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## Primes and ideals

Probably the first important result in algebraic number theory is the following. Let $K$ be a finite field extension of $\mathbb{Q}$. Let $\mathcal{O}_K$ be the ring of algebraic integers in $K$.

Theorem: The ideals of $\mathcal{O}_K$ factor uniquely into prime ideals.

This is the “correct” generalization of the fact that $\mathbb{Z}$, as well as some small extensions of it such as $\mathbb{Z}[ \imath ], \mathbb{Z}[\omega]$, have unique factorization of elements. My goal, in this next series of posts, is to gain some intuition about this result from a geometric perspective, wherein one thinks of a ring as the ring of functions on some space. Setting up this perspective will take some time, and I want to do it slowly. Let’s start with the following simple questions.

• What is the right notion of prime?
• Why look at ideals instead of elements?

In this series I will assume the reader is familiar with basic abstract algebra but may not have a strong intuition for it.

Primes

First a little motivation and review. $\mathbb{Z}$ has a division algorithm, hence is a Euclidean domain and a unique factorization domain. A starting observation here is that to some extent this is true of certain rings that one can define by extending $\mathbb{Z}$. For example, $\mathbb{Z}[\imath]$, the Gaussian integers, form a Euclidean domain with the norm $N(a + bