Math 129: Topics in Number Theory

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 Ga�l, Attila Peth�, and Michael E. Pohst.
    On the resolution of index form equations in quartic number fields.
    J. Symbolic Comp., 16:563--584, 1993.

GPP96
    Istvan Ga�l, Attila Peth�, 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

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.

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