at El Paso), they proved the so-called Weinberg s conjecture, which states that there is only one lattice order on matrix rings over the field of rational number in which the identity matrix is positive The present work continues the study of lattice-ordered matrix algebras in a more general setting More precisely, this article considers lattice-ordered matrix algebras over certain lattice-ordered fields, which include totally ordered fields as a special case PUBLIC INTEREST STATEMENT Given a positive integer n, a n by n matrix is a square array consisting of n rows and n columns of numbers We could add two matrices, multiply two matrices, and compare two matrices by certain rules Matrices with those operations consist of a very important research area in mathematics, and it has rich applications to other areas in mathematics, physics, etc The present research continues the study of matrices with the operations and order It considers matrices with the operations and order in a more general setting to generalize some important known results The article exposes more properties of such matrices to better understand them 2015 The Author(s) This open access article is distributed under a Creative Commons Attribution (CC-BY) 40 license Page 1 of 7

entrywise lattice order (Ma, 2000) The reader is referred to Steinberg (2010) for more information on research activities in this area The same problem for matrices over a lattice-ordered field (l-field) has not been studied In this article, we consider n n matrix l-algebra over certain Archimedean l-fields We first provide an example to show that the n n matrix l-algebra M n (K) over an l-field K may not be isomorphic to the M n (K) with entrywise lattice order even it contains a positive n-cycle Example 11 Consider the l-field K = Q[ 2]={α + β 2 α, β Q} R with coordinatewise lattice order, that is, α + β 2 0 if α 0 and β 0 in Q Now define a matrix a =(a ij ) M n (K) to be positive if each a ij is greater than or equal to zero with respect to the usual total order in R Since the coordinatewise lattice order on K is contained in the total order induced from R, M n (K) is an l-algebra over K and the identity matrix is positive under this order We notice that this lattice order is not the entrywise lattice order on M n (K) over the l-field K, for instance, the matrix ( 2 1)I is positive, where I is the identity matrix, but 2 1 is not positive in K with respect to the coordinatewise lattice order The n n matrix e = e 12 + e 23 + + e n 1, n + e n1 is an n-cycle as defined below, where e ij are standard matrix units It is clear that e is positive in the above lattice order defined on M n (K) The main result of this article is a proof that for a certain l-field K, if the l-algebra M n (K) over K contains a positive n-cycle e such that I + e + + e n 1 is also a d-element on K then the l-algebra (K) with the entrywise lattice order We will briefly review few definitions and results, the reader is referred to Birkhoff and Pierce (1956) and Steinberg (2010) for general information on l-rings and undefined terminologies We call the permutation matrix e i1 i + + e 2 in 1 i + e n in i as an n-cycle in the n n matrix ring, where 1 e ij are the standard matrix units, that is, ijth entry of e ij is 1, and other entries are zero For an l-field K, the l-ring M n (K) is called a lattice-ordered algebra (l-algebra) if for any α K and a M n (K), α 0, a 0 αa 0 A positive element x of M n (K) is called a d-element on M n (K) (on K) if for any a, b M n (K), a b = 0 xa xb = ax bx = 0 in M n (K) (if for any u, v K, u v = 0 ux vx = 0 in M n (K)), and a positive element α K is called a d-element on K (on M n (K)) if for any u, v K, u v = 0 αu αv = 0 in K (if for any a, b M n (K), a b = 0 αa αb = 0 in M n (K)) It is well known that if u, u 1 K are both positive then u is a d-element both on M n (K) and on K, and if a, a 1 M n (K) are both positive, then a is a