Jingjing MaNo Descriptionhttps://hdl.handle.net/10657.1/24132024-09-14T09:54:28Z2024-09-14T09:54:28Z91Lattice-ordered matrix algebras containing positive cyclesMa, Jingjinghttps://hdl.handle.net/10657.1/24532020-09-04T08:00:39Z2013-01-01T00:00:00Zdc.title: Lattice-ordered matrix algebras containing positive cycles
dc.contributor.author: Ma, Jingjing
dc.description.abstract: It is shown that if a lattice-ordered n × n (n ≥ 2) matrix ring over a totally ordered integral domain or division ring containing a positive n-cycle, then it is isomorphic to the lattice-ordered n × n matrix ring with entrywise lattice order.
2013-01-01T00:00:00ZRecognition of Lattice-ordered matrix algebrasMa, Jingjinghttps://hdl.handle.net/10657.1/24512020-09-04T08:00:38Z2013-01-01T00:00:00Zdc.title: Recognition of Lattice-ordered matrix algebras
dc.contributor.author: Ma, Jingjing
dc.description.abstract: For an ℓ-unital ℓ-ring R, various recognition criteria are given for R to be isomorphic to a matrix ℓ-ring over an ℓ-unital ℓ-ring with the entrywise order.
2013-01-01T00:00:00ZLattice -ordered matrix rings over totally ordered ringsMa, Jingjinghttps://hdl.handle.net/10657.1/24502020-09-04T08:00:37Z2014-01-01T00:00:00Zdc.title: Lattice -ordered matrix rings over totally ordered rings
dc.contributor.author: Ma, Jingjing
dc.description.abstract: For an nxn matric algebra over a totally ordered integral domain, necessary and sufficient conditions are derived such that the entrywise lattice order on it is the only lattice order (up to an isomorphism) to make it into a lattice ordered algebra in which the identity matrix is positive. The conditions are then applied to particular integral domains. In the second part of the paper we consider nxn matrix rings containing a positive n-cycle over totally ordered rings. Finally, a characterization of lattice-ordered matrix rings with the entrywise lattice order is given.
2014-01-01T00:00:00ZPositive derivations on Archimedean d-ringsMa, Jingjinghttps://hdl.handle.net/10657.1/24492020-09-04T08:00:37Z2014-01-01T00:00:00Zdc.title: Positive derivations on Archimedean d-rings
dc.contributor.author: Ma, Jingjing
dc.description.abstract: For an Archimedean d-ring R and a positive derivation D on R it is shown that D(R) is a subset of N(R), where N(R) is the l-radical of R.
2014-01-01T00:00:00ZDirected partial orders on real quaternionsMa, Jingjinghttps://hdl.handle.net/10657.1/24482020-09-04T08:00:36Z2015-01-01T00:00:00Zdc.title: Directed partial orders on real quaternions
dc.contributor.author: Ma, Jingjing
dc.description.abstract: It is shown that a division ring of real quaternions can be made into a partially ordered ring with a directed partial order.
2015-01-01T00:00:00ZMatrix l-Algebras over l-fieldsMa, Jingjinghttps://hdl.handle.net/10657.1/24472020-09-04T08:00:36Z2015-01-01T00:00:00Zdc.title: Matrix l-Algebras over l-fields
dc.contributor.author: Ma, Jingjing
dc.description.abstract: It is shown that if a matrix ℓ-algebra Mn(K) over certain ℓ-fields K contains a positive n-cycle e such that I+e+⋯+en−1 is a d-element on K then it is isomorphic to the ℓ-algebra Mn(K) over K with the entrywise lattice order.
2015-01-01T00:00:00ZPartial Orders on C=D + Di and H=D +Di +Dj + DkMa, Jingjinghttps://hdl.handle.net/10657.1/24462020-09-04T08:00:35Z2015-01-01T00:00:00Zdc.title: Partial Orders on C=D + Di and H=D +Di +Dj + Dk
dc.contributor.author: Ma, Jingjing
dc.description.abstract: Let \(D\) be a totally ordered integral domain. We study partial orders on the rings \(C = D + Di\) and \(H = D + Di + Dj + Dk\), where \(i^{2} = j^{2} = k^{2} = -1\).
2015-01-01T00:00:00ZDirected Partial Orders on Complex Numbers ad Quaternions IIMa, Jingjinghttps://hdl.handle.net/10657.1/24452020-09-04T08:00:33Z2015-01-01T00:00:00Zdc.title: Directed Partial Orders on Complex Numbers ad Quaternions II
dc.contributor.author: Ma, Jingjing
dc.description.abstract: .Suppose that F is a partially ordered field with a directed partial order and K is a non-archemedean totally ordered subfield of F with K+=F+∩K. In this note, directed partial orders are constructed for complex numbers and quaternions over F. It is also shown that real quaternions cannot be made into a directed algebra over the real field with the total order.
Let T be a non-archimedean totally ordered field and CT=T+Ti be the field of complex numbers over T, where i2=−1, and HT=T+Ti+Tj+Tk be the division algebra of quaternions over T. In [4], a general method of constructing directed partial orders on CT and HT has been developed. The purpose of this note is to generalize the results in [4] to complex numbers and quaternions over non-archimedean partially ordered fields with a directed partial order.
We review a few definitions and the reader is referred to [2, 4] for undefined terminology and background on partially ordered rings and lattice-ordered rings (ℓ-rings). For a partially ordered ring R, the positive cone is defined as R+={r∈R|r≥0}. The partial order on a partially ordered ring R is called directed if each element of R can be written as a difference of two positive elements of R. A partially ordered ring (algebra) with a directed partial order is called a directed ring (algebra). In the following, we always assume that F is a partially ordered field with a directed partial order and K is a non-archimedean totally ordered subfield of F, so F+∩K=K+.
Following result will be used in the proof of main results in the paper.
Lemma 1
Suppose that R is a partially ordered ring with the property that for any r∈R, 2r≥0 implies that r≥0. Then for any x,y∈R+, and a,b∈R, −x≤a≤x and −y≤b≤y implies that −xy≤ab≤xy.
Proof
From −x≤a≤x and −y≤b≤y, we have
(x+a)(y−b)≥0⇒xy+ay−xb−ab≥0,
(1)
and
(x−a)(y+b)≥0⇒xy−ay+xb−ab≥0.
(2)
Adding (1) and (2), we have 2(xy−ab)≥0, so xy−ab≥0 and xy≥ab.
From −x≤a≤x and −y≤b≤y, we also have
(x+a)(y+b)≥0⇒xy+ay+xb+ab≥0,
(3)
and
(x−a)(y−b)≥0⇒xy−ay−xb+ab≥0.
(4)
Adding (3) and (4), we have 2(xy+ab)≥0, so xy+ab≥0, that is, ab≥−xy. □
We notice that since F contains a totally ordered subfield K, for any a∈F, 2a≥0 implies that a≥0. In fact, since 0<2∈K, 0<12∈K, so a=12(2a)≥0.
Theorem 1
Take 0≤x,y≤1 in F with x≠0 or y≠0 such that x−1>0 if x≠0 and y−1>0 if y≠0, define the positive cone Px,y of CF=F+Fi as follows.
Px,y={a+bi|a∈F+,−xa≤nb≤ya in F for all positive integers n}.
Then Px,y is a directed partial order on CF such that (CF,Px,y) is a partially ordered algebra over F and Px,y∩F=F+.
Proof
It is clear that Px,y∩−Px,y={0}, Px,y+Px,y⊆Px,y and F+Px,y⊆Px,y. Suppose that a+bi,c+di∈Px,y. Then a,c∈F+, and for all positive integers n,
−a≤−xa≤nb≤ya≤a, −c≤−xc≤nd≤yc≤c.
We show that (a+bi)(c+di)=(ac−bd)+(ad+bc)i∈Px,y.
From −a≤b≤a, −c≤d≤c and Lemma 1, we have bd≤ac, that is, ac−bd∈F+.
For any positive integer n, 3nd≤yc,3nb≤ya and a,c∈F+ implies that 3nad≤y(ac),3nbc≤y(ac). Since −a≤3b≤a and −c≤d≤c, Lemma 1 implies that 3bd≤ac; so 3ybd≤y(ac). Hence
3nad+3nbc+3ybd≤y(ac)+y(ac)+y(ac)=3y(ac).
