On the equivalence of matrices over commutative rings modulo ideals
We study connections between wildness of the problem of classifying the matrices over an integral domain up to equivalence modulo an ideal and properties of the set of prime elements of the domain.
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irk-123456789-1523372019-06-11T01:25:15Z On the equivalence of matrices over commutative rings modulo ideals Bondarenko, V.M. Tylyshchak, A.A. Stoika, M.V. We study connections between wildness of the problem of classifying the matrices over an integral domain up to equivalence modulo an ideal and properties of the set of prime elements of the domain. 2014 Article On the equivalence of matrices over commutative rings modulo ideals / V.M. Bondarenko, A.A. Tylyshchak, M.V. Stoika // Algebra and Discrete Mathematics. — 2014. — Vol. 17, № 1. — С. 12–19 . — Бібліогр.: 3 назв. — англ. 1726-3255 2010 MSC:15A21, 15B33. http://dspace.nbuv.gov.ua/handle/123456789/152337 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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We study connections between wildness of the problem of classifying the matrices over an integral domain up to equivalence modulo an ideal and properties of the set of prime elements of the domain. |
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Bondarenko, V.M. Tylyshchak, A.A. Stoika, M.V. |
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Bondarenko, V.M. Tylyshchak, A.A. Stoika, M.V. On the equivalence of matrices over commutative rings modulo ideals Algebra and Discrete Mathematics |
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Bondarenko, V.M. Tylyshchak, A.A. Stoika, M.V. |
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Bondarenko, V.M. |
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On the equivalence of matrices over commutative rings modulo ideals |
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On the equivalence of matrices over commutative rings modulo ideals |
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On the equivalence of matrices over commutative rings modulo ideals |
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On the equivalence of matrices over commutative rings modulo ideals |
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On the equivalence of matrices over commutative rings modulo ideals |
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on the equivalence of matrices over commutative rings modulo ideals |
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Інститут прикладної математики і механіки НАН України |
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On the equivalence of matrices over commutative rings modulo ideals / V.M. Bondarenko,
A.A. Tylyshchak, M.V. Stoika // Algebra and Discrete Mathematics. — 2014. — Vol. 17, № 1. — С. 12–19 . — Бібліогр.: 3 назв. — англ. |
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Algebra and Discrete Mathematics |
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AT bondarenkovm ontheequivalenceofmatricesovercommutativeringsmoduloideals AT tylyshchakaa ontheequivalenceofmatricesovercommutativeringsmoduloideals AT stoikamv ontheequivalenceofmatricesovercommutativeringsmoduloideals |
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1837498440144125952 |
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Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 17 (2014). Number 1. pp. 12 – 19
c© Journal “Algebra and Discrete Mathematics”
On the equivalence of matrices over
commutative rings modulo ideals
Vitaliy M. Bondarenko, Alexander A. Tylyshchak,
Myroslav V. Stoika
Communicated by V. V. Kirichenko
Abstract. We study connections between wildness of the
problem of classifying the matrices over an integral domain up to
equivalence modulo an ideal and properties of the set of prime
elements of the domain.
Introduction
The paper is devoted to the problem of classifying up to equivalence
the matrices over commutative rings not being principal ideal ones.
Let K be a commutative ring and J 6= K be its ideal. For matrices
P and Q over K, the notation P ≡ Q (mod J) means that P − Q is a
matrix with entries from J .
Two m × n matrices A and B over K is said to be equivalent modulo
J if, for some invertible m × m matrix Y and invertible n × n matrix X,
B ≡ Y −1AX (mod J).
Analogously, n × n matrices A and B over K is said to be similar modulo
J if, for some invertible n × n matrix X,
B ≡ X−1AX (mod J).
2010 MSC: 15A21, 15B33.
Key words and phrases: commutative ring, prime element, equivalent matrices
modulo an ideal, perfect (I, J)-matrix, wildness.
V. M. Bondarenko, A. A. Tylyshchak, M. V. Stoika 13
When J = 0, we have the classical definitions of equivalent and similar
matrices.
