Is there any way to tell the number of distinct ways to factor $a\in\mathcal{O}_k$ (up to units, of course) when $k$ is not a PID? A simple investigation in $\mathbb{Q}(\sqrt{-5})$ with integer ring $\mathcal{O}=\mathbb{Z}[\alpha]$ and Galois group $G=\langle\sigma\rangle$, with $\alpha = \sqrt{-5}$, gives, for example (and this is incredibly basic, but in case we have algebra people not familiar with the computations of number theorists or outside interested parties to motivate the question).

$6=\begin{cases} 2\cdot 3 \\ (1+\alpha)(1-\alpha)\end{cases}\qquad (*)$

Since $\mathbb{Z}[\alpha]$ is integrally closed, we can look mod 2 and mod 3 to find factorizations for $(2)$ and $(3)$, which gives $m_\alpha(x)\equiv (x+1)^2\mod 2$ and $m_\alpha(x)\equiv (x-1)(x-2)\mod 3$, and so $(2)$ ramifies as $(2,1+\alpha)^2=\mathfrak{p}_1^2=\mathfrak{p}_2^2$ (the reason for using two labels for the same ideal will be clear later) and since $N_{k/\mathbb{Q}}(1+\alpha)=6$, we know $(2,1+\alpha)$ is not principal, so $2$ is an irreducible integer, and similarly for $(3)=(3,\alpha-1)(3,\alpha -2)=(3,1-\alpha)(3,1+\alpha)=\mathfrak{p}_3\mathfrak{p}_4$ to show 3 is irreducible, and $1\pm\alpha$ are irreducible by norm calculations as well.

So in this case we'd like to see that $(*)$ produces all of the factorizations of $6$ into irreducibles. So we need only look at ideals above 2 and 3. Since $h_k\,(=|\text{Pic}(\mathcal{O}_k)|)$ any pairing of two of the four (we need to count the ramifying prime twice, of course, and this is why we double counted it earlier) ideals we've produced multiply to an irreducible in some factorization of 6 so we see that of the ${4\choose 2}=3$ ways to do this, but clearly matching 1 with 3 is the same as matching 1 with 4 because $\mathfrak{p}_1=\mathfrak{p}_2$.

I'd like to know if there is a general way to tell the total number of factorizations of a given number in an integer ring, $\mathcal{O}_k$ for a number field, $k$. If this is too hard, then I'm also willing to settle on number of factorizations for a rational integer where we are allowed to use integers from $\mathcal{O}_k$ in the factorization.

Obviously the combinatorics of the problem mean that there is not just one number for the entire field, simply looking at the number 12 in the example above illustrates why this is. However some easy reductions:

If we look at the ideal factorization, we can always ignore inert primes, as they clearly have no contribution to making more factorizations, moreover we can ignore principal primes, but these are harder to detect with simple tricks if the norm of the element given to us by a factorization in some reduction with a coprime conductor (this also illustrates the problem when we cannot find a ring with trivial conductor in our integer ring). If the class number (again, I don't know if algebraists use the same terminology, so I'll clarify: this is the number $h_k=|\text{Pic}(\mathcal{O}_k)|$) is 2, then we know we can select non-principal ideals and their product is principal and that the product ideal is generated by an irreducible element without any further information about the exact structure of $\text{Pic}(\mathcal{O}_k)$.

One might argue that one need only pair inverse (non-trivial) classes of primes together to get more information, but as far as I know detecting the exact ideal class of a non-trivial element when $h_k>2$ is a non-trivial feat, but then this is not my prime area of expertise, and I would love to be proven wrong on that front.

EDIT: It seems reasonable that if you have an ideal $\mathfrak{a}=\prod\mathfrak{p}_i^{e_i}$ that if you could detect the ideal classes of the $\mathfrak{p}_i$ you could manage to pair them up based on the abelian group decomposition of the class group into a direct sum of cyclic groups and reducing the question to finding group inverses with minimal exponents for the representing ideal powers, but it's not clear if this would be a prescription for every possibility or not.


You are asking for a formula for the number of decompositions $\mathfrak{m}=\mathfrak{a}\mathfrak{b}$, where $\mathfrak{a}$ and $\mathfrak{b}$ are principal ideals. Considering the group of characters $\widehat{\text{Pic}(\mathcal{O}_k)}$, the number of these decompositions is $$ \frac{1}{h_k^2}\sum_{\chi,\psi\in\widehat{\text{Pic}(\mathcal{O}_k)}}\sum_{\mathfrak{m}=\mathfrak{a}\mathfrak{b}}\chi(\mathfrak{a})\psi(\mathfrak{b}).$$ The inner sum is an explicit multiplicative function in $\mathfrak{m}$, so we have an explicit formula as an average of $h_k^2$ multiplicative functions. I doubt it gets simpler than that.


It's ironic that you begin your query with an example in a ring with class number 2 because this is precisely the case that's been worked out in detail.

Inspired by a 1960 result of Leonard Carlitz, Robert Valenza published a paper in 1990 entitled ``Elasticity of factorization in number fields'' which works out something akin to what you're asking about. http://www.ams.org/mathscinet-getitem?mr=1072466 He shows that the number of ways one can factor an element into irreducibles grows something like $h_k/2$ - assuming that the ring $\mathcal O_k$ is not a PID. Unfortunately, this is a lot more subtle than it sounds: for example, if $\mathcal O_k = \mathbb Z[\sqrt{-5}]$ is the classic example with class number $h_k = 2$, then the element $6 = 2 \cdot 3 = (1 + \sqrt{-5}) \cdot (1 - \sqrt{-5})$ can only be factored in to 2 irreducibles. In general, if

$ a = \displaystyle \prod_{i=1}^n \varpi_i = \prod_{j=1}^m \pi_j$

has different factorizations, then the ratio $n/m \leq h_k/2$. Valenza's theorem asserts that we always have $m = n$ if and only if $h_k \leq 2$. As another nice example, Valenza considers the ring $\mathcal O_k = \mathbb Z[(1+\sqrt{-23})/2]$ with class number $h_k = 3$. We observes that

$27 = 3^3 = (2 + \sqrt{-23}) \cdot (2 - \sqrt{-23})$

so that $n/m = h_k/2 = 3/2$. In general, if $Z_n \hookrightarrow \text{Pic}(\mathcal O_k)$ then Valenza explains how to exhibit a factorization with $m = 2$ and $n = |Z_n|$ .

My graduate student Alex Barrios once wrote up a very nice summary of Valenza's result; it can be found here: http://www.math.purdue.edu/~egoins/seminar/12-10-05.pdf

  • $\begingroup$ This is really fantastic, I'll take a look at the write-up for sure! $\endgroup$ – Adam Hughes Nov 5 '13 at 18:58

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