Is the class number of $K$ the number of factorizations an element of $\mathcal{O}_K$ can have?

Question

Consider the number field $K = \mathbf{Q}(\sqrt{-5})$, which has ring of integers $\mathcal{O}_K = \mathbf{Z}[\sqrt{-5}]$. It is known that the class number of $K$ is $2$. It is also true that you can write the element $6$ as a product of irreducible elements in two different ways:

$$ 6 = (1+\sqrt{-5})(1-\sqrt{-5}) = (2)(3).$$

I am noticing that the class number is $2$ and the number of different ways to factor $6$ is also $2$; is this a coincidence, or is this true in general?

That is, if a number field $K$ has class number equal to $k$, can we always find an element $x \in \mathcal{O}_K$ that can be factored into irreducibles in $k$ different ways? Is the class number of a number field $K$ equal to the maximum number of ways an element in $\mathcal{O}_K$ can be factored into irreducibles?

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(Here, I am considering two factorizations to be the same if they are equal up to units; e.g: $6 = 2\cdot 3$ and $6 = (-2)*(-3)$ are the "same" factorization in $\mathbf{Z}$.)

Answer 1

If the class group is not 1 then there are infinitely many non-principal prime ideals, from which we get a sequence of distinct prime ideals $P_j$ all equivalent in the class group. Let $n$ be the order of $P_1$, then for all $m$ and $l\in n\ldots nm$, $$\underbrace{\prod_{j=1}^{mn} P_j}_{(a_m)} = \underbrace{(P_l \prod_{j=1}^{n-1} P_j)}_{(b_{m,l})}\underbrace{(\prod_{j=n,j\ne l}^{nm} P_j)}_{(c_{m,l})}$$ Then $b_{m,l}$ is irreducible and it doesn't divide any of the $c_{m,i}$, so you get at least $n(m-1)+1$ factorizations of $a_m$.

Answer 2

For class number 2 there is an interpretation, found by Carlitz (A characterization of algebraic number fields with class number two, Proc. Amer. Math. Soc. 11 (1960), 391-392) here: a number field $K$ has class number 2 if and only if (i) $\mathcal O_K$ is not a UFD and (ii) all irreducible factorizations of an element have the same number of irreducible factors. For example, since $\mathbf Q(\sqrt{-5})$ has class number 2, all irreducible factorizations of 6 will involve 2 factors because you found one such factorization with 2 factors (don't confuse the two roles of 2 there, which is what led to your question). But $\mathbf Q(\sqrt{-14})$ has class number 4, which is bigger than 2, and that means some element will have two irreducible factorizations with different numbers of factors. For instance, $$ 81 = 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 = (5+2\sqrt{-14})(5-2\sqrt{-14}) $$ show us irreducible factorizations of 81 with 4 and 2 irreducible factors.

A characterization of number fields with a specific ideal class group $G$ was given in 1983 by Rush (An arithmetic characterization of algebraic number fields with a given class group, Math. Proc. Cambr. Phil. Soc. 94 (1983), 23-28). See here. It is not as simple as the description Carlitz gave for class number 2.

Answer 3

As a concrete example of having more factorizations than the class number, consider first the numbers $6, 14, 21$ in $\mathbb Z[\sqrt{-5}]$. We have

$6=2×3=(1+\sqrt{-5})(1-\sqrt{-5})$

$14=2×7=(3+\sqrt{-5})(3-\sqrt{-5})$

$21=3×7=(1+2\sqrt{-5})(1-2\sqrt{-5})=(4+\sqrt{-5})(4-\sqrt{-5})$

Note that $21$ already has three factorizations, each involving exactly two until factors. Now form their LCM and using the above factorizations render

$42=2×3×7=2×\color{blue}{21}=3×\color{blue}{14}=7×\color{blue}{6}$

where the blue numbers each have the factorizations given above. Thereby $42$ has at least five distinct factorizations (modulo units and order) in this class-2 domain.

Doubling up

At least for imaginary quadratic fields, the class number is more closely connected with how to modify the domain so as to make it UF. In the case of class 2, but not higher classes, augmenting the original integer domain with a single additional lattice in the complex plane suffices to gain UF -- if we give up closure with respect to addition.

Let's see how this works in the case of $\mathbb Z[\sqrt{-5}]$. We form the union of this set with algebraic integers having the form $a\sqrt2+b\sqrt{-10}$ with $a,b$ rational, that is the products of $\sqrt2$ with elements of $\mathbb Z[(1+\sqrt{-5})/2]$. For instance, $(\sqrt2+\sqrt{-10})/2$ whose minimal polynomial equation is $x^4+4x^2+9=0$ belongs in this second set.

When we overlay the two lattices together we lose closure with respect to addition, but the union is closed with respect to multiplication and now it is UF. Watch what happens now with the numbers I noted above:

$6=2×3=(\sqrt2)^2\cdot\left(\dfrac{\sqrt2+\sqrt{-10}}2\right)\left(\dfrac{\sqrt2-\sqrt{-10}}2\right)$

$6=(1+\sqrt{-5})(1-\sqrt{-5})=(\sqrt2)\left(\dfrac{\sqrt2+\sqrt{-10}}2\right)\cdot(\sqrt2)\left(\dfrac{\sqrt2-\sqrt{-10}}2\right)$

$14=2×7=(\sqrt2)^2\cdot\left(\dfrac{3\sqrt2+\sqrt{-10}}2\right)\left(\dfrac{3\sqrt2-\sqrt{-10}}2\right)$

