# Distinct primitive factorizations over integers of number fields

I am curious about the following.

Let $K$ be a number field. For any $a \in \mathcal{O}_K$ in its ring of integers, let $N(a)$ be zero if there exist elements $b, c \in \mathcal{O}_K \setminus \mathcal{O}_K^*$ such that $a = bc$ and $b$ has at least two different (that is, nonassociate) factorizations into irreducible factors (over $\mathcal{O}_K$), otherwise let $N(a)$ be the number of different factorizations of $a$ into irreducible factors. Is the set $\{N(a) | a \in \mathcal{O}_K\}$ unbounded for some $K$? Is it bounded for some $K$ that has nontrivial ideal class group?

Similarly interesting is the same question with $N(a)$ defined to be zero if and only if $a=bc$ and both $b,c$ have at least two different factorizations. I have asked this question earlier on math.stackexchange forum (https://math.stackexchange.com/questions/712482/on-number-of-different-factorizations-over-integers-of-a-number-field), but have not received an answer.

• You should definitely give a look at A. Geroldinger and F. Halter-Koch's monograph on the factorization theory of (abelian cancellative) monoids: Non-Unique Factorizations: Algebraic, Combinatorial and Analytic Theory, Pure and Applied Mathematics 278, Chapman & Hall/CRC, 2006. Apr 9, 2014 at 18:41

The quantity in question can be bounded in terms of the class number only. I will show that it is at most the number of partitions of $h^2$, where $h$ is the class number. (This could be improved in several ways.)

Let $a \in \mathcal{O}_K$. Suppose $N(a) > 1$. Clearly $a$ is neither irreducible nor invertible.

Let $c$ be an irreducible dividing $a$. Then $c^{-1}a$ has an (essentially) unique factorization, since otherwise $N(a)=0$.

Let $a= c_1\dots c_n = c_1'\dots c_m'$ be two essentially distinct factorizations, which exist by assumption. By the just made observation we have that $(c_i) \neq (c_j')$ for all $i,j$; and this is true for any two distinct factorizations.

Let $(c_1) = P_1 \dots P_s$ be a factorization into prime ideals. There exists a subset $J$ of $\{1, \dots , m\}$ of cardinality at most $s$ such that $c_1$ divides $\prod_{j \in J} c_j' = d$. (This follows by the uniqueness of the factorization into prime ideals and picking, if necessary, for each $P_j$ a $c_j'$ that contains it in its factorization.)

Now, consider a factorization into irreducibles of $d$ that contains $c_1$, say $d=c_1c_2'' \dots c_k''$. Then $c_1c_2'' \dots c_k'' \prod_{j \notin J} c_j'$ is a factorization of $a$. This factorization contains $c_1$ and is thus (essentially) equal to $c_1\dots c_n$. Since $(c_i) \neq (c_j')$, this is only possible if $\prod_{j \notin J} c_j'$ is empty, that is $J = \{1, \dots , m\}$. Yet this means $m \le s$.

Now, note that the number of prime ideals in a factorization of an irreducible is at most the class number $h$. [See the other answer for details.]

This tells us two things. First, that $m \le s \le h$, and then that the total number $q$ of prime ideals (with multiplicity) in the factorization into prime ideal of $(a)$ is at most $mh \le h^2$.

Since there certainly cannot be more essentially distinct factorizations of $a$ than there can be partitions of a set of cardinality $q$ we see that $N(a)$ is at most the number of partitions of $h^2$.

I discussed something like this in this answer (TCS stack exchange).

Your function will be bounded in terms of the class number $h$: let $a\in\cal{O}_K$, and let $(a) = \prod_{i=1}^{r}\frak{p}_i^{e_i}$ be its unique factorization into primes. Then by the pigeon-hole principle, any subproduct of length at least $h$ contains a further non-trivial subproduct which is trivial in the class group but minimal, hence represents an irreducible element of $\cal{O}_K$.

Now for your problem we can assume that no $\frak{p}_i$ is principal. Suppose first that $r\geq 3$. If $\sum_i e_i \leq 6h$ there is a bounded number of ways to partition the factorization, so clearly a bounded numbers of factorizations into integers of $a$ and any of its divisors. We show that if $r \geq 6h$ then $N(a)=0$. Order the primes in increasing order of $e_i$. Then $\frak{p}_1,\frak{p}_2$ jointly occur less than $2/3$ of the time. Since the $\frak{p}_i$ are non-principal and since $r/3 \geq 2h$, for each $i=1,2$ we have an ideal $\frak{b}_i$, a product of at most $h-1$ primes distinct from both the fixed primes, such that $\frak{p}_i\frak{b}_i$ is principal, and such that $(b) = (\frak{p}_1\frak{b}_1)(\frak{p}_2\frak{b}_2)$ is a subproudct of the factorization of $(a)$. Note that this is a factorization of $b$ in $\cal{O}_K$, but note that if we instead started with $\frak{p}_1\frak{p}_2$, and completed that to a principal ideal, we'd get a different factorization.

