[math-fun] Unique factorization (connection to music?) question
Victor Miller victorsmiller at gmail.com According to SAGE the field Q(2^{1/12}) has class number 1, and so does have unique factorization. In addition Z[2^{1/12}] is the maximal order in this field.
--so: for which N does Q(2^{1/N}) have class number 1? http://oeis.org/A005848 = {1, 3, 4, 5, 7, 8, 9, 11, 12, 13, 15, 16, 17, 19, 20, 21, 24, 25, 27, 28, 32, 33, 35, 36, 40, 44, 45, 48, 60, 84} claims to answer the same question for Q(Nth root of unity) duplicating the Main Theorem of J. Myron Masley and Hugh L. Montgomery: Cyclotomic fields with unique factorization, J. Reine Angew. Math. 286/287 (1976) 248-256 which shows that all such N obey N<=84 and gives the full list. Amazingly enough, all the historically-used music-scale sizes {4,5,6,7,8,12} happen to be on this list (asterisk: M&M explain that if k is odd and on their list, then 2k automatically is also, so therefore they omit mentioning all 2k with k odd). Which actually is not so amazing since ALL numbers 1-12 are on the M&M list.
According to SAGE the class number of Q(2^(1/n)) is 1 for n=2,...,18. For 19 it's still computing with a message: warning Zimmert's bound is large (1259129) certification will take a long time. Victor On Wed, Dec 3, 2014 at 2:47 PM, Warren D Smith <warren.wds@gmail.com> wrote:
Victor Miller victorsmiller at gmail.com According to SAGE the field Q(2^{1/12}) has class number 1, and so does have unique factorization. In addition Z[2^{1/12}] is the maximal order in this field.
--so: for which N does Q(2^{1/N}) have class number 1?
http://oeis.org/A005848 = {1, 3, 4, 5, 7, 8, 9, 11, 12, 13, 15, 16, 17, 19, 20, 21, 24, 25, 27, 28, 32, 33, 35, 36, 40, 44, 45, 48, 60, 84} claims to answer the same question for Q(Nth root of unity) duplicating the Main Theorem of J. Myron Masley and Hugh L. Montgomery: Cyclotomic fields with unique factorization, J. Reine Angew. Math. 286/287 (1976) 248-256 which shows that all such N obey N<=84 and gives the full list. Amazingly enough, all the historically-used music-scale sizes {4,5,6,7,8,12} happen to be on this list (asterisk: M&M explain that if k is odd and on their list, then 2k automatically is also, so therefore they omit mentioning all 2k with k odd). Which actually is not so amazing since ALL numbers 1-12 are on the M&M list.
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I've heard a rumor, some years ago, that N up to 30 had been checked, with all 2^1/N being UFDs. I verified twenty years ago that 2^1/5 makes a Euclidean number field, but I don't know of anyone pushing on to 2^1/6. Note for the uninitiated: When number theorists use the phrase "Number Field", they mean a ring, not a field. The number field generated by sqrt2 includes the algebraic integers A+Bsqrt2 with A & B ordinary integers, but does not include 1/2, or 1/3, or 1/sqrt2. This terminological nuance (to use a polite phrase) helps us separate out our special clique. Rich -----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Victor Miller Sent: Wednesday, December 03, 2014 1:25 PM To: math-fun Subject: [EXTERNAL] Re: [math-fun] Unique factorization (connection to music?) question According to SAGE the class number of Q(2^(1/n)) is 1 for n=2,...,18. For 19 it's still computing with a message: warning Zimmert's bound is large (1259129) certification will take a long time. Victor On Wed, Dec 3, 2014 at 2:47 PM, Warren D Smith <warren.wds@gmail.com> wrote:
Victor Miller victorsmiller at gmail.com According to SAGE the field Q(2^{1/12}) has class number 1, and so does have unique factorization. In addition Z[2^{1/12}] is the maximal order in this field.
--so: for which N does Q(2^{1/N}) have class number 1?
http://oeis.org/A005848 = {1, 3, 4, 5, 7, 8, 9, 11, 12, 13, 15, 16, 17, 19, 20, 21, 24, 25, 27, 28, 32, 33, 35, 36, 40, 44, 45, 48, 60, 84} claims to answer the same question for Q(Nth root of unity) duplicating the Main Theorem of J. Myron Masley and Hugh L. Montgomery: Cyclotomic fields with unique factorization, J. Reine Angew. Math. 286/287 (1976) 248-256 which shows that all such N obey N<=84 and gives the full list. Amazingly enough, all the historically-used music-scale sizes {4,5,6,7,8,12} happen to be on this list (asterisk: M&M explain that if k is odd and on their list, then 2k automatically is also, so therefore they omit mentioning all 2k with k odd). Which actually is not so amazing since ALL numbers 1-12 are on the M&M list.
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Rich Schroeppel wrote:
Note for the uninitiated: When number theorists use the phrase "Number Field", they mean a ring, not a field. The number field generated by sqrt2 includes the algebraic integers A+Bsqrt2 with A & B ordinary integers, but does not include 1/2, or 1/3, or 1/sqrt2. This terminological nuance (to use a polite phrase) helps us separate out our special clique.
I wonder whether this nuance is characteristic of some sub-clique rather than of number theorists as a whole. Both the algebraic number theory textbooks I just checked (Lang and Frohlich&Taylor) define a "number field" to mean a finite extension of the rationals. So (for what little it's worth) do Wikipedia and MathWorld. Everything on the first page of hits for "number field" on MathOverflow uses the term to refer to the field rather than its ring of integers. So, in fact, does everything else I can easily lay my hands on. (Some of them prefer the term "algebraic number field", and some of those explicitly say something like "algebraic number field, or number field for short".) I don't have much in the way of actual scholarly articles written by actual number theorists to hand, but I tried the experiment of looking at recent articles in the Journal of Number Theory; all I can see is their abstracts, but they too seem to be unanimous in using the term to mean the field rather than its ring of integers. -- g
This is not quite the case in my experience. What I've seen is that a number field is defined as any finite extension field of Q in C. Rather, when people talk about factorization, primes, etc. with respect to a number field, they are talking about those things in the number field's ring of algebraic integers (those elements of that number field that are roots of some integer polynomial that's monic. --Dan
On Dec 3, 2014, at 2:58 PM, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
Note for the uninitiated: When number theorists use the phrase "Number Field", they mean a ring, not a field. The number field generated by sqrt2 includes the algebraic integers A+Bsqrt2 with A & B ordinary integers, but does not include 1/2, or 1/3, or 1/sqrt2. This terminological nuance (to use a polite phrase) helps us separate out our special clique.
Hi, Rich. Thanks for the thank you (I think), but was surprised you didn't mention this. --Dan
On Dec 3, 2014, at 3:26 PM, Dan Asimov <dasimov@earthlink.net> wrote:
This is not quite the case in my experience. What I've seen is that a number field is defined as any finite extension field of Q in C.
Rather, when people talk about factorization, primes, etc. with respect to a number field, they are talking about those things in the number field's ring of algebraic integers (those elements of that number field that are roots of some integer polynomial that's monic.
--Dan
Note for the uninitiated: When number theorists use the phrase "Number Field", they mean a ring, not a field. The number field generated by sqrt2 includes the algebraic integers A+Bsqrt2 with A & B ordinary integers, but does not include 1/2, or 1/3, or 1/sqrt2. This terminological nuance (to use a polite phrase) helps us separate out our special clique.
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I apologize if I'm posting the same comment that I already posted a couple of months ago. (I can't recall if I actually posted it or just intended to do so.) But: I don't think number theorists mean a non-field ring when they say "number field". Rather, they mean an actual field K that is a finite extension of the rationals Q (an "algebraic" extension of Q). But then attention is often directed to the ring O(K) of algebraic integers of K. O(K) is defined as all members of K that are roots of monic polynomials with coefficients in Z. It is not immediately obvious that for all algebraic extensions K of Q, the set O(K) is always closed under addition and multiplication -- i.e., is actually a ring -- but it is. --Dan
Note for the uninitiated: When number theorists use the phrase "Number Field", they mean a ring, not a field. The number field generated by sqrt2 includes the algebraic integers A+Bsqrt2 with A & B ordinary integers, but does not include 1/2, or 1/3, or 1/sqrt2. This terminological nuance (to use a polite phrase) helps us separate out our special clique.
As far as I know "number field" refers to the field. When you talk about class numbers these are of "orders" in the number field -- finitely generated sub-algebras over Z (not Q). Rich's example with Q(sqrt(2)) is the maximal order. Victor On Wed, Dec 3, 2014 at 3:46 PM, Schroeppel, Richard <rschroe@sandia.gov> wrote:
I've heard a rumor, some years ago, that N up to 30 had been checked, with all 2^1/N being UFDs. I verified twenty years ago that 2^1/5 makes a Euclidean number field, but I don't know of anyone pushing on to 2^1/6.
Note for the uninitiated: When number theorists use the phrase "Number Field", they mean a ring, not a field. The number field generated by sqrt2 includes the algebraic integers A+Bsqrt2 with A & B ordinary integers, but does not include 1/2, or 1/3, or 1/sqrt2. This terminological nuance (to use a polite phrase) helps us separate out our special clique.
