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In order took for units of $ \mathbb{Z} [ \sqrt[3]{3}] $ I am using a generalized Euclidean algorithm on three numbers. If $x \leq y \leq z$ then :

$$ (x,y,z) \to \text{ sort } ( x, y ,z -y ) $$

The three numbers I will use are $1, \sqrt[3]{3},\sqrt[3]{9}$. Then if $z-y$ is the biggest number, I wrote A, if it's in the middle I wrote B, and otherwise it was last and I wrote C.

Unfortunately this doesn't seem to terminate. the algorithmic starting with $1, \sqrt[3]{2},\sqrt[3]{4}$ repeats itself after 19 steps.

Does the sequence starting with $1, \sqrt[3]{3},\sqrt[3]{9}$ repeat itself eventually?


this is different from asking if the continued fraction of $\sqrt[3]{3}$ repeats as Hale Trotter and Serge Lang showed this is probably not the case

Data dump for $(1, \sqrt[3]{2},\sqrt[3]{4} )$... Brun observes that step 19 is proportional to step 1!

The first step reads $ \sqrt[3]{4} - \sqrt[3]{2} \leq \sqrt[3]{2} \leq \sqrt[3]{4} $ the inequalities get more interesting and shocking as you read down the list. One problem with Brun's algorithm is he doesn't give a condition when it terminates. You merely observe...

c (0, -1, 1) (1, 0, 0) (0, 1, 0)

c (-1, 1, 0) (0, -1, 1) (1, 0, 0)
a (-1, 1, 0) (0, -1, 1) (1, 1, -1)
a (-1, 1, 0) (0, -1, 1) (1, 2, -2)
c (1, 3, -3) (-1, 1, 0) (0, -1, 1)
b (1, 3, -3) (1, -2, 1) (-1, 1, 0)
a (1, 3, -3) (1, -2, 1) (-2, 3, -1)
a (1, 3, -3) (1, -2, 1) (-3, 5, -2)
b (1, 3, -3) (-4, 7, -3) (1, -2, 1)
c (5, -9, 4) (1, 3, -3) (-4, 7, -3)
a (5, -9, 4) (1, 3, -3) (-5, 4, 0)
a (5, -9, 4) (1, 3, -3) (-6, 1, 3)
c (-7, -2, 6) (5, -9, 4) (1, 3, -3)
b (-7, -2, 6) (-4, 12, -7) (5, -9, 4)
c (9, -21, 11) (-7, -2, 6) (-4, 12, -7)
c (3, 14, -13) (9, -21, 11) (-7, -2, 6)
c (-16, 19, -5) (3, 14, -13) (9, -21, 11)
c (6, -35, 24) (-16, 19, -5) (3, 14, -13)
b (6, -35, 24) (19, -5, -8) (-16, 19, -5)

c (-35, 24, 3) (6, -35, 24) (19, -5, -8)
a (-35, 24, 3) (6, -35, 24) (13, 30, -32)
a (-35, 24, 3) (6, -35, 24) (7, 65, -56)
c (1, 100, -80) (-35, 24, 3) (6, -35, 24)
b (1, 100, -80) (41, -59, 21) (-35, 24, 3)
a (1, 100, -80) (41, -59, 21) (-76, 83, -18)
a (1, 100, -80) (41, -59, 21) (-117, 142, -39)
b (1, 100, -80) (-158, 201, -60) (41, -59, 21)
c (199, -260, 81) (1, 100, -80) (-158, 201, -60)
a (199, -260, 81) (1, 100, -80) (-159, 101, 20)
a (199, -260, 81) (1, 100, -80) (-160, 1, 100)
c (-161, -99, 180) (199, -260, 81) (1, 100, -80)
b (-161, -99, 180) (-198, 360, -161) (199, -260, 81)
c (397, -620, 242) (-161, -99, 180) (-198, 360, -161)
c (-37, 459, -341) (397, -620, 242) (-161, -99, 180)
c (-558, 521, -62) (-37, 459, -341) (397, -620, 242)
c (434, -1079, 583) (-558, 521, -62) (-37, 459, -341)
b (434, -1079, 583) (521, -62, -279) (-558, 521, -62)