Since 3∈K+ and K is totally ordered, 13∈K+⊆F+, so 3(nad+nbc+ybd)≤3y(ac) implies that nad+nbc+ybd≤y(ac). Therefore n(ad+bc)≤y(ac−bd) for all positive integers n. Similarly −x(ac−bd)≤n(ad+bc). We have proved that (a+bi)(c+di)∈Px,y. Thus Px,y is a partial order on CF with respect to which CF is a partially ordered algebra. Clearly Px,y∩F=F+.
We verify that Px,y is a directed partial order on CF. Suppose that y≠0. A similar argument could be used in the case x≠0. Let a+bi∈CF. Since F is directed, a is a difference of two elements in F+, so a is a difference of two elements in Px,y since F+⊆Px,y. Consider bi=b1i−b2i, where b1,b2>0 in F. Since K is a non-archimedean totally ordered field, there exists z∈K+ such that n1≤z for all positive integers n. Let w=y−1b1. Then w,wz∈F+ and for all positive integers n, −x(wz)≤0≤nb1≤b1z=y(wz), that is, wz+b1i∈Px,y. Thus b1i=(wz+b1i)−wz is a difference of two positive elements in CF. Similarly b2i is a difference of two positive elements. Hence bi is a difference of two positive elements in CF, so a+bi is a difference of two positive elements. Therefore Px,y is directed. □
Remark 1
If the partial order on F is a lattice order, then 0<x<1 implies that x−1>0 since x<1 implies that x is an f-element [3, Theroem 1.20(2)].
Next we consider quaternions over F. Recall that HF=F+Fi+Fj+Fk as a vector space over F with the multiplication as follows.
=(a0+a1i+a2j+a3k)(b0+b1i+b2j+b3k)(a0b0−a1b1−a2b2−a3b3)+(a0b1+a1b0+a2b3−a3b2)i+(a0b2+a2b0+a3b1−a1b3)j+(a0b3+a3b0+a1b2−a2b1)k,
where ai,bi∈F.
Theorem 2
Take 0<x≤1 in F such that x−1>0, define the positive cone Px of HF as follows.
Px={a0+a1i+a2j+a3k | a0∈F+ and −xa0≤na1≤xa0,−xa0≤na2≤xa0, −xa0≤na3≤xa0 in F for all positive integers n}.
Then Px is a directed partial order on HF such that (HF,Px) is a partially ordered algebra over F and Px∩F=F+.
Proof
It is clear that Px∩−Px={0}, Px+Px⊆Px and F+Px⊆Px. We show that PxPx⊆Px. Suppose that a0+a1i+a2j+a3k,b0+b1i+b2j+b3k∈Px. We check that (a0+a1i+a2j+a3k)(b0+b1i+b2j+b3k)∈Px. Since a0+a1i+a2j+a3k,b0+b1i+b2j+b3k∈Px, we have for all positive integers n,
a0≥0, −xa0≤na1≤xa0, −xa0≤na2≤xa0, −xa0≤na3≤xa0,
and
b0≥0, −xb0≤nb1≤xb0, −xb0≤nb2≤xb0, −xb0≤nb3≤xb0.
We first check that a0b0−a1b1−a2b2−a3b3≥0 in F.
From −a0≤−xa0≤3a1≤xa0≤a0, −b0≤−xb0≤b1≤xb0≤b0 and Lemma 1, we have a0b0≥3a1b1, that is, a0b0−3a1b1≥0. Similarly, a0b0−3a2b2≥0 and a0b0−3a3b3≥0. Adding those three inequalities together, we obtain 3a0b0−3a1b1−3a2b2−3a3b3≥0. Then multiplying the both sides by 13 we get a0b0−a1b1−a2b2−a3b3≥0.
For simplicity, let w=x(a0b0−a1b1−a2b2−a3b3). We next show that for all positive integers n,
−w≤n(a0b1+a1b0+a2b3−a3b2)≤w.
