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to the forefront
of international research
and to stabilize this position.
2. To strengthen cooperation
with international research centers:
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Harish-Chandra Research Institute,
Allahabad, India
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Washington and Lee University,
Lexington, Virginia, USA
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Université Mohammed Premier,
Oujda, Morocco
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Australian National University,
Canberra, Capital Territory
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Aichi University of Education,
Nagoya, Japan
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University of Calgary,
Alberta, Canada
International Conferences:
January 21 - 25, 2019:
Asia-Australia Algebra Conference 2019
Western Sydney University, Parramatta City campus
Sydney, New South Wales, Australia
April 25 - 27, 2019:
Conference on Algebra, Number Theory and Their Applications
Université Mohammed Premier, Faculté des Sciences
Oujda, Region Oriental, Morocco
Daniel C. Mayer's 3 Lectures:
1. Proving the Conjecture of Scholz
2. Differential principal factors
3. Pure Metacyclic Fields
July 1 - 5, 2019:
31st Journées Arithmétiques
Istanbul University, Faculty of Science
JA 2019, Istanbul, Turkey
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Open Problem and Conjecture concerning
Real Quadratic Number Fields with 3-Class Group of Type (3,3):
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ASSUMPTIONS.
Let K = Q(d1/2) be a real quadratic field
with 3-class group Cl3(K) ∼ C(3)×C(3),
that is, with abelian type invariants (3,3).
Denote by N1,N2,N3,N4 the four unramified cyclic cubic extensions of K
within the Hilbert 3-class field F31(K) of K,
which exist according to the Artin reciprocity law of class field theory.
Let Ji : Cl3(K) → Cl3(Ni) be the natural extension homomorphisms of 3-classes, for 1 ≤ i ≤ 4.
Their kernels are called the capitulation kernels of K in N1,N2,N3,N4.
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CONJECTURE.
The situation with four total capitulation kernels,
ker(Ji) = Cl3(K) , for 1 ≤ i ≤ 4,
cannot occur for identical 3-class groups Cl3(Ni) ∼ C(3)×C(3), for 1 ≤ i ≤ 4.
(In terms of the Artin transfer pattern:
The structure (τ(K),κ(K)) ∼ ((11,11,11,11),(0000)) is forbidden.)
HINT.
To discourage unnecessary waste of time, we point out the following:
For each 1 ≤ i ≤ 4, let Li be one of the three isomorphic totally real cubic subfields of Ni.
Since we necessarily would have Cl3(Li) ∼ C(3), for 1 ≤ i ≤ 4,
the class number relation would not be violated by the problematic situation:
h3(Ni) = 9 = (1/9) × 9 × 32 = (Q/32) × h3(K) × h3(Li)2,
for 1 ≤ i ≤ 4,
where Q = 1 denotes the coinciding index of subfield units for all extensions N1,N2,N3,N4.
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REMARK.
The simplest situation with four total capitulation kernels,
which actually occurs, is the following:
a single distinguished 3-class group Cl3(N1) ∼ C(9)×C(9) and
Cl3(Ni) ∼ C(3)×C(3), for 2 ≤ i ≤ 4
(In terms of the Artin transfer pattern:
The admissible structure is (τ(K),κ(K)) ∼ ((22,11,11,11),(0000)).)
The situation occurs indeed, for instance, in the case d = 62501.
Since we have Cl3(L1) ∼ C(9) and Cl3(Li) ∼ C(3), for 2 ≤ i ≤ 4,
the class number relation for the first extension in the actual situation is:
h3(N1) = 81 = (1/9) × 9 × 92 = (Q/32) × h3(K) × h3(L1)2.
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Principal Investigator and
Project Leader of several
International Scientific Research Lines:
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