d-element on M n (K) but may not be a d-element on K For instance, the element e in Example 11 is not a d-element on K We refer the reader to Steinberg (2010) for more information on properties of d-elements 2 Main result Let K be an l-field with a positive identity element 1 It is well known that 1 is a basic element in the sense that for any a, b K +, a, b 1 a b or b a Define F ={x K x is an f -element of K} Then F is the largest totally ordered subfield of K (Schwartz, 1986, Theorem 4) In the following, we always assume that K is an Archimedean l -field and finite-dimensional as a vector space over F Then K has a vl-basis {u 1 } (Schwartz, 1986, Corollary, p 186), that is, K = Fu 1 + +Fu m and α 1 u 1 + +α m u m 0, where α i F, if and only if each α i 0 Moreover, we assume that each u i is a d-element on K Then u 1 i > 0 since 1 = 1 = u 1 i u i = u 1 i u i implies that u 1 i = u 1 i > 0 A simple example of this situation is K = Q[ 2]={α + β 2 α, β Q} with the coordinatewise lattice order Then u 1 = 1, u 2 = 2 is a vl-basis of K over F It is clear that F = Q and u 1, u 2 are d-elements on K and on any l-algebra M n (K) Suppose that K is an Archimedean l-field with 1 > 0 and suppose K is finite-dimensional over F with a vl-basis that consists of d-elements of K Consider the n n matrix algebra A = M n (K) over K with n 2 Suppose that A is an l-algebra over K Then A is also an l-algebra over F since the Page 2 of 7

lattice order on K extends the total order of F Since A contains no nonzero nilpotent ideals, A is Archimedean over F (Birkhoff & Pierce, 1956, Corollary 1, p 51) Since A is also finite-dimensional over F, it implies that A is a direct sum of maximal convex totally ordered subspaces over F (Conrad, 1961, Theorem 73, p 232) Let {u 1 } be a vl-basis of K over F in which each u i is a d-element of K We will assume that u 1 = 1, the identity element of K The identity matrix of M n (K) is denoted by I The following is the main result of the paper Theorem 21 Suppose that A = M n (K) is an l-algebra over K If A contains a positive n-cycle e and I + e + + e n 1 is a d-element on K then the l-algebra A (K) over K with the entrywise lattice order Proof As we discussed in the previous paragraph, A is a direct sum of maximal convex totally ordered subspaces of A over F Since the identity matrix I = e n > 0, I is a sum of disjoint basic elements Suppose that 0 < a I is a basic element Since u 1 k > 0 for each u k, every u k a is also a basic element of A For k = 1,, m, define M k =(u k a) and H k e j Then each M k is a maximal convex totally ordered subspace of A over F since A is a f-module over F, and each H k is a convex l-subspace of A over F We divide the proof of Theorem 21 into several lemmas Lemma 22 H i ={0} for 1 i, j m and i j Proof Clearly the sets {e s M i e t 1 s, t n} and {e s M j e t 1 s, t n} of maximal convex totally ordered subspaces over F are either identical or disjoint (Conrad, 1961, Lemma 31) Thus either H i = H j or H i ={0} Suppose that H i {0} Then H i = H j, so u j a H i and hence u j a e r M i e s for some 1 r, s n since u j a is basic Thus, u j a and e r (u i a)e s are comparable On the other hand, u i u j = 0 and I + e + + e n 1 is a d-element on K, so u i (I + e + + e n 1 ) u j (I + e + + e n 1 )=0 Since u i a u i I, e r (u i a)e s e r (u i I)e s = u i e r+s u i (I + e + + e n 1 ), and also u j a u j I u j (I + e + + e n 1 ) Thus e r (u i a)e s u j a = 0, which is a contradiction Therefore, we must have H i ={0} for i j Lemma 23 A = Proof Let f k i, j=1 ei (u k a)e j H k, k = 1,, m Then by Lemma 22, {f 1,, f m } is a disjoint set, that is, f i f j = 0, i j For each f k, since ef k = f k e = f k, we have f k = v k (I + e + + e n 1 ), 0 v k K, k = 1,, m Indeed, since e n = I and e m I for 1 m < n, the characteristic polynomial of e is λ n 1, so e has n distinct eigenvalues and hence the eigenspace for each eigenvalue is of one-dimensional From