Consider n(a0b1+a1b0+a2b3−a3b2)≤w first. We divide the calculations into several steps. Let n be a positive integer.
1.
Since 7nb1≤xb0 and a0≥0, we have 7na0b1≤xa0b0.
2.
Since 7na1≤xa0 and b0≥0, we have 7na1b0≤xa0b0.
3.
Since −a0≤7na2≤a0 and −xb0≤b3≤xb0, Lemma 1 implies that 7na2b3≤xa0b0.
4.
Since −a0≤7na3≤a0 and −xb0≤b2≤xb0, Lemma 1 implies that 7na3b2≥−xa0b0, so −7na3b2≤xa0b0.
5.
Since −a0≤7a1≤a0 and −b0≤b1≤b0, Lemma 1 implies that 7a1b1≤a0b0, so 7xa1b1≤xa0b0. Similarly, 7xa2b2≤xa0b0 and 7xa3b3≤xa0b0.
Adding inequalities in the above (1) to (5) together, we have
7na0b1+7na1b0+7na2b3−7na3b2+7xa1b1+7xa2b2+7xa3b3≤7xa0b0,
and hence, by multiplying both sides of this inequality by 17,
na0b1+na1b0+na2b3−na3b2+xa1b1+xa2b2+xa3b3≤xa0b0.
It follows that n(a0b1+a1b0+a2b3−a3b2)≤x(a0b0−a1b1−a2b2−a3b3).
Similarly we prove −w≤n(a0b1+a1b0+a2b3−a3b2). Let n be a positive integer.
1.
Since −xb0≤7nb1 and a0≥0, we have that −xa0b0≤7na0b1.
2.
Since −xa0≤7na1 and b0≥0, we have that −xa0b0≤7na1b0.
3.
Since −a0≤7na2≤a0 and −xb0≤b3≤xb0, Lemma 1 implies that −xa0b0≤7na2b3.
4.
Since −a0≤7na3≤a0 and −xb0≤b2≤xb0, Lemma 1 implies that 7na3b2≤xa0b0, so −xa0b0≤−7na3b2.
5.
Since −a0≤7a1≤a0 and −b0≤b1≤b0, we have −a0≤7a1≤a0 and −b0≤−b1≤b0, so Lemma 1 implies that −a0b0≤−7a1b1, so −xa0b0≤−7xa1b1. Similarly, −xa0b0≤−7xa2b2 and −xa0b0≤−7xa3b3.
Adding up the above inequalities, we have
−7xa0b0≤7na0b1+7na1b0+7na2b3−7na3b2−7xa1b1−7xa2b2−7xa3b3,
and hence
−xa0b0≤na0b1+na1b0+na2b3−na3b2−xa1b1−7xa2b2−xa3b3,
that is, −x(a0b0−a1b1−a2b2−a3b3)≤n(a0b1+a1b0+a2b3−a3b2).
By similar calculations, we have
−w≤n(a0b2+a2b0+a3b1−a1b3)≤w,
and
−w≤n(a0b3+a3b0+a1b2−a2b1)≤w.
Therefore PxPx⊆Px, so (HF,Px) is a partially ordered algebra over F. It is straightforward to verify that Px∩F=F+.
Finally we show that Px is a directed partial order on HF. Take a∈F. Since F is directed, a=b−c, where b,c∈F+. Since F+⊆Px, a is a difference of two positive elements in (HF,Px). Consider ai=bi−ci. Since K is a non-archimedean totally ordered field, there exists z∈K+ such that n1≤z for all positive integers n. Let v=x−1b. Then v,vz∈F+ and for all positive integers n, nb≤bz=x(vz), that is, vz+bi∈Px. Thus bi=(vz+bi)−vz is a difference of two positive elements in (HF,Px). Similarly, ci is also a difference of two positive elements in (HF,Px), and hence ai=bi−ci is a difference of two positive elements in (HF,Px). The same argument may be used to show that aj and ak are also a difference of two positive elements in (HF,Px). It follows that each element in HF is a difference of two positive elements in (HF,Px), that is, Px is directed. This completes the proof of Theorem 2. □
In [1], Birkhoff asked if H=HR can be made into a directed algebra over R with the usual total order, where R is the field of real numbers. In the following, we show that the answer is no. More generally, for any totally ordered subfield T of R, HT cannot be a directed algebra over T.
