Math 129: Topics in Number Theory

Challenge Questions

These are problems that occured to me while preparing this course, but to which I don't know the answer. If you know the answer to any of these, please tell me.

  1. From: [email protected]
    To: [email protected]
    Date: 2005-04-05 11:23 am
    
    hello, sir.
    i'm a student of other university.
    now, i study the basic algebraic number theory.
    (and i see your lecture note)
    
    anyway, one question...
    i know that some biquadratic number field has a ring of integer.
    
    for example, ring of integer of Q( sqrt(5), sqrt(7) ) be Z( (1+sqrt(5))/2, 
    sqrt(7) ).
    but i don't know a primitive element of ring of integer.
    Z( (1+sqrt(5))/2, sqrt(7) ) =Z(alpha)    ,       alpha = ?
    
    help....
    
    and i'm sorry...
    
    i'm Korean...and i wrote in poor English.
    
    --------
    My Response:
    
    
    I think you are asking for a generator for the ring of integers of
    Q(sqrt(5),sqrt(7)).  Some rings of integers have a generator and some
    *don't*.  I don't know whether, in this particular case, yours has a
    generator or not, and the obvious checks don't tell me.
    -------
    
    CHALLENGE: Answer his question.
    
    Solution By Dustin Clausen on April 6, 2005:
    From: "Dustin Clausen" 
    Date: 2005-04-06 10:02 pm
    
    Hello Professor Stein,
    
    I see you've got a new challenge problem up!  Magma can solve this one
    for you using the IndexFormEquation command.  Define K to be
    Q(sqrt(5),sqrt(7)) (somehow.. I couldn't figure out how to get Magma
    to treat this as an absolute extension without explicitly calculating
    the minimal polynomial of sqrt(5)+sqrt(7), which I did in PARI).  Then
    just use
    
    O:=MaximalOrder(K);
    IndexFormEquation(O,1);
    
    and it should return an empty list, confirming that there is no a in O
    such that O=Z[a].  Furthermore, if you contine testing various
    IndexFormEquation(O,n), you'll see that there are a,b with [O:Z[a]]=12
    and [O:Z[b]]=23, so O has no essential disciminant divisors.  I wonder
    if there is some other nice obstruction to rings of integers being
    generated by one element?  I tried to follow the link on the Magma
    site to the paper that discusses the algorithm used for
    IndexFormEquation, but I can't read German very well.
    
    Regards,
    Dustin
    
    ----
    From: Claus Fieker 
    To: William Stein 
    CC: [email protected]
    Date: 2005-04-14 09:32 pm
    
    Willam (and Dustin),
    
    While I am unfortunately by no means an expert on index form equations I
    can adress some of the problems easily:
    
    To set the field up:
    
    x := Polynomial([0,1]);
    K1 := NumberField([x^2-5, x^2-7]);
    K2 := SimpleExtension(K1);
    
    now K2 is a absolute simple extension.
    
    And indeed:
    
    > IndexFormEquation(MaximalOrder(K2), 1);
    []
    > 
    
    as you expected. In general however, I think that the implementation
    that Klaus Wildanger did (his PhD Thesis is the German reference)
    requires you to input a maximal equation order (a solution to
    IndexFormEquation(?, 1);) in order to find all other solutions (up to
    equivalence). For degree 4 problems the algorithm is different and is
    based on simultanious Thue-Equations (papers by Gaal and Pohst):
    
    GPP93
        Istvan GaŠl, Attila PethŲ, and Michael E. Pohst.
        On the resolution of index form equations in quartic number fields.
        J. Symbolic Comp., 16:563--584, 1993.
    
    GPP96
        Istvan GaŠl, Attila PethŲ, and Michael E. Pohst.
        Simultaneous representation of integers by a pair of ternary quadratic forms -- With an application to index form equations in quartic number fields.
        J. Number Th., 57:90--104, 1996.
    
    which are luckily in English...
    
    Greetings
     Claus
    
  2. We gave examples in class to show that the ring of integers of a number field need not be generated by one element. Is the ring of integers always generated by at most two elements?
    Solution By Dustin Clausen on March 27, 2005:
    If O_K is generated by n elements, it's a quotient of Z[X_1,...,X_n].
    If 2 splits completely in K, then there are d = deg K quotients of O_K
    isomorphic to F_2, each one giving rise to a distinct homomorphism
    Z[X_1,...,X_n] --> F_2, of which there are clearly only 2^n (one
    choice for each generator).
    
    So it all rests on finding, for any n, a number field F of degree n in
    which 2 splits competely.  I claim that if p is large enough and p
    splits completely in Q(a) where a is a root of x^n-2, then p==1 (mod
    n) and we can take F to be the unique subfield of Q(\zeta_p) of degree
    n.  By Chebatorev's Density Theorem, there will be infinitely many
    such p to choose from, so we're golden.
    
    Proof: By "large enough" I mean it doesn't divide the index [O_{Q(a)} : 
    Z[a]].  In such a case we can apply Dedekind's Theorem, so p splits 
    completely in Q(a) if and only if x^n-2 splits into linear factors (mod p).  
    The quotient of any two disctinct roots gives an n-th root of 1 mod p, hence 
    there are n n-th roots of 1 (mod p), hence p==1 (mod n).  Also, we deduce 
    2^((p-1)/n) == 1 (mod p) from Fermat's Little Theorem.  Hence the frobenius 
    at 2 in F is trivial, so 2 splits completely in F, QED.
    

    Added by William Stein:
    I just looked for the first explicit example of a field K such that O_K cannot 
    be generated as a ring by two elements using the strategy explained above.
    
    If we can find a number field K of degree 5 in which 2 splits completely,
    then O_K cannot be generated by 2 elements.  This is because, by the CRT, 
       O_K/(2) = F_2 + F_2 + F_2 + F_2 + F_2  (direct sum)
    which is not a quotient of F_2[x,y], as there are only 4 homomorphisms
    from F_2[x,y] to F_2.  
    
    So, take n=5 in Clausen's construction, and consider L=Q(2^(1/5)).  
    The smallest prime p that splits completely is p=151.  Then Clausen
    argues that if we let K be the unique degree 5 extension of Q contained in 
    Q(zeta_{151}), then 2 splits completely in K.  Indeed, we have 
       2^((151-1)/5) == 1 (mod 151),
    so Frob_2 is trivial in K.   Thus the degree 5 subfield of Q(zeta_{151}) is an 
    example of a field whose integer ring is not generated by 2 elements.  
    
    This is the field defined by a root of
      f = x^5+x^4-60*x^3-12*x^2+784*x+128
    which is the only totally real field unramified outside 151 at
    
    http://math.la.asu.edu/~jj/numberfields/
    
    Indeed, checking with PARI (or whatever) we see that 2 does split completely 
    in the above field.  
    
    William