The goal of this blog post is to provide an overview of the general theory of divisors, as described in a book Number Theory by Borevich and Shafarevich. The point of this post is for it to be somewhat expository, so it will avoid the longer proofs, sometimes just sketching the ideas.

Throughout, by a “ring” we will mean an integral domain, i.e. commutative ring with unity without zero divisors

## Unique factorization

Let $$R$$ be an arbitrary ring. Recall that we call an element $$r\in R$$ irreducible if $$r$$ is not zero, not a unit and whenever we write $$r=ab$$ with $$a,b\in R$$, then one of $$a,b$$ is a unit. We say that $$R$$ has unique factorization, or that it is a unique factorization domain (UFD) if every nonzero element of $$R$$ can be written as a product of a unit and some number of irreducible elements, and this expression is unique up to ordering and unit multiples, i.e. whenever we have $$u_1r_1\dots r_n = u_2q_1\dots q_m$$ with $$u_1,u_2$$ units and $$r_1,\dots,r_n,q_1,\dots,q_m$$ irreducibles, then $$n=m$$ and there is a bijection between $$r_i$$ and $$q_j$$ which maps $$r_i$$ to some its unit multiple.

In general, there is little to no reason to expect $$R$$ is a UFD. A famous example of a ring which doesn’t have unique factorization is $$\mathbb Z[\sqrt{-5}]$$ – $$2\cdot 3=(1+\sqrt{-5})(1-\sqrt{-5})$$ can be verified to be an example of nonunique factorization as defined above.

The unique factorization is a very useful tool. For example, in a UFD, if a product $$ab$$ of relatively prime elements is a perfect $$n$$th power, then, up to unit multiples, both $$a$$ and $$b$$ are $$m$$th powers as well. Hence we would like to have something in the spirit of unique factorization available in a greater range of rings. Is that possible?

## Salvaging unique factorization

An idea to salvage rings which are not UFDs is to somehow embed the elements of the ring in some larger structure in which unique factorization does hold. At the same, it would be desirable for it to have as little redundancy as possible.

Because at this point we are interested only in unique factorization, which is purely multiplicative property of a ring, we will only require this larger structure to have multiplication. Also, for convenience purposes, we will ignore the zero element – its multiplicative behaviour is prefectly well understood anyways.

Our “dream structure” would then be a so called commutative semigroup (which differs from a (commutative) group in that we don’t require inverses) $$\mathcal R$$, into which the semigroup $$R\setminus\{0\}$$ would be mapped. There is an obvious way in which we can define divisibility in $$\mathcal R$$, and we can speak of elements of $$\mathcal R$$ dividing elements of $$R$$. Because of that, the elements of $$\mathcal R$$ are called divisors of $$R$$. We will denote the divisor corresponding to $$a\in R\setminus\{0\}$$ by $$(a)$$ and we will call such divisors principal.

There come two properties we will want this mapping to satisfy: first, we want multiplication to be preserved, i.e. $$(a)(b)=(ab)$$ (that is, we require it to be a homomorphism of semigroups), and we will want the divisibility (and indivisibility) to be preserved, i.e. $$a$$ divides $$b$$ in $$R$$ iff $$(a)$$ divides $$(b)$$ in $$\mathcal R$$. The first of these properties implies $$(1)$$ is a multiplicative identity when it comes to multiplying by $$(a)$$. We want it to be multiplicative identity in the whole semigroup.

Any divisor $$\frak a$$ of $$R$$ induces a subset of $$R\setminus\{0\}$$, namely the set of elements it divides, which we will denote by $$\overline{\frak a}$$. In $$R$$, if $$a$$ and $$b$$ are divisible by $$c$$, then so are $$a+b$$ and $$a-b$$. With notation above, this can be phrased as: $$\overline{(c)}$$ is closed under addition and subtraction. This property shall be required for all divisors: $$\overline{\frak a}$$ is closed under addition and subtraction.

One more property is that we will want a divisor $$\frak a$$ to be completely characterized by $$\overline{\frak a}$$ (so that we don’t have any redundant divisors). That is, we want $$\frak a\neq\frak b$$ to imply $$\overline{\frak a}\neq\overline{\frak b}$$. This has one more effect – unit multiples in $$R$$ are being ignored. Indeed, one can now verify that $$(a)=(b)$$ iff $$a$$ and $$b$$ are unit multiples of each other. Thanks to this, it is particularly easy to state unique factorization, in a way akin to $$\mathbb Z$$: first define a prime divisor to be a divisor $$\frak p$$ such that, whenever represented as a product of two divisors, one of them is the unit $$(1)$$. Then we can state unique factorization as: Every divisor $$\frak a$$ can be represented as a product $$\frak p_1\dots\frak p_n$$ in a unique way up to a permutation of factors. Using more difficult words, this means that $$\mathcal R$$ is a free commutative semigroup generated by the prime divisors $$\frak p$$.

To sum up, we will define a theory of divisors for a ring $$R$$ to be a free commutative semigroup $$\mathcal R$$ together with a semigroup homomorphism $$R\setminus\{0\}\rightarrow\mathcal R,a\mapsto (a)$$ such that:

1. for $$a,b\in R\setminus\{0\}$$, $$a$$ divides $$b$$ in $$R$$ iff $$(a)$$ divides $$(b)$$ in $$\mathcal R$$,
2. for $$a,b\in R\setminus\{0\}$$ are divisible by divisor $$\frak a$$, then so are $$a+b,a-b$$ (provided they are in $$R\setminus\{0\}$$), and
3. if $$\overline{\frak a}=\overline{\frak b}$$, then $$\frak a=\frak b$$ for all divisors $$\frak a,\frak b$$.

## Uniqueness

It is far from clear whether a theory of divisors exists for a given ring or not, and if so, whether it is (“essentially”) unique. The latter of these questions turns out to be relatively easy to answer – a theory of divisors, if exists, is unique. More precisely, if we have two theories of divisors, $$\mathcal R_1$$ together with a map $$a\mapsto (a)_1$$ and $$\mathcal R_2$$ together with a map $$a\mapsto (a)_2$$, then there is an isomorphisms of these two semigroups sending $$(a)_1$$ to $$(a)_2$$. We now sketch the proof of this fact.

Let $$\frak p\in\mathcal R_1$$ be prime. We shall show there is a prime divisor $$p’\in\mathcal R_2$$ such that $$\overline{\frak p’}\subseteq\overline{\frak p}$$ (the $$\overline{\frak p},\overline{\frak p’}$$ are the sets of elements divisible by $$\frak p,\frak p’$$ in respective theories of divisors). Suppose that there is no such prime divisor. Choose any $$\beta$$ divisible by $$\frak p$$. Factor $$\beta(=(\beta)_2)$$ as $$\frak p_1^{k_1}\dots\frak p_r^{k_r}$$ in $$\mathcal R_2$$. Choose $$\beta_i\in\overline{\frak p_i}\setminus\overline{\frak p}$$. Then $$\beta_1^{k_1}\dots\beta_r^{k_r}$$ is divisible by $$\beta$$, but not $$\frak p$$, which is a contradiction.

Similarly, there is a prime $$\frak q\in\mathcal R_1$$ such that $$\overline{\frak q}\subset\overline{\frak p’}\subseteq\overline{\frak p}$$. We now claim $$\overline{\frak q}=\overline{\frak p}$$. Otherwise, choosing $$\alpha$$ divisible by $$\frak q$$ but not $$\frak{pq}$$, we would have $$\alpha\in\overline{\frak q}\setminus\overline{\frak p}$$. Hence $$\overline{\frak p}=\overline{\frak p’}$$.

Matching $$\frak p\in\mathcal R_1$$ with $$\frak p’\in\mathcal R_2$$ such that $$\overline{\frak p}=\overline{\frak p’}$$ gives a bijection between prime divisors in both theories, which we easily extend multiplicatively to an isomorphism. We just need to check it preserves principal divisors. To avoid technical details, we omit this part of the proof.

## Valuations

Unique factorization in $$\mathcal R$$ can be also stated in the following way: for any divisor $$\frak a$$ there are uniquely defined integers $$\nu_{\frak p}(\frak a)$$ for each prime divisor $$\frak p$$ such that

$$\displaystyle\frak a=\prod_{\frak p}\frak p^{\nu_{\frak p}(\frak a)}$$.

We can also define functions $$\nu_{\frak p}$$ on $$R\setminus\{0\}$$ by $$\nu_{\frak p}(a)=\nu_{\frak p}((a))$$. By defining $$\nu_{\frak p}(a/b)=\nu_{\frak p}(a)-\nu_{\frak p}(b)$$ and checking this function is well-defined, we can extend it to the field of fractions $$K$$ of $$R$$ with zero excluded. This function now has the following properties:

1. the image of $$K\setminus\{0\}$$ under $$\nu_{\frak p}$$ is $$\mathbb Z$$
2. $$\nu_{\frak p}(ab)=\nu_{\frak p}(a)+\nu_{\frak p}(b)$$, and
3. $$\nu_{\frak p}(a+b)\geq\min\{\nu_{\frak p}(a),\nu_{\frak p}(b)\}$$ with equality if $$\nu_{\frak p}(a)=\nu_{\frak p}(b)$$.