Rich
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Victor Miller Sent: Wednesday, December 03, 2014 1:25 PM To: math-fun Subject: [EXTERNAL] Re: [math-fun] Unique factorization (connection to music?) question
According to SAGE the class number of Q(2^(1/n)) is 1 for n=2,...,18. For 19 it's still computing with a message: warning Zimmert's bound is large (1259129) certification will take a long time.
Victor
On Wed, Dec 3, 2014 at 2:47 PM, Warren D Smith <warren.wds@gmail.com> wrote:
Victor Miller victorsmiller at gmail.com According to SAGE the field Q(2^{1/12}) has class number 1, and so does have unique factorization. In addition Z[2^{1/12}] is the maximal order in this field.
--so: for which N does Q(2^{1/N}) have class number 1?
http://oeis.org/A005848 = {1, 3, 4, 5, 7, 8, 9, 11, 12, 13, 15, 16, 17, 19, 20, 21, 24, 25, 27, 28, 32, 33, 35, 36, 40, 44, 45, 48, 60, 84} claims to answer the same question for Q(Nth root of unity) duplicating the Main Theorem of J. Myron Masley and Hugh L. Montgomery: Cyclotomic fields with unique factorization, J. Reine Angew. Math. 286/287 (1976) 248-256 which shows that all such N obey N<=84 and gives the full list. Amazingly enough, all the historically-used music-scale sizes {4,5,6,7,8,12} happen to be on this list (asterisk: M&M explain that if k is odd and on their list, then 2k automatically is also, so therefore they omit mentioning all 2k with k odd). Which actually is not so amazing since ALL numbers 1-12 are on the M&M list.
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Apparently my clique is smaller than I realized. Thank you for the correction! --Rich ------- Quoting Victor Miller <victorsmiller@gmail.com>:
As far as I know "number field" refers to the field. When you talk about class numbers these are of "orders" in the number field -- finitely generated sub-algebras over Z (not Q). Rich's example with Q(sqrt(2)) is the maximal order.
Victor
On Wed, Dec 3, 2014 at 3:46 PM, Schroeppel, Richard <rschroe@sandia.gov> wrote:
I've heard a rumor, some years ago, that N up to 30 had been checked, with all 2^1/N being UFDs. I verified twenty years ago that 2^1/5 makes a Euclidean number field, but I don't know of anyone pushing on to 2^1/6.
Note for the uninitiated: When number theorists use the phrase "Number Field", they mean a ring, not a field. The number field generated by sqrt2 includes the algebraic integers A+Bsqrt2 with A & B ordinary integers, but does not include 1/2, or 1/3, or 1/sqrt2. This terminological nuance (to use a polite phrase) helps us separate out our special clique.
Rich
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Victor Miller Sent: Wednesday, December 03, 2014 1:25 PM To: math-fun Subject: [EXTERNAL] Re: [math-fun] Unique factorization (connection to music?) question
According to SAGE the class number of Q(2^(1/n)) is 1 for n=2,...,18. For 19 it's still computing with a message: warning Zimmert's bound is large (1259129) certification will take a long time.
Victor
On Wed, Dec 3, 2014 at 2:47 PM, Warren D Smith <warren.wds@gmail.com> wrote:
Victor Miller victorsmiller at gmail.com According to SAGE the field Q(2^{1/12}) has class number 1, and so does have unique factorization. In addition Z[2^{1/12}] is the maximal order in this field.
--so: for which N does Q(2^{1/N}) have class number 1?
http://oeis.org/A005848 = {1, 3, 4, 5, 7, 8, 9, 11, 12, 13, 15, 16, 17, 19, 20, 21, 24, 25, 27, 28, 32, 33, 35, 36, 40, 44, 45, 48, 60, 84} claims to answer the same question for Q(Nth root of unity) duplicating the Main Theorem of J. Myron Masley and Hugh L. Montgomery: Cyclotomic fields with unique factorization, J. Reine Angew. Math. 286/287 (1976) 248-256 which shows that all such N obey N<=84 and gives the full list. Amazingly enough, all the historically-used music-scale sizes {4,5,6,7,8,12} happen to be on this list (asterisk: M&M explain that if k is odd and on their list, then 2k automatically is also, so therefore they omit mentioning all 2k with k odd). Which actually is not so amazing since ALL numbers 1-12 are on the M&M list.
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participants (6)
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Dan Asimov -
Gareth McCaughan -
rcs＠xmission.com -
Schroeppel, Richard -
Victor Miller -
Warren D Smith