The above definitions can be naturally generalized to any numbers of
matrices, and in particular (in the case of similarity) for matrix represen-
tations of K-algebras.
Definition 1. Let I, J 6= K be ideals of K and let Σ = K〈x, y〉 denotes
the free associative K-algebra generated by x, y. An m×n matrix M over
Σ is said to be (I, J)-perfect if from the equivalence of the matrices M ⊗T
and M ⊗ T ′ over K modulo I, where T, T ′ are matrix representations
of Σ over K, it follows that T and T ′ are equivalent modulo J . We say
that the problem of classifying the matrices over K up to equivalence
modulo I is wild over K modulo J , or simply wild modulo J , if there is
an (I, J)-perfect representation over Σ.
Note that we do not require that K/J is a field, i. e. the notion of
wildness is introduced by us not only concerning fields, as before for
various classification problems, but also concerning rings (more detail see
Section 5).
From now on K denotes, unless otherwise stated, an integral domain.
Its non-unit element c is called a prime element, or simple a prime, if
from c|ab for some a, b ∈ K it follows that c|a or c|b. By different prime
elements of K we mean pairwise non-associated ones.
Now we formulate the main results of this paper. In all statements, J
denotes (as above) an ideal of K different from the ring itself.
Theorem 1. Let K be an integral domain and J contains two different
primes. Then the problem of classifying up to equivalence the matrices
over K is wild modulo J .
Theorem 2. Let K be an integral domain, t1, t2 ∈ J be different primes
and t ∈ J be non-zero. Then the problem of classifying the matrices over
K up to equivalence modulo t1t2tK is wild modulo J .
Theorem 3. Let K be an integral domain, t1 ∈ J be a prime and t ∈ J
be non-zero. If J contains a prime t2 different from t1, then the problem
of classifying the matrices over K up to equivalence modulo t2
1
tK is wild
modulo J .
From Theorems 2 and 3 it follows the following corollary.
Corollary 1. Let K be an integral domain, and t1, t2, t3 ∈ J be primes
that are not all equal. Then the problem of classifying the matrices over
K up to equivalence modulo t1t2t3K is wild modulo J .
14 Equivalence of matrices over commutative rings
1. Proof of Theorem 1
We need the following lemma.
Lemma 1. Let p, q be different prime elements from J and let
p2a + q2b + pqc = 0 (1)
for some a, b, c ∈ K. Then a, b, c ∈ J .
Proof. From (1) it follows that, firstly,
q(qb + pc) = −p2a
and, secondly,
p(pa + qc) = −q2b
whence respectively q|a and p|b. So a = qa′ ∈ J and b = pb′ ∈ J for some
a′, b′ ∈ K. Then the equality (1) is equivalent to the equality
pq(pa′ + qb′ + c) = 0
whence pa′ + qb′ + c = 0 and consequently c ∈ J .
It is natural to identify a matrix representations T of Σ = K〈x, y〉
over K with the ordered pair of matrices T (x), T (y); if these matrices
are of size n × n, we say that T is of K-dimension n. Then, for a matrix
M over K〈x, y〉 (see above the definition of wildness), the matrix M ⊗ T
with T of K-dimension m is obtained from the matrix M by change x
and y on the matrices T (x) and T (y), and a ∈ K on the scalar matrix
aEn, where En is the identity n × n matrix.
Consider the following matrix M (of size 1 × 1) over Σ:
M = p2x + q2y + pq.
We prove that the matrix M is (0, J)-perfect.
Let T = (A, B) and T ′ = (A′, B′) be matrix representations of Σ over
K of a K-dimension n. Then
M ⊗ T = p2A + q2B + pqEn,
M ⊗ T ′ = p2, A′ + q2B′ + pqEn.
V. M. Bondarenko, A. A. Tylyshchak, M. V. Stoika 15
Assume that the matrices M ⊗ T and M ⊗ T ′ (over K) are equivalent,
i. e. there exists invertible n × n matrices X and Y over K such that
(M ⊗ T )X = Y (M ⊗ T ′). So we have the equality
(p2A + q2B + pqEn)X = Y (p2A′ + q2B′ + pqEn)
or, equivalently,
p2(AX − Y A′) + q2(BX − Y B′) + pq(X − Y ) = 0.