For properties 2, 3 it is easy to verify them for $$R\setminus\{0\}$$, and then extend them to $$K\setminus\{0\}$$. A function with these three properties is called a valuation. It is customary to additionally define $$\nu_{\frak p}(0)=\infty$$, so that these properties still hold on all of $$K$$.

No two valuations $$\nu_{\frak p},\nu_{\frak q}$$ are the same for distinct $$\frak p,\frak q$$. Hence the prime divisors can be identified with a subset $$V$$ of the set of valuations on $$K$$. The set of valuations corresponding to these divisors further satisfies the following properties:

1. for a fixed $$a\in K\setminus\{0\}$$, $$\nu(a)=0$$ for all but finitely $$\nu\in V$$,
2. for $$a\in K$$, $$a\in R$$ iff $$\nu(a)\geq 0$$ for all $$\nu\in V$$, and
3. for any $$\nu_1,\dots,\nu_m\in V$$ and nonnegative integers $$k_1,\dots,k_m$$ there is $$a\in R$$ such that $$\nu_i(a)=k_i$$.

Property 1 is clear. For property 2, write $$a=b/c, b,c\in R\setminus\{0\}$$ and note that this property is then equivalent to $$\nu_{\frak p}(b)\geq\nu_{\frak p}(c)$$ for all prime divisors $$\frak p$$ iff $$c\mid b$$, which is easy to see (recall property 1 in the definition of theory of divisors). For property 3, consider valuations corresponding to prime divisors $$\frak p_1,\dots,\frak p_m$$. Consider $$a_i\in\overline{\frak p_1^{k_1+1}\dots\frak p_i^{k_i}\dots\frak p_m^{k_m+1}}\setminus\overline{\frak p_1^{k_1+1}\dots\frak p_i^{k_i+1}\dots\frak p_m^{k_m+1}}$$. Then $$a=a_1+\dots+a_m$$ has the desired properties.

It is not hard to show the converse: if the set $$V$$ of valuations on a field $$K$$ satisfies the mentioned three properties, then $$R$$ has a theory of divisors which then gives rise to $$V$$ when considering valuations corresponding to its prime divisors. For that reason, the search for theories of divisors is reduced to a search for certain sets of valuations. As it won’t cause confusion, we will call $$V$$ a theory of divisors as well.

## Divisors, UFDs and integral closure

We give, somewhat belatedly, an example of theory of divisors. Suppose $$R$$ is a UFD. Because of units, we can’t just take $$\mathcal R$$ to be $$R\setminus\{0\}$$. Instead, for $$a\in R\setminus\{0\}$$, we define $$(a)$$ to be the set of all its unit multiples, and take $$\mathcal R$$ to be the set of all such sets. Already the notation suggests the mapping $$R\setminus\{0\}\rightarrow\mathcal R$$. Since $$R$$ is a UFD, it’s not difficult to see this gives us a theory of divisors.

We can also give an example of a ring which does not have a theory of divisors. This can be done because every ring with a theory of divisors must be integrally closed in its field of fractions $$K$$, i.e. suppose $$a\in K$$ is a root of a monic polynomial with coefficients in $$R$$. Then $$a\in R$$. To see why this is true, suppose

$$a^n+d_{n-1}a^{n-1}+\dots+d_1a+d_0=0$$

with $$d_i\in R$$. If $$a\not\in R$$, there is a valuation $$\nu\in V$$ such that $$k=\nu(a)<0$$. Then $$\nu(a^n)=kn$$ and $$\nu(d_ia^i)\geq \nu(a^i)=ki>kn$$. Then, by property 3 in the definition of valuation, it follows that

$$\nu(0)=\nu(a^n+d_{n-1}a^{n-1}+\dots+d_1a+d_0)=\min\{\nu(a^n),\nu(d_{n-1}a^{n-1}),\dots,\nu(d_1a),\nu(d_0)\}=kn$$,

which is clearly wrong. This proves $$R$$ is integrally closed.

An example of non-integrally closed ring is $$\mathbb Z[\sqrt{-3}]$$, since $$\frac{1}{2}+\frac{1}{2}\sqrt{-3}$$ is a root of $$x^2-x+1$$.

One last useful fact is somewhat of a converse to the first example – we can show that if a ring has a theory of divisors, then it is a UFD, provided the theory of divisors has finitely many prime divisors $$\frak p_1,\dots,\frak p_m$$. To see why, just note that there is an element $$\pi_i\in R$$ which is divisible by $$\frak p_i$$ but not any other $$\frak p_j$$, essentially thanks to property 3 of valuations forming a theory of divisors. Using these elements we can replace a factorization into prime divisors by a factorization into $$\pi_i$$.

## Coming soon

In the next blog post we will establish a number of results regarding extending valuations and theories of divisors to finite field extensions. In particular, we will show any ring of algebraic integers in a number field has a theory of divisors.