c (-1079, 583, 217) (434, -1079, 583) (521, -62, -279)
a (-1079, 583, 217) (434, -1079, 583) (87, 1017, -862)
a (-1079, 583, 217) (434, -1079, 583) (-347, 2096, -1445)
c (-781, 3175, -2028) (-1079, 583, 217) (434, -1079, 583)
b (-781, 3175, -2028) (1513, -1662, 366) (-1079, 583, 217)
a (-781, 3175, -2028) (1513, -1662, 366) (-2592, 2245, -149)
a (-781, 3175, -2028) (1513, -1662, 366) (-4105, 3907, -515)
b (-781, 3175, -2028) (-5618, 5569, -881) (1513, -1662, 366)
c (7131, -7231, 1247) (-781, 3175, -2028) (-5618, 5569, -881)
a (7131, -7231, 1247) (-781, 3175, -2028) (-4837, 2394, 1147)
a (7131, -7231, 1247) (-781, 3175, -2028) (-4056, -781, 3175)
c (-3275, -3956, 5203) (7131, -7231, 1247) (-781, 3175, -2028)
b (-3275, -3956, 5203) (-7912, 10406, -3275) (7131, -7231, 1247)
c (15043, -17637, 4522) (-3275, -3956, 5203) (-7912, 10406, -3275)
c (-4637, 14362, -8478) (15043, -17637, 4522) (-3275, -3956, 5203)
c (-18318, 13681, 681) (-4637, 14362, -8478) (15043, -17637, 4522)
c (19680, -31999, 13000) (-18318, 13681, 681) (-4637, 14362, -8478)
b (19680, -31999, 13000) (13681, 681, -9159) (-18318, 13681, 681)
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  • $\begingroup$ caac baab caac bccccb or possible I switched abc for the case $1, \sqrt[3]{2},\sqrt[3]{4}$ this was checked (by hand!) by Viggo Brun and I also did on my computer but I can't find it on my phone $\endgroup$
    – cactus314
    Aug 5, 2016 at 16:52
  • $\begingroup$ Sorry about that previous comment being largely unrelated to your actual question. Autopilot kicked in :-) And the beef was also included in Will's answer. $\endgroup$ Aug 5, 2016 at 19:14
  • $\begingroup$ @JyrkiLahtonen why do you only get two letters? Brun himself gets a period length of 18 $\endgroup$
    – cactus314
    Aug 5, 2016 at 20:32
  • $\begingroup$ I don't know. $x-z$ was never larger than $y$. $\endgroup$ Aug 5, 2016 at 20:34
  • $\begingroup$ @JyrkiLahtonen I have fixed my glaring typo $\endgroup$
    – cactus314
    Aug 5, 2016 at 20:36

3 Answers 3

4
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After some fiddling: I seem to have the triples that give $1$ as my matrix determinant as an infinite cyclic group. Each triple is $(a,b,c)$ such that the polynomial $a^3 + 3 b^3 + 9 c^3 - 9abc=1.$ Evidently it is correct to get these in an infinite cyclic group, the other "units" are just $(-a,-b,-c).$ Multiplication is what you expect, $$ (a,b,c)\cdot (d,e,f) = (ad+3bf+3ce, ae+bd+3ce, af+be+cd). $$

The generator is $(-2,0,1),$ or $9^{1/3} - 2.$ The generator that is larger than $1$ as a real number is then its group inverse, $4 + 3 \cdot 3^{1/3} + 2 \cdot 9^{1/3}.$

====================================================

  -7           15778480 10940187  7585502
  -6            1263601   876132   607476
  -5             101194    70164    48649
  -4               8104     5619     3896
  -3                649      450      312
  -2                 52       36       25
  -1                  4        3        2
   0                  1        0        0
   1                 -2        0        1
   2                  4        3       -4
   3                  1      -18       12
   4                -56       72      -23
   5                328     -213      -10
   6              -1295      396      348
   7               3778      252    -1991
   8              -6800    -6477     7760
   9              -5831    36234   -22320
  10             120364  -139428    38809
  11            -659012   395283    42746
  12            2503873  -662328  -744504
  13           -6994730  -908856  3992881
  14           11262892 13796355-14980492

==================================================

It appears that the abelian group of units of an order of a number field is finitely generated, this being Dirichlet's unit theorem. In practice, I think this means that finding a handful of units will allow describing all units in $\mathbb Z [ \alpha].$ Hmmmm. Keith says some things that I misinterpreted. Evidently the correct group is $\pm u^{\mathbb Z},$ where we can take $u > 1$ as a real number. And $9^{1/3} \approx 2.08,$ so we are okay in that detail.

Here are a few

               -56      72     -23
                -2       0       1
                 1     -18      12
                 1       0       0
                 4       3      -4
                 4       3       2
                52      36      25
               328    -213     -10



? m = a * id + b * x + c * x^2
%5 = 
[a b c]

[3*c a b]

[3*b 3*c a]

? p = matdet(m)
%6 = a^3 - 9*c*b*a + (3*b^3 + 9*c^3)
? 
? x
%7 = 
[0 1 0]

[0 0 1]

[3 0 0]
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  • $\begingroup$ can you discuss your print-outs a bit more now...? I am at my computer and maybe can retrieve some data -- or later tonight $\endgroup$
    – cactus314
    Aug 5, 2016 at 20:14
  • $\begingroup$ @cactus314, right, each triple is $(a,b,c)$ such that the polynomial $a^3 + 3 b^3 + 9 c^3 - 9abc=1.$ Evidently it is correct to get these in an infinite cyclic group, the other "units" are just $(-a,-b,-c).$ Multiplication is what you expect, $$ (a,b,c)\cdot (d,e,f) = (ad+3bf+3ce, ae+bd+3ce, af+be+cd). $$ $\endgroup$
    – Will Jagy
    Aug 5, 2016 at 20:19
  • $\begingroup$ lastly, how did you obtain this wonderful list? $\endgroup$
    – cactus314
    Aug 5, 2016 at 20:22
  • $\begingroup$ @cactus314 C++ with GMP, just put in a bound on $|a|, |b|, |c|.$ Got quicker after I realized we needed $a \equiv 1 \pmod 3.$ Then the cyclic group thing is very fast, once you pick a likely generator and find its inverse. $\endgroup$
    – Will Jagy
    Aug 5, 2016 at 20:29
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The theorem in question is the Dirichlet Unit Theorem, which says that if there are $r_1$ real embeddings of your field and $2r_2$ complex embeddings, that is, $r_2$ conjugate pairs of complex embeddings, then the group of units is a finite cyclic group plus a free group of rank $r=r_1+r_2-1$. For this field, $r_1=r_2=1$, and of course the group of roots of unity in the field is just $\{\pm1\}$. So the units are, as Will Jagy says, $\{\pm u^n\}$, for a well chosen “fundamental unit” $u$.

It would seem that except for the quadratic case, it’s easy to say how many units there are, but hard to find them. In this case, we’re lucky that $3^{2/3}-2$ is an obvious unit, and it certainly looks small. Seems to me that I’ve seen that it is indeed fundamental, but I don’t have the tools or the smarts to prove it.

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  • $\begingroup$ I was hoping some version of CF algorithm was in the books where cubic Dirichlet Unit Theorem looked just like quadratic. With internet I can find lots of sources written in confusing and arcane language. $\endgroup$
    – cactus314
    Aug 6, 2016 at 13:48
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The algorithm does find units but maybe does not repeat:

c 6 ( 0, -1,  1) ( 1, 0,  0) ( 0,   1,  0)
c 2 (-1,  1,  0) ( 0,-1,  1) ( 1,   0,  0)
c 4 ( 1,  1, -1) (-1, 1,  0) ( 0,  -1,  1)
c 4 ( 1, -2,  1) ( 1, 1, -1) (-1,   1,  0)
c 1 (-2,  0,  1) ( 1,-2,  1) ( 1,   1, -1)
b 1 (-2,  0,  1) ( 0, 3, -2) ( 1,  -2,  1)
c 4 ( 1, -5,  3) (-2, 0,  1) ( 0,   3, -2)
a 4 ( 1, -5,  3) (-2, 0,  1) ( 2,   3, -3)
c 1 ( 4,  3, -4) ( 1,-5,  3) (-2,   0,  1)
a 1 ( 4,  3, -4) ( 1,-5,  3) (-3,   5, -2)
b 1 ( 4,  3, -4) (-4, 10,-5) ( 1,  -5,  3)
b 1 ( 4,  3, -4) ( 5,-15, 8) (-4,  10, -5)
a 1 ( 4,  3, -4) ( 5,-15, 8) (-9,  25,-13)
a 1 ( 4,  3, -4) ( 5,-15, 8) (-14, 40,-21)
c 20 (-19, 55, -29) (4, 3, -4) (5, -15, 8)
c 1 ( 1, -18, 12) (-19, 55, -29) (4, 3, -4)
a 1 ( 1, -18, 12) (-19, 55, -29) (23, -52, 25)

I am reading that $-2 + \sqrt[3]{3^2}$ is a unit, so is $4 + 3\sqrt[3]{3} - 4\sqrt[3]{9}$ and $1 - 18 \sqrt[3]{3} + 12\sqrt[3]{9}$... so we are finding them even if the algorithm might not repeat itself.


Conversely, I can observe the limit of any sequence if it does repeat involves the solution to a cubic, since our word ccccc bcaca bbaac ccaaa ... can be converted into the product of matrices:

$$ \left[ \begin{array}{rrr} 1 & & \\ & 1 & \\ & -1 & 1 \\ \end{array} \right] \text{ and } \left[ \begin{array}{rrr} & & 1 \\ 1 & & \\ & 1 & \\ \end{array} \right] \text{ or } \left[ \begin{array}{rrr} & 1 & \\ & & 1 \\ 1 & & \\ \end{array} \right] \text{ or } \left[ \begin{array}{rrr} 1 & & \\ & 1 & \\ & & 1 \\ \end{array} \right] $$ leaving in the identity matrix (leave everything the same) for good measure. If our sequence repeats itself our limiting proportion will be eigenvectors of some $3 \times 3 $ matrix, effectively solving that cubic.

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