CoCalc Shared Fileswww / 129-05 / challenges.html
Author: William A. Stein
Math 129: Topics in Number Theory

# 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.

 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  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