Imaginary Quadratic Field – How to Visualize Class Number

algebraic-number-theorynt.number-theory

Let me detail the title of the question. I'm trying to give students an intuition of what the class number is.

Let $K=\mathbb{Q}(\sqrt{-d})$, with $d>0$ a square-free integer, be a quadratic imaginary field. Let $\mathcal{O}_K$ be its ring of integers. It is of the form $\mathbb{Z}[\tau]$ with $\tau=\sqrt{-d}$ or $\tau=\frac{1+\sqrt{-d}}{2}$ depending on the value of $d$ mod $4$.

So let us think of $\mathcal{O}_K$ as the lattice of $\mathbb{C}$ generated by $1$ and $\tau$. Then ideals should correspond to sublattices of $\mathcal{O}_K$ and two of them should define the same class in the class group if one can pass from one to another by multiplying by an element of $\alpha$, isn't it?

Could anybody help me to make this analogy precise? For instance, how can one see that $\mathbb{Q}(i)$ has class number 1 but $\mathbb{Q}(\sqrt{-5})$ doesn't just by looking at the corresponding lattices? The (non-equivalent) decompositions

$2.3=(1+\sqrt{5}i)(1-\sqrt{5}i)$

suggest to consider the lattices $\mathbb{Z}\cdot 2+\mathbb{Z}\cdot(1+\sqrt{5}i)$ and $\mathbb{Z}\cdot 3+\mathbb{Z}\cdot(1-\sqrt{5}i).$ Is that what I have to do?

Thanks!

Best Answer

This is an interesting question that I've wondered about myself, so I can't really answer it properly but I'll make a couple elementary observations. First, for a lattice in ${\mathbb Z}[\tau]\subset{\mathbb C}$ to be an ideal just means that multiplication by $\tau$ takes the lattice to itself. For example, for the Gaussian integers this says the lattice is invariant under 90 degree rotation about the origin. It is easy to see that this means the lattice is a square lattice with a basis consisting of two of its shortest vectors. (This is really just a disguised version of the Euclidean algorithm for Gaussian integers.) Thus every ideal is principal in this case.

Back to the general case, another observation is that if two ideals are in the same ideal class, then the two lattices are related by an orientation-preserving similarity, that is, rotation and rescaling, which is what multiplying an ideal by an element of ${\mathbb Z}[\tau]$ does to a lattice. (It seems the converse should be true as well). For example in the case $\tau =\sqrt5i$ the principal ideal class consists of rectangular lattices similar to ${\mathbb Z}[\tau]$ itself, and the other ideal class (the class number is 2 here) consists of lattices similar to the lattice $(2,1+\sqrt5i)$ which is skewed rather than rectangular. It is enlightening to draw a picture to see how this lattice is invariant under multiplication by $\sqrt5i$.

Another thing that can be viewed geometrically is the correspondence between ideal classes and equivalence classes of binary quadratic forms of fixed discriminant. An ideal, viewed as a lattice, determines a quadratic form by restricting the usual norm (squared) $x^2 + y^2$ to the lattice, then renormalizing suitably to make equivalent ideals have equivalent quadratic forms. A textbook that explains this, to some extent at least, is Advanced Number Theory by Harvey Cohn.

It would be interesting to work out some more examples to see what the different similarity classes of lattice-ideals look like, especially in cases when the class number is larger than 2. Is the structure of the ideal class group somehow visible in how the similarity classes are related?

Related Solutions

Understanding the Class Number in Number Theory

This is from Buell, Binary Quadratic Forms. From page 84, the class number for a negative discriminant $\Delta$ is about $$\frac{\sqrt{|\Delta|}}{\pi},$$ which comes from an $L$-function calculation on page 83.

Let's see, on page 101, he points out that for negative field discriminants, class group and narrow class group are identical. Then on page 103, the group of classes of binary quadratic forms is isomorphic to the narrow class group. So that works out.

I don't know about surjectivity of class numbers. I imagine so. See OEIS

I wrote a little program up to 1000, here it is up to 111. The first number that achieves a given class number tends to be squarefree, an exception being h=104. EDIT: I've run this up to 4000 so far. To get a class number $h,$ there was always some $k$ with $k < 4 h^2.$ The largest $h$ where $h^2$ did not suffice was $h=677,$ with smallest $k = 601247.$ For $678 \leq h \leq 4000,$ there was always some $k \leq 0.751517... h^2,$ equality at $h=857, k=551951.$

    1       3 = 3
    2      15 = 3 * 5
    3      23 = 23
    4      39 = 3 * 13
    5      47 = 47
    6      87 = 3 * 29
    7      71 = 71
    8      95 = 5 * 19
    9     199 = 199
   10     119 = 7 * 17
   11     167 = 167
   12     231 = 3 * 7 * 11
   13     191 = 191
   14     215 = 5 * 43
   15     239 = 239
   16     399 = 3 * 7 * 19
   17     383 = 383
   18     335 = 5 * 67
   19     311 = 311
   20     455 = 5 * 7 * 13
   21     431 = 431
   22     591 = 3 * 197
   23     647 = 647
   24     695 = 5 * 139
   25     479 = 479
   26     551 = 19 * 29
   27     983 = 983
   28     831 = 3 * 277
   29     887 = 887
   30     671 = 11 * 61
   31     719 = 719
   32     791 = 7 * 113
   33     839 = 839
   34    1079 = 13 * 83
   35    1031 = 1031
   36     959 = 7 * 137
   37    1487 = 1487
   38    1199 = 11 * 109
   39    1439 = 1439
   40    1271 = 31 * 41
   41    1151 = 1151
   42    1959 = 3 * 653
   43    1847 = 1847
   44    1391 = 13 * 107
   45    1319 = 1319
   46    2615 = 5 * 523
   47    3023 = 3023
   48    1751 = 17 * 103
   49    1511 = 1511
   50    1799 = 7 * 257
   51    1559 = 1559
   52    1679 = 23 * 73
   53    2711 = 2711
   54    2759 = 31 * 89
   55    4463 = 4463
   56    1991 = 11 * 181
   57    2591 = 2591
   58    2231 = 23 * 97
   59    2399 = 2399
   60    2159 = 17 * 127
   61    3863 = 3863
   62    2471 = 7 * 353
   63    2351 = 2351
   64    2519 = 11 * 229
   65    3527 = 3527
   66    3431 = 47 * 73
   67    3719 = 3719
   68    2831 = 19 * 149
   69    3119 = 3119
   70    3239 = 41 * 79
   71    5471 = 5471
   72    3311 = 7 * 11 * 43
   73    2999 = 2999
   74    4151 = 7 * 593
   75    4703 = 4703
   76    3071 = 37 * 83
   77    6263 = 6263
   78    5111 = 19 * 269
   79    4391 = 4391
   80    5183 = 71 * 73
   81    3671 = 3671
   82    3839 = 11 * 349
   83    3911 = 3911
   84    4031 = 29 * 139
   85    4079 = 4079
   86    6767 = 67 * 101
   87    5279 = 5279
   88    4199 = 13 * 17 * 19
   89    6311 = 6311
   90    5951 = 11 * 541
   91    4679 = 4679
   92    4991 = 7 * 23 * 31
   93    5351 = 5351
   94    7367 = 53 * 139
   95    6959 = 6959
   96    6071 = 13 * 467
   97    5519 = 5519
   98    6191 = 41 * 151
   99    5591 = 5591
  100    7991 = 61 * 131
  101    5879 = 5879
  102    9383 = 11 * 853
  103   13799 = 13799
  104    9359 = 7^2 * 191
  105    6719 = 6719
  106    7631 = 13 * 587
  107    8231 = 8231
  108    5759 = 13 * 443
  109    5711 = 5711
  110    7751 = 23 * 337
  111   15359 = 15359

[Math] Why is $(\mathbb{Z}/3\mathbb{Z})^3$ not a class group of an imaginary quadratic number field

The only proof that I know that $(\mathbb{Z}/3\mathbb{Z})^3$ does not appear as the class group of a quadratic imaginary number field is by brute force search. Roughly, the idea is that since class numbers of quadratic fields increase with the discriminant, there are only finitely many such fields of any given class number. We can then enumerate all class groups of a given size, and check to see which class groups appear. Most recently, Watkins's computation of all quadratic imaginary number fields with class number up to 100 provides in particular a complete list of all class groups of order 27, and so by simple enumeration one can observer that $(\mathbb{Z}/3\mathbb{Z})^3$ does not appear. I should also mention in this regard the tremendous amount of data available in a recent article of Jacobson, Ramachandran, and Williams.

[Edit: As to another of your questions, the same technique shows 6 other class groups of order less than 100 do not appear as quadratic imaginary number fields, of orders 27,32,54,64,64,81,81 respectively, a fact I have stolen from https://math.stackexchange.com/a/11025/20762 (and now updated based on ABC's answer below).]

Lacking any particularly compelling arithmetic reasons which proscribe certain finite abelian groups from appearing as class groups of quadratic imaginary number fields, it's worth thinking about some heuristics. An argument of Soundararajan gives that, on average, the number of quadratic imaginary number fields of class number $p^n$ is roughly $p^n$ (or better, asymptotically between $p^n/n$ and $np^n$). On the other hand, the number of finite abelian groups of order $p^n$ is governed by the partition theory of $n$, and hence a standard asymptotic suggests there are $\frac{1}{4\sqrt{3}n}e^{\pi\sqrt{2n/3}}$ such groups. So roughly, there are increasingly more number fields of order $p^n$ than there are possible class groups of that size, so one might perhaps expect that (again, without any obvious arithemtical obstruction) that there are only finitely many examples like $(\mathbb{Z}/3\mathbb{Z})^3$ that do not appear, at least among $p$-groups (and especially for large $p$).

For real quadratic fields, there is no such bound pushing class numbers higher, and I can't see any reason not to expect every finite abelian group to appear as a class group of such a field. Of course, this result is nowhere near known.

Best Answer

Related Solutions

Understanding the Class Number in Number Theory

[Math] Why is $(\mathbb{Z}/3\mathbb{Z})^3$ not a class group of an imaginary quadratic number field

Related Question