ef k = f k, each column in f k is an eigenvector of e to 1 Then each column in f k is a scalar multiple of the vector v, each of whose components is equal to 1 Similarly, each row of f k is a scalar multiple of v t, the transpose of v Thus, f k = v k (I + e + + e n 1 ) for some 0 v k K Let v k 0 = w k in K Then since I + e + + e n 1 is a d-element on K, 0 = f k 0 = v k (I + e + + e n 1 ) 0 = w k (I + e + + e n 1 ), so w k = 0, and hence v k > 0 for each k Similarly we show that v k v l = 0 for k l, and so {v 1,, v m } is a disjoint set in K Since each positive element in K is a positive linear combination of u 1, each Page 3 of 7

v i is a strictly positive multiple by a scalar in F of exactly one of u 1, so without loss of generality, we may assume that v i = α i u i, where 0 <α i F, i = 1,, m Suppose that A Then there exists a maximal convex totally ordered subspace M of A over F which is not in the sum of Let 0 < x M and g i, j=1 ei xe j Then eg = ge = g and by a similar argument as before g = v(i + e + + e n 1 ) for some 0 < v K, and v v i = 0 in K for i = 1, m Hence v u i = 0 in K for i = 1,, m, so v = 0, which is a contradiction Therefore, A = Lemma 24 For each k = 1,, m, H k e j is a direct sum Proof To prove that H k e j is a direct sum, we show that any two summands are different, and thus they must have zero intersection (Conrad, 1961) For 1 s, t n with s n or t n, suppose first that e s M k e t = M k Then for any 0 x M k, e s xe t and x are comparable If x > e s xe t then x > e s xe t > e 2s xe 2t > > e ns xe nt = x, which is a contradiction Similarly x e s xe t Thus, e s xe t = x for all x M k, so e s ge t = g for all g H k For any y M l, l k, since y and u l a are comparable, u k u 1 y l and u k a are comparable, so e s (u k u 1 l y)et = u k u 1 y by l es M k e t = M k and previous arguments Hence, e s ye t = y for all y M l, so e s he t = h for all h H l, l k Since A = by Lemma 23, we have e s fe t = f for all f A In particular, e s+t = I, so s + t = n, and hence s n and t n Therefore, e s f = e s fe n = e s fe t e s = fe s for any f A, that is, e s is in the center of A with 1 s < n, which is a contradiction Hence, for 1 s, t n with s n or t n, e s M k e t M k Now for 1 i 1, i 2, j 1, j 2 n, suppose that e i 1 Mk e j 1 = e i 2Mk e j 2 If i1 < i 2 then M k = e i 2 i 1 Mk e j 2 j 1 if j 2 j 1, which is a contradiction by previous paragraph If j 2 < j 1 then M k = e i 2 i 1 Ae j 2 +n j 1, which is again a contradiction Thus, i 1 i 2 Similarly, i 2 i 1, and hence i 1 = i 2 Similarly, j 1 = j 2 Therefore, H k e j is a direct sum of n 2 maximal convex totally ordered subspaces e i M k e j, 1 i, j n We notice that since dim F A = mn 2 and A is a direct sum of mn 2 totally ordered subspaces over F, each M i, i = 1,, m, is one-dimensional over F, so M i = F(u i a) for each i = 1,, m Lemma 25 (F) over F with the entrywise lattice order Proof First, we notice that I cannot be a basic element since in that case, I = a M 1, so e = ei = Ie implies that em 1 = M 1 e, which is a contradiction by Lemma 25 Suppose that I = a + + +a p, where a,,, a p are disjoint basic elements and p 1 For each i = 1,, n 1, I = e i e n i = e i ae n i + +e i a p e n i implies that each e i ae n i is an f-element for i = 1,, n, and e i ae n i is also a basic element since e n = I We claim that I = a + eae n 1 + + e n 1 ae For i = 1,, n 1, since e i ae n i < I is basic, there exists a j for some j = 1,, p such that e i ae n i and a j are comparable We show that e i ae n i = a j Since e i ae n i and a j are comparable, e i ae n i a v = 0 for any 1 v p and v j Otherwise e i ae n i and a v will be comparable since both are basic elements, so a v are comparable, a contradiction Then from a + + +a p = I = e i ae n i + +e i a p e n i, =(e i ae n i ) = a j, which implies that a v and a j we have e i ae n i a j, and similarly a j e i ae n i Thus e i ae n i = a j, so each e i ae n i, i = 1,, n 1, appears in the sum I = a + + +a p Next, we show that each a r, 1 r p, is equal to e i ae n i for some 1 i n 1 We first notice that since e i ae n i, i = 0,, n 1, appear in the sum for I, they are disjoint f -element, so ae i a = 0 for i = 1,, n 1 Since a r A is basic, a r = e s xe t for some 0 < x M w and 1 s, t n, 1 w m Then that a r is idempotent implies that e s xe t e s xe t = e s xe t, so xe s+t x = x Suppose that s + t = n + v, where 0 v < n Thus xe v x = x From M w = F(u w a), we have x = α(u w a) for some Page 4 of 7

0 <α F, then αu w (ae v a)=a, and hence v = 0, otherwise ae v a = 0 It follows from αu w a = a that αu w = 1, an identity element of K Therefore, a r = e s ae n s Since I = a + eae n 1 + + e n 1 ae is a sum of disjoint elements, each e i ae n i is an idempotent f -element and e i ae n i e j ae n j = 0 for i j Thus, i, j=1 F(ei ae j ) is an l-algebra over F For 1 i, j n, define c ij = e i ae n j Then {c ij 1 i, j n} is a disjoint set of basic elements and also a set of n n matrix units It follows that Fc is a direct sum as a vector lattice over F Define φ:h M i, j=1 ij 1 n (F) by n α c i, j=1 ij ij n α e, where α i, j=1 ij ij ij F Then φ is an l-isomorphism between two l-algebras over F, so (F) with the entrywise lattice order Now for i = 1,, m, H i i, j=1 ei M i e j and M i = F(u i a) implies that H i = u i, i = 1,, m Therefore, the l-algebra A = + H 2 = u 1 + u 2 + +u m (K) over K with the entrywise lattice order This completes the proof of Theorem 21 Example 1 shows that the condition I + e + + e n 1 is a d-element on K" couldn t be omitted in Theorem 21 However, if K is a totally ordered field then this condition is automatically satisfied, so it is not necessary By Schwartz (1986, Theorem 10), under the assumption for the l-field K in Theorem 21, the lattice order on K is uniquely extendible to a total order Although we believe that if I + e + + e n 1 is not a d-element on K then the lattice order on M n (K) is defined by the unique total order on K just like the situation in Example 11, we lack the ability to prove it In the next, we will show that this fact is true for the simplest case 3 2 2 case Let K be an Archimedean l-field which is a two-dimensional extension of F and let {1, u} be a vl -basis of K over F and let u be a d-element Suppose that M 2 (K) is a 2 2 matrix l-algebra over K and that e = e 12 + e 21 0 By Theorem 21, we know that if I + e is a d-element on K then M 2 (K) is isomorphic to the l-algebra M 2 (K) with the entrywise lattice order In this section, we show that if I + e is not a d-element on K then the lattice order on M 2 (K) is the lattice order defined in Example 11 Theorem 31 Let M 2 (K) be an l-algebra over K in which e = e 12 + e 21 is positive, but I + e is not a d -element on K Then the lattice order on M 2 (K) is defined by declaring a matrix a =(a ij ) M 2 (K) to be positive if each a ij 0 in K, where is the unique total order that extends the lattice order on K Proof As before, suppose that F is the unique largest totally ordered subfield of K and {1, u} is an vl -basis of K over F where u is a d-element From the discussion in the previous section, M 2 (K) is a direct sum of totally ordered subspaces of K over F We first show that I e = 0 Let I e = z Then ez = ze = e e 2 = e I = z, so z = v(i + e) for some v K Since 0 z I, z is an f-element of M 2 (K), so I e = z implies that z 2 = z ze = z z = z Thus, z = v(i + e) implies that 2v 2 = v, and hence either v = 0 or 2v = 1 If 2v = 1 then I + e = 2z 2I implies that e I, which is a contradiction Thus, we must have v = 0, so I e = z = 0 Next, we show that I cannot be a basic element Assume, on the contrary, that I is basic Therefore, e is also basic Suppose that M 2 T k, where each T i is a totally ordered subspace of M 2 (K) over F and k 2 We may assume that I T 1 and e T 2 We consider two cases (1) k = 2 Take 0 < a T 1 Then e a = 0 implies that I ea = I ae = 0, so ae, ea T 2 Hence, ae and ea are comparable since T 2 is totally ordered If ae < ea then a = ae 2 < (ea)e < e(ea) =a, which is a contradiction Similarly, ea ae Thus, we must have ae = ea, so for any element x T 1, ex = xe Similarly for each y T 2, ey = ye Therefore, for each matrix w M 2 (K), ew = we, which is a contradiction Page 5 of 7