By applying Lemma 1 to all scalar equalities of the last matrix equality,
we have that
AX − Y A′ ≡ 0 (mod J),
BX − Y B′ ≡ 0 (mod J),
X − Y ≡ 0 (mod J),
and consequently AX ≡ XA′ (mod J) and BX ≡ XB′ (mod J), as
claimed.
Theorem 1 is proved.
2. Proof of Theorem 2
We need the following lemma.
Lemma 2. Let a, b, c ∈ K such that
t2
1a + t2
2b + t1t2c ≡ 0 (mod t1t2tK). (2)
Then a, b, c ∈ J .
Proof. The comparison (2) means that there exists u ∈ K such that
t2
1a + t2
2b + t1t2c = t1t2 tu. (3)
From (3),
t2(t2b + t1c − t1tu) = −t2
1a (4)
whence t2|a and therefore a ∈ J . Let a = t2a′. Then we have from (4)
(after reducing by t2 and elementary transformations) that
t1(c + t1a′ − tu) = −t2b
whence t1|b and t2|c + t1a′ − tu; consequently b, c ∈ J .
16 Equivalence of matrices over commutative rings
Consider the following matrix M (of size 1 × 1) over Σ:
M = t2
1x + t2
2y + t1t2.
We prove that the matrix M is (t1t2tK, J)-perfect.
Let T = (A, B) and T ′ = (A′, B′) be matrix representations of Σ over
K of a K-dimension n. Then
M ⊗ T = t2
1A + t2
2B + t1t2En,
M ⊗ T ′ = t2
1A′ + t2
2B′ + t1t2En.
Assume that the matrices M ⊗ T and M ⊗ T ′ (over K) are equivalent
modulo t1t2tK, i. e. there exists invertible n × n matrices X and Y over
K such that (M ⊗ T )X ≡ Y (M ⊗ T ′) (mod t1t2tK). So we have the
comparison
(t2
1A + t2
2B + t1t2En)X ≡ Y (t2
1A′ + t2
2B′ + t1t2En) (mod t1t2tK)
or, equivalently,
t2
1(AX − Y A′) + t2
2(BX − Y B′) + t1t2(X − Y ) ≡ 0 (mod t1t2tK).
By applying Lemma 2 to all scalar comparisons of the last matrix
one, we see that (as in the proof of Theorem 1) AX ≡ XA′ (mod J), and
BX ≡ XB′ (mod J), as claimed.
Theorem 2 is proved.
3. Proof of Theorem 3
We need the following lemma.
Lemma 3. Let a, b, c ∈ K such that
t2
1a + t2
2b + t1t2c ≡ 0 (mod t2
1tK). (5)
Then a, b, c ∈ J .
Proof. The comparison (5) means that there exists u ∈ K such that
t2
1a + t2
2b + t1t2c = t2
1 tu. (6)
From (6),
t1(t1a + t2c − t1tu) = −t2
2b (7)
V. M. Bondarenko, A. A. Tylyshchak, M. V. Stoika 17
whence t1|b and therefore b ∈ J . Let b = t1b′. Then we have from (7)
(after reducing by t1 and elementary transformations) that
t1(a − tu) = −t2(c + t2b′)
whence t2|a − tu and t1|c + t2b′; consequently a, c ∈ J .
We take as a (t2
1
tK, J)-perfect matrix M over Σ the matrix of the
same form as in the proof of Theorem 2 with t2 to be any prime element
different from t1 (it exists by the condition of the theorem). Then the
proof of Theorem 3 is analogously to that of Theorem 2, but it is need to
use Lemma 3 instead of Lemma 2.
4. The case of factorial rings
A factorial ring K is an integral domain in which there exists a system
of prime elements P such that every non-zero element x ∈ K admits a
unique representation
x = ε
∏
p∈P
psp ,
where ε is invertible and the integral exponents sp ≥ 0 are non-zero for
only a finite number of elements. The number l(x) =
∑
p∈P sp is called
the length of x.