Now if $r=1$ or $r=2$ and $e_1 = 1$ there is only one factorization up to units, so it remains to consider the case $r=2$, $2\leq e_1 \leq e_2$. Again if $e_2 \leq 2h$, the function $N$ is bounded for combinatorial reasons, and otherwise find minimal $1\leq f \leq h-1$ and $0 \leq g \leq h-2$ such that $\frak{p}_1 \frak{p}_2^f$ and $\frak{p}_1^2 \frak{p}_2^g$ are principal (hence irreducible) [note that necessarily $g\leq 2f$]. Let $(b) = \frak{p}_1^2 \frak{p}_2^{2f}$, again a subproduct of $(a)$. Then this has the distinct factorizations $$(b) = (\frak{p}_1\frak{p}_2^f)(\frak{p}_1\frak{p}_2^f) = (\frak{p}_1^2\frak{p}_2^g)(\frak{p}_1^2\frak{p}_2^{2f-g})$$

• Thank you for the answer. However, I struggle to see that $\mathfrak{b_1, b_2}$ as above must indeed exist and would be thankful for a couple more details if they do. In fact, the way I read it, it seems that, for instance, if the class group is cyclic of order $4$ and the ideals $\mathfrak{p_1, p_2}$ are generators of the class group and their product is a principal ideal, while $\mathfrak{p_3,...,p_r}$ are ideals that have order $2$ in the class group, then $\mathfrak{p_1p_2p_3^2...p_r^2}$ is principal, but no such $\mathfrak{b_i}$ exist. Apr 13, 2014 at 17:49
• You're right. I'll have to fix the argument. Apr 14, 2014 at 3:05

I'd like to add that, after reading the preceeding answers, it becomes quite clear that $N : \mathcal{O}_K \rightarrow \mathbb{N}_{\geq 0}$, as defined in the second part of the question, is also bounded.

To see that, observe first that if

$$(a) = \prod_{i = 1}^r{\mathfrak{p}_i}^{e_i},$$

where none of the $\mathfrak{p}_i$ is principal and $r \geq (2h - 1)(h-1) + 1$, then $(a)$ will have at least two (essentially) distinct factorizations. Indeed, there exists an element in the class group of $K$, that is not the identity element, such that the corresponding equivalence class coincides with the equivalence classes of at least $2h$ of the factors of $(a)$, say $\mathfrak{p'}_1, ..., \mathfrak{p'}_{2h}$, whose exponents in the factorization of $(a)$ are $e'_1 \leq e'_2 ... \leq e'_{2h}$, respectively. Let $l$ denote the order of these ideals in the class group. Then the products $\mathfrak{p'}_1\mathfrak{p'}_2...\mathfrak{p'}_l, \mathfrak{p'}_1\mathfrak{p'}_{l+1}...\mathfrak{p'}_{2l}$ are principal, genrated by (essentially) distinct irreducible elements $a_1, a_2$. Note that $a_1^{e'_1}, a_2^{e'_1} | a$, but $a_1^{e'_1}a_2^{e'_1} \nmid a$. Thus $a$ has distinct factorizations (this is essentially Lior's argument, with a minor correction).

One can notice that, further, $N(a)$ must also be zero if $r \geq (2h - 1)(h-1) + 1 + 2h$, since, after selecting $\mathfrak{p'}_1, ..., \mathfrak{p'}_{2h}$ as above, one can form a principal ideal $(b) = \mathfrak{p'}_1 ...\mathfrak{p'}_{2h}$ that has distinct factorizations (for the same reason: $\mathfrak{p'}_1\mathfrak{p'}_2...\mathfrak{p'}_l = (b_1), \mathfrak{p'}_1\mathfrak{p'}_{l+1}...\mathfrak{p'}_{2l} = (b_2)$, the elements $b_1, b_2$ are distinct, irreducible, $b_1, b_2 | b$, but $b_1b_2 \nmid b$ ). Then $c = b^{-1}a$ will also have distinct factorizations, as the first observation implies. Hence $N(a) = 0$.

Thus we may restrict ourselves to the case $r \leq (2h - 1)(h - 1) + 2h$. Let us call an irreducible algebraic integer absolutely irreducible if the corresponding principal ideal is a power of some prime ideal. If some $a$ now has "many" (that is, an arbitrarily large number) distinct factorizations, then there must exist an irreducible element $a'$ that is not absolutely irreducible and whose "large" power divides $a$ (since the number of distinct irreducible, nonassociate divisors of $a$ is now bounded in terms of $h$). One can then take $b$ to be a "large" power of $a'$, so that there also remains a "large" power of $a'$ to divide $c = b^{-1}a$. Then both $b, c$ will have distinct factorizations: one that contains as factors as many as possible $a'$'s, another - that contains as many as possible absolutely irreducible factors. Hence $N(a)$ is zero if the number of distinct factorizations of $a$ is large enough.