(2) k 3 Define H = 2 i, j=1 ei T 1 e j and N = 2 i, j=1 ei T 3 e j Then H = T 1 T 2 Since I, e are basic elements, it is straightforward to check that ui, ue are also basic elements If ui T 3 then u(i + e) N Since H N, H N ={0} On the other hand, since I + e is not a d-element on K, 0 (I + e) u(i + e) H N, which is a contradiction Thus, we must have ui H, and hence u(i + e) H Since each element in K can be written as α + βu with α, β F, that I + e and u(i + e) are in H implies that v(i + e) H for any v K Take 0 < a T 3 and form b = 2 i, j=1 ei ae j N Then be = eb = b implies that b = v(i + e) for some 0 < v K Then b H, which contradicts the fact that H N ={0} Hence, I cannot be a basic element Since I is not basic, I = + + a m, where,, a m are disjoint basic elements with 2 m 4 since e,, ea m are also disjoint and M 2 (K) is eight-dimensional over F Since a i I, each a i is an f-element, and hence a i a j = 0 for i j implies that a i a j = a i a j a i a j = 0 Thus, each a i is idempotent From I = + + a m, we have I = e e + + ea m e with ea i e ea j e = 0 for i j since e is a d-element on M 2 (K) Thus, each ea s e = a t for some t If s = t then ea s = a s e and since a s is idempotent, a s = 1 (I + e), 2 which is a contradiction since a i is basic We claim that m = 2 If m = 3 then we have I = + a 2 + a 3 = e e + ea 2 e + ea 3 e Since e e, we may assume e e = a 2 Then ea 2 e = implies that ea 3 e = a 3, which is a contradiction Suppose that m = 4 Then {, a 2, a 3, a 4, e, ea 2, ea 3, ea 4 } is disjoint and M 2 (K) is eight-dimensional over F, so {, a 2, a 3, a 4, e, ea 2, ea 3, ea 4 } is a vl-basis of M 2 (K) over F It is straightforward to check that u is a basic element, and hence u = α or u = β(e ) for some 0 <α, β F In the first case, we have u = α, which is a contradiction So u = β(e ), and hence β 2 = βe(βe )=uβ(e )=u 2 Hence u 2 = β 2, so u = β or β, which is again a contradiction Therefore, we must have m 4, so m = 2 Thus, we have I = + a 2, where, a 2 are disjoint basic elements Then as we have discussed before e e = a 2 and I = + e e Let M 2 T k, where k 4 and each T i is a maximal convex totally ordered subspace of M 2 (K) over F Assume that T 1, and define H = 2 i, j=1 ei T 1 e j = T 1 + et 1 + T 1 e + et 1 e Then I + e H Since u is basic, u T l for some T l If T l is not in {T 1, et 1 e, et 1 e} then H N ={0}, where N = 2 i, j=1 ei T l e j On the other hand, u T l implies that u(i + e) N, then 0 (I + e) u(i + e) H N, which is a contradiction Thus, T l is one of T 1, et 1 e, et 1 e, so u is in one of {T 1, et 1 e, et 1 e} If u e then u = ua 2 a e 1 1 = 0, a contradiction If e u then ue, so = a 2 ua ea = 0 Similarly u 1 1 1 is not comparable with e and e e, so u T 1 Since and u are linearly independent over F is two-dimensional over F, and hence et 1 e, and et 1 e are all two-dimensional over F Hence, M 2 et 1 T 1 e et 1 e as a vector lattice over F Let us consider the structure of T 1 first Since, u are linearly independent over F, T 1 ={α + βu α, β F} =K Then clearly T 1 is a field that is isomorphic to K under the mapping v v for any v K Since T 1 is totally ordered, K will be totally ordered if we define an order on K by saying that for any v K, v 0 if v 0 in T 1 Take v K, if v 0 in K then since M 2 (K) is an l-algebra over K, we have v 0 in M 2 (K), so v 0 in K Thus, the total order on K extends the lattice order on K, and hence = since there is a unique total order on K that extends the lattice order on K Hence, for any v K, v 0 if and only if v 0 in M 2 (K) Since M 2 T 1 e et 1 et 1 e = K K( e) K(e ) K(e e), if we define the mapping φ: M 2 (K) M 2 (K): for any a M 2 (K) ( ) a11 a a = 1 + 2 ( e)+a 21 (e )+a 22 (e e) 12, a 21 a 22 then clearly φ is an algebra isomorphism over K We also have a 0 1 0, 2 ( e) 0, a 21 (e ) 0, a 22 (e e) 0 1 0, 2 0, a 21 0, a 22 0 in T 1 1 0, 2 0, a 21 0, a 22 0 in K Page 6 of 7