Theorem 3
H cannot be a directed algebra over R with the usual total order.
Proof
We will suppose that H is a directed algebra over R and derive a contradiction.
We first show that if w=a0+a1i+a2j+a3k>0 in H, then a0>0 in R. Note that w2−2a0w=−(a20+a21+a22+a23) by direct calculation, and that if 0<a in R and 0<z in H, then 0<az in H because H is a partially ordered algebra over R that has no divisor of zero.
Now suppose by the way of contradiction that a0≤0 in R. Then −a0≥0 in R, and thus since w>0 in H, w2−2a0w≥0 in H. That is, −(a20+a21+a22+a23)≥0 in H, and hence −(a20+a21+a22+a23)w>0 in H. But since (a20+a21+a22+a23)≥0 in R, (a20+a21+a22+a23)w≥0 in H as well, and therefore (a20+a21+a22+a23)w=0. So since w≠0 and H is a division ring, we must have a20+a21+a22+a23=0. But then a0=a1=a2=a3=0 because a0,a1,a2, and a3 are all in R, and hence w=a0+a1i+a2j+a3k=0, a contradiction. It follows that a0>0 in R.
Since the partial order on H is directed, there exists z=a+bi+cj+dk>0 in H with z∉R. For instance, i=z1−z2, where z1,z2 are positive in H. Clearly z1,z2 cannot be both in R. The argument above shows that a>0 in R. Then since R is totally ordered, a−1>0 in R and hence a−1z=1+(a−1b)i+(a−1c)j+(a−1d)k>0 in H. Suppose that a−1b=s,a−1c=t,a−1d=u. Then we have w=1+si+tj+uk>0 in H and w∉R. For simplicity, let v=si+tj+uk. Then v2=−(s2+t2+u2), so −v2∈R+. Therefore −v2w≥0 in H, and hence w3−v2w>0 in H. Since
1+3v+2v2==(1+2v)(1+v)=(1+2v+v2−v2)w(w2−v2)w=w3−v2w,
we have 1+3v+2v2>0 in H. Let w1=1+3v+2v2. Then w1>0 in H, and hence (w1−2v2)w=w1w−2v2w>0 in H. Since
1+4v+3v2=(1+3v)(1+v)=(1+3v+2v2−2v2)w=(w1−2v2)w
1+4v+3v2>0 in H. If we continue this procedure, we get that for any positive integer n, (1+nv)(1+v)=1+(n+1)v+nv2>0 in H. Therefore since the real part of a positive element in H must be positive in R, we must have 0≤1+nv2 for all positive integers n, so −nv2≤1 for all positive integers n. Then −v2=0, so v2=0 since R is archimedean with respect to the total order. Hence s=t=u=0, and w∈R, a contradiction of our observation above that w∉R.
Therefore H cannot be a directed algebra over R with the total order.
2015-01-01T00:00:00ZDirected Partial Orderson Complex Numbers and Quaternions over Non-Archimedean Linearly Ordered FieldsMa, Jingjinghttps://hdl.handle.net/10657.1/24442020-09-04T08:00:30Z2016-01-01T00:00:00Zdc.title: Directed Partial Orderson Complex Numbers and Quaternions over Non-Archimedean Linearly Ordered Fields
dc.contributor.author: Ma, Jingjing
dc.description.abstract: Let 'F' be a non-archimedean linearly ordered field, and 'C' and 'H' be the field of complex numbers and the division algebra of quaternions over 'F', respectively. In this paper, a class of directed partial orders on 'C' are constructed directly and concretely using additive subgroup of 'F+'. This class of directed partial orders includes those given in Rump and Wang (J. Algebra 400, 1-7, 2014), and Yang (J. Algebra 295 (2), 452-457, 2006) as special cases and we conjecture that it covers all directed partial orders on 'C' such that 1>0. It turns out that this construction also works very well on 'H'. We note that none of these directed partial orders is a lattice order on 'C' or 'H'.
2016-01-01T00:00:00Z