
The inspiration.
During the start and the completion of the halfunit project,
I was definitely feeling much internal inspiration and motivation
by the spirit of my honourable academic teacher
Alexander Aigner (May 18, 1909  June 7, 1988).
Here I express my respect and gratitude to him
for helping me to succeed in this enterprise.



*

Preliminaries.
It is well known that the maximal order O_{1} of a quadratic number field K = Q(d^{1/2})
contains halfintegers of the shape (x + y*d^{1/2})/2 with coprime rational integers x,y in Z,
if and only if the discriminant d of K is congruent 1 modulo 4, d = 4*n + 1 (n in Z).
In the other cases, d = 0(mod 4) with d/4 = 2(mod 4) or d/4 = 3(mod 4),
the ring O_{1} of algebraic integers of K consists only of elements of the shape x + y*d^{1/2} with x,y in Z.
The proof does not use more than
High School Algebra.

*

The start in 1988.
However, it is not well known which conditions must be satisfied in the case d = 1(mod 4) of a real quadratic field K
in order that the fundamental unit e of K is a halfinteger, i. e., a so called halfunit
[1, § 49, p. 174, ff.].
A first general restriction is provided by the prime residue class group U(O_{1}/F) of the conductor F = 2*O_{1}
of the suborder O_{2} = Z*1 + Z*d^{1/2} in the maximal order O_{1} = Z*1 + Z*(1 + d^{1/2})/2,
with unit groups U_{2} and U_{1}, respectively ([2, p. 42, Proposition 4.1]):
for d = 1(mod 8), 2 splits in K, U(O_{1}/F) = C_{1}, and (U_{1}:U_{2}) = 1, whereas
for d = 5(mod 8), 2 remains inert in K, U(O_{1}/F) = C_{3}, and (U_{1}:U_{2}) = 1 or 3.
Thus, there are certainly no halfunits, if d = 1(mod 8),
but there may exist halfunits, i. e., (U_{1}:U_{2}) = 3, if d = 5(mod 8).


*

A necessary and sufficient condition for the existence of halfunits in the case d = 5(mod 8)
is provided by the formula for the ring class number modulo 2 of K:
[F_{2}K] = 3 * [F_{1}K] / (U_{1}:U_{2}),
which yields the following
Theorem 1. (for the proof see [2, p. 50, Theorem 4.4])
The 3ring class field F_{2} modulo 2 of K
is a cyclic cubic extension of the Hilbert 3class field F_{1} of K
if and only if there are no halfunits in K:
[F_{2}F_{1}] = 3 < == > (U_{1}:U_{2}) = 1

A drawback of this theorem is that testing the class field theoretic condition
is even harder than directly testing the index of the unit groups by means of continued fractions.
However, for real quadratic fields with class number coprime to 3, i. e., with 3class rank zero,
the above Theorem 1 changes to the following convenient criterion
Theorem 2. (for the proof see [2, p. 50, Theorem 4.4])
There are no halfunits in K if and only if
there exists an absolute (nonGalois) cubic extension LQ with discriminant 4*d.


*

The completion in 2008.
Chapter 4 of the tract [2] had a considerable influence on the development of my theory of
discriminantal multiplicities of pring class fields over quadratic fields in my paper [3].
This theory reached its highlights by the introduction of the concepts of pring spaces and pdefects
in my presentation
Wiener Kongress 2001
.
Now it turned out that the impact of pring spaces V_{p}(f) mod f is not limited to counting dihedral fields of degree 2p
but can also be specialized to the problem of the existence of halfunits by taking p = 3 and f = 2.
Since f = 2 is a prime conductor, the 3ring space V_{3}(2) mod 2 is
at most of codimension 1 in the F_{3}vector space V_{3} of nontrivial generators of ideal cubes,
i. e., the 3defect of 2 is bounded by 0 <= d_{3}(2) <= 1.
The following Theorem 3 extends Theorem 1 by distinguishing three cases
in terms of 3deficient elements mod 2 of V_{3}.
Theorem 3. (for the proof see [4])
Let K be a real quadratic field with discriminant d = 5(mod 8) and 3class rank r.
Denote the 3ring class field modulo 2 of K by F_{2}
and the Hilbert 3class field of K by F_{1}.
Further let e be the fundamental unit of K
and a_{1},...,a_{r} the other generating ideal cubes of the
(r+1)dimensional F_{3}vector space V_{3} = < e,a_{1},...,a_{r} >.
1. (Case with a 3deficient unit mod 2)
There exist halfunits in K if and only if
F_{2} coincides with F_{1}.
(Then e in O_{1} but a_{1},...,a_{r} in O_{2}.)
2. Halfunits do not exist in K if and only if
a) (Case with a 3deficient nonunit mod 2)
either
F_{2} is a nonsplit extension of F_{1} over K
(Then e in O_{2} and exactly one of a_{1},...,a_{r} in O_{1}
and there exist 3^{r1} cyclic nonic extensions N~K with conductor 2.)
b) (Case without 3deficient ideal cubes mod 2)
or
F_{2} is a split extension of F_{1} over K.
(Then e,a_{1},...,a_{r} in O_{2} and there exist
3^{r} nonisomorphic cubic extensions LQ with discriminant 4*d.)