Hence, the l-algebra M 2 (K) is isomorphic to the l-algebra M 2 (K) with the lattice order defined by (a ij ) 0 if each a ij 0 in K This completes the proof Finally, we provide an example to show that the assumption that u is a d-element cannot be omitted in Theorem 31 Example 32 Let K = Q[ 2]={α + β 2 α, β Q} and u = 1 + 2 Then {1, u} is a basis of vector space K over Q Define an element α + βu to be positive if α 0 and β 0 Since u 2 = 1 + 2u, K becomes an l-field in which u is not a d-element since 1 u = 0 but u u 2 = u (1 + 2u) =u 0 Another lattice order may be defined by the positive cone P = Q + 1 + Q + 2 Then K + P Now, we define a lattice order on M 2 (K) by calling a matrix a =(a ij ) positive if each a ij P Since K + P, M 2 (K) is an l-algebra over K We notice that e = e 12 + e 21 0, and I + e is not a d-element on K since 1 u = 0 but (I + e) u(i + e) =I + e 0 However, the lattice order on M 2 (K) is not defined by the usual total order extension of K + Acknowledgement The author thanks professor Yuehui Zhang (Shanghai Jiao Tong University) for valuable comments during the preparation of this paper Funding The authors received no direct funding for this research Author details Jingjing M E-mail: ma@uhcledu 1 Department of Mathematics, University of Houston-Clear Lake, 2700 Bay Area Boulevard, Houston, TX 77059, USA Citation information Cite this article as: Matrix l-algebras over l-fields, Jingjing Ma, Cogent Mathematics (2015), 2: 1053660 References Birkhoff, G, & Pierce, R S (1956) Lattice-ordered rings Anais da Academia Brasileira de Ciências, 28, 41 69 Conrad, P (1961) Some structure theorems for lattice-ordered groups Transactions of the American Mathematical Society, 99, 212 240 Ma, J (2000) Lattice-ordered matrix algebras with the usual lattice order Journal of Algebra, 228, 406 416 Ma, J, & Wojciechowski, P (2002) A proof of Weinberg s conjecture on lattice-ordered matrix algebras Proceedings of the American Mathematical Society, 130, 2845 2851 Schwartz, N (1986) Lattice-ordered fields Order, 3, 179 194 Steinberg, S (2010) Lattice-ordered rings and modules New York, NY: Springer Weinberg, E (1966) Scarcity on lattice-ordered matrix algebras Pacific Journal of Mathematics, 19, 561 571 2015 The Author(s) This open access article is distributed under a Creative Commons Attribution (CC-BY) 40 license You are free to: Share copy and redistribute the material in any medium or format Adapt remix, transform, and build upon the material for any purpose, even commercially The licensor cannot revoke these freedoms as long as you follow the license terms Under the following terms: Attribution You must give appropriate credit, provide a link to the license, and indicate if changes were made You may do so in any reasonable manner, but not in any way that suggests the licensor endorses you or your use No additional restrictions You may not apply legal terms or technological measures that legally restrict others from doing anything the license permits Cogent Mathematics (ISSN: 2331-1835) is published by Cogent OA, part of Taylor & Francis Group Publishing with Cogent OA ensures: Immediate, universal access to your article on publication High visibility and discoverability via the Cogent OA website as well as Taylor & Francis Online Download and citation statistics for your article Rapid online publication Input from, and dialog with, expert editors and editorial boards Retention of full copyright of your article Guaranteed legacy preservation of your article Discounts and waivers for authors in developing regions Submit your manuscript to a Cogent OA journal at wwwcogentoacom Page 7 of 7