From the above we have the following theorem.
Theorem 4. Let K be a factorial ring and J 6= K be its ideal having
at least two primes. Then, for any element v ∈ J of length l(v) > 2, the
problem of classifying the matrices over K up to equivalence modulo vK
is wild modulo J .
Indeed, if l(v) > 2 then v = t1t2t, where t1, t2 are prime and t is not
invertible. And consequently the statement of the theorem follows from
Theorem 2 if t1 6= t2 and from Theorem 3 if t1 = t2.
5. The case of Noether domain
The equivalence of matrices over Noether factorial and local rings
were studied in [3]. We formulate here some consequences of our theorem
in the case when the ring K is Noether.
Recall that according the main idea of wildness, a classification matrix
problem is called wild if it involves the problem of classifying (up to
18 Equivalence of matrices over commutative rings
similarity) the pairs of matrices over a field k; see precise definitions in
[1, 2] (in particular, for a case with the ring of p-adic numbers [2, pp.
70-71]). The first authors proposed to generalize the traditional definition
of wild matrix problems allowing to take k being a commutative ring
(it is especially actual for matrix problems over non-Noether rings). An
example of the realization of this idea is the definition of wildness modulo
J of the problem of classifying the matrices over K up to equivalence
modulo I (see Definition 1).
Definition 2. The problem of classifying the matrices over K up to
equivalence modulo I is said to be wild over a field, or simply wild, if it
wild modulo a maximal ideal J .
It is easy to see that from Theorems 1–3 and Corollary 1 we have
respectively the following statements.
Theorem 5. Let K be a Noether domain and p, q be its different primes
such that pK +qK 6= K. Then the problem of classifying up to equivalence
the matrices over K is wild.
Theorem 6. Let K be a Noether domain and t1, t2, t be its non-zero
elements such that t1K + t2K + tK 6= K. If t1 and t2 are different primes,
then the problem of classifying the matrices over K up to equivalence
modulo t1t2tK is wild.
Theorem 7. Let K be a Noether domain and t1, t2, t be its non-zero
elements such that t1K + t2K + tK 6= K. If t1 and t2 are different primes,
then the problem of classifying the matrices over K up to equivalence
modulo t2
1
tK is wild.
Corollary 2. Let K be a Noether domain and t1, t2, t3 be its primes that
are not all equal, and such that t1K + t2K + t3K 6= K. Then the problem
of classifying the matrices over K up to equivalence modulo t1t2t3K is
wild.
References
[1] Ju. A. Drozd, Tame and wild matrix problems, Matrix problems, Akad. Nauk Ukrain.
SSR, Inst. Mat., Kiev, 1977, pp. 104-114 (in Russian).
[2] Ju. A. Drozd, Tame and wild matrix problems, Representations and quadratic forms,
Akad. Nauk Ukrain. SSR, Inst. Mat., Kiev, 1979, pp. 39-74 (in Russian).
[3] P. M. Gudivok, On the equivalence of matrices over commutative rings, Infinite
groups and related algebraic structures, Infinite groups and related algebraic struc-
tures, Akad. Nauk Ukrainy, Inst. Mat., Kiev, 1993, pp. 431-437 (in Russian).
V. M. Bondarenko, A. A. Tylyshchak, M. V. Stoika 19
Contact information
V. M. Bondarenko Institute of Mathematics, Tereshchenkivska 3,
01601 Kyiv, Ukraine
E-Mail: vit-bond@imath.kiev.ua
URL: http://www.imath.kiev.ua
A. A. Tylyshchak Mathematical Faculty, Uzhgorod National Univ.,
Universytetsyka str., 14, 88000 Uzhgorod,
Ukraine
E-Mail: alxtlk@gmail.com
M. V. Stoika Humanities and Natural Sciences Faculty, Uzh-
gorod National Univ., Universytetsyka str., 14,
88000 Uzhgorod, Ukraine
E-Mail: stoyka−m@yahoo.com
Received by the editors: 19.02.2014
and in final form 19.02.2014.
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