*

Now we illustrate Theorem 3
by a table which shows the minimal discriminants of all known configurations,
ordered by increasing 3class rank of the real quadratic fields K.
For 3class rank zero, the nonsplit extension case cannot occur.
The letter w denotes the generating surd d^{1/2} of K.
3class rank  Gal(F_{1}K)  Discriminant  Fundamental unit  Gal(F_{2}K)  Generators of ideal cubes  3ring space mod 2  3defect of 2 
0  1  5  e = (1 + w)/2 in O_{1}  1    O  1 
0  1  37  e = 6 + w in O_{2}  (3)    [e]  0 
1  (3)  229  e = (15 + w)/2 in O_{1}  (3)  a = 2391 + 158*w in O_{2}  [a]  1 
1  (3)  1765  e = 42 + w in O_{2}  (9)  a = (303 + 7*w)/2 in O_{1}  [e]  1 
1  (3)  7053  e = 4703 + 56*w in O_{2}  (3,3)  a = 86 + w in O_{2}  [e,a]  0 
2  (3,3)  130397  e = (6861 + 19*w)/2 in O_{1}  (3,3)  a = 8521509330 + 23598409*w in O_{2} b = 42513400012 + 117731327*w in O_{2}  [a,b]  1 
2  (3,3)  62501  e = 250 + w in O_{2}  (9,3)  a = 8110567 + 32442*w in O_{2} b = (251 + w)/2 in O_{1}  [e,a]  1 
2  (3,3)  4562765  e = 123715750188465719 + 57917651345508*w in O_{2}  (3,3,3)  a = 7259 + 10*w in O_{2} b = 576737 + 270*w in O_{2}  [e,a,b]  0 
3  (9,3,3)  925219837  e in O_{2}  (9,3,3,3)  a in O_{2}, b in O_{2}, c in O_{2}  [e,a,b,c]  0 
Remark:
In the cases with 3defect d_{3}(2) = 1,
the first generating ideal cube a = (x + y*w)/2 encountered in
solving the corresponding Diophantine norm form equation x^{2}  d*y^{2} = 4*q^{3}
is not always an element of O_{2} but frequently of O_{1}.
Then e*a, e^{2}*a, b*a or b^{2}*a is in general an element of O_{2}.
Examples:
For d = 229, we have a = (27 + w)/2, e*a = (317 + 21*w)/2, e^{2}*a = 2391 + 158*w.
For d = 130397, we have a = (363 + w)/2, e*a = (2484043 + 6879*w)/2, e^{2}*a = 8521509330 + 23598409*w
and b = (1807 + 5*w)/2, e*b = (12392771 + 34319*w)/2, e^{2}*b = 42513400012 + 117731327*w.
For d = 62501, we have a = (267 + w)/2, b*a = (64759 + 259*w)/2, b^{2}*a = 8110567 + 32442*w.

*

Finally we conclude with Class Field Diagrams for the minimal discriminants of all known configurations.
    K = Q(5^{1/2}) = F_{1} = F_{2}     
         
    Q     


    N_{1} = F_{2}     
         
    K = Q(37^{1/2}) = F_{1}     
         
    Q     

    N^_{1} = F_{1} = F_{2}     
         
    K = Q(229^{1/2})     
         
    Q     

    N~_{1} = F_{2}     
         
    N^_{1} = F_{1}     
         
    K = Q(1765^{1/2})     
         
    Q     

    F_{2}     
 /   /   \   \  
N^_{1} = F_{1}   N_{1}     N_{2}   N_{3} 
 \   \   /   /  
    K = Q(7053^{1/2})     
         
    Q     

    F_{1} = F_{2}     
 /   /   \   \  
N^_{1}   N^_{2}     N^_{3}   N^_{4} 
 \   \   /   /  
    K = Q(130397^{1/2})     
         
    Q     

        F_{2}     
     /   /   \   \  
    F_{1}   N~_{1}     N~_{2}   N~_{3} 
 /   /   \  \  \   /   /  
N^_{1}   N^_{2}     N^_{3}   N^_{4}     
 \   \   /   /      
    K = Q(62501^{1/2})         
             
    Q         

      F_{2}       
/              \ 
F_{1}       ...       
\  \  \  \  /  /               
N^_{1}  N^_{2}  N^_{3}  N^_{4}  N_{1}  N_{2}  N_{3}  N_{4}  N_{5}  N_{6}  N_{7}  N_{8}  N_{9} 
\  \  \  \  \  \    /  /  /  /  /  / 
      K = Q(4562765^{1/2})       
             
      Q       
Final Note:
Some of these minimal discriminants occurred in other context already.
First, d = 62501 as the smallest example of
family principal factorization type (Alpha_{1},Alpha_{1},Alpha_{1},Alpha_{1})
.
Second, d = 4*4562765 = 18251060 and d = 4*925219837 = 3700879348 as the minimal discriminants with
multiplicity 9 and 27
.