$14=(3+\sqrt{-5})(3-\sqrt{-5})=(\sqrt2)\left(\dfrac{3\sqrt2+\sqrt{-10}}2\right)\cdot(\sqrt2)\left(\dfrac{3\sqrt2-\sqrt{-10}}2\right)$

$21=3×7=\left(\dfrac{\sqrt2+\sqrt{-10}}2\right)\left(\dfrac{\sqrt2-\sqrt{-10}}2\right)\cdot\left(\dfrac{3\sqrt2+\sqrt{-10}}2\right)\left(\dfrac{3\sqrt2-\sqrt{-10}}2\right)$

$21=(1+2\sqrt{-5})(1-2\sqrt{-5})= -1\left(\dfrac{\sqrt2-\sqrt{-10}}2\right)\left(\dfrac{3\sqrt2-\sqrt{-10}}2\right)\cdot(-1)\left(\dfrac{\sqrt2+\sqrt{-10}}2\right)\left(\dfrac{3\sqrt2+\sqrt{-10}}2\right)$

$21=(4+\sqrt{-5})(4-\sqrt{-5})=\left(\dfrac{\sqrt2+\sqrt{-10}}2\right)\left(\dfrac{3\sqrt2-\sqrt{-10}}2\right)\cdot\left(\dfrac{\sqrt2-\sqrt{-10}}2\right)\left(\dfrac{3\sqrt2+\sqrt{-10}}2\right)$

The numbers $6,14,21$ are now factored into irreducibles that differ only by units and order, and so will their LCM $42$.

Class 2 imaginary quadratic fields can be overlaid with one additional lattice to form a quasiperiodic set that becomes UF with respect to multiplication. This property is connected with the fact that there are exactly two reduced quadratic forms associated with such domains (such as $x^2+5y^2$ and $2x^2+2xy+3y^2$), where the second quadratic form corresponds to the lattice we added. Classes greater than two require more extensive overlays to get similar results.

Rendering the Carlitz interpretation

KCd mentions the Carlitz interpretation of class 2: with class 2 all nonunique factorizations of a given number have the same number of non-unit irreducibles, such as $6=2×3$ and $6=(1+\sqrt{-5})(1-\sqrt{-5})$ both having two of them. This emerges from the unique factorization in the double lattice. To see how, consider the unique factorization

$6=(\sqrt{2})^2\left(\dfrac{\sqrt2+\sqrt{-10}}2\right)\left(\dfrac{\sqrt2-\sqrt{-10}}2\right)$

which has four non-units from the augmented lattice I used. To retrieve a factorization in $\mathbb Z[\sqrt{-5}]$, we must multiply together pairs of factors from the second lattice, the products of such pairs being in the original class 2 domain if the second lattice is chosen properly, for instance in this case

$(\sqrt{2})\left(\dfrac{\sqrt2+\sqrt{-10}}2\right)=1+\sqrt{-5}$

Thus four factors from the second lattice will generate half that many, or two, when we go back into $\mathbb Z[\sqrt{-5}]$. Then the two factorizations

$6=2×3=(1+\sqrt{-5})(1-\sqrt{-5})$

correspond to different pairings of the four unique factors from the double lattice, both of which must produce two $\mathbb Z[\sqrt{-5}]$ non-units. Similarly the unique double-lattice factorization

$6699=3×7×11×29=\left(\dfrac{\sqrt2+\sqrt{-10}}2\right)\left(\dfrac{\sqrt2-\sqrt{-10}}2\right)\left(\dfrac{3\sqrt2+\sqrt{-10}}2\right)\left(\dfrac{3\sqrt2-\sqrt{-10}}2\right)(11)(3+2\sqrt{-5})(3-2\sqrt{-5})$

(where natural primes $\equiv11\bmod 20$ are irreducible in the double lattice) will give five non-units when rendered back into $\mathbb Z[\sqrt{-5}]$: two from pairing the factors of $3×7=21$ which are in the second lattice, and the other three factors ($11$ and the factors of $29$) which remain unchanged:

$6699=(3×7×11)(3+2\sqrt{-5})(3-2\sqrt{-5})$

$6699=(1+2\sqrt{-5})(1-2\sqrt{-5})(11)(3+2\sqrt{-5})(3-2\sqrt{-5})$

$6699=(4+\sqrt{-5})(4-\sqrt{-5})(11)(3+2\sqrt{-5})(3-2\sqrt{-5})$

Answer 4

A less-elementary, but occasionally useful, characterization of having class group $H$ (for a Dedekind domain), is that if we localize so as to kill off ideals generating the ideal class group, then the resulting ring is (still Dedekind and) has class number one (and is a PID). So the number of generators of the class group is the minimum number of prime ideals that must be localized-away to obtain a PID.

EDIT: for example, for $\mathfrak p$ a non-principal ideal, but with $\mathfrak p^n=\alpha\cdot \mathfrak o$ principal (here the classgroup is finite), localize the integers $\mathfrak o$ just a little by adjoining $1/\alpha$. Then there are $\beta_1,\ldots,\beta_n\in\mathfrak p$ whose product is the unit $\alpha$ in $\mathfrak o[1/\alpha]$. So all $\beta_i's$ are units in that localized ring, so $\mathfrak p\cdot \mathfrak o[1/\alpha]=\mathfrak o[1/\alpha]$ has become principal.

In fact, every ideal in $\mathfrak o$ that was equivalent to $\mathfrak p$ in $\mathfrak o$ becomes principal, as well.

Indeed, this may be similar to the device mentioned by @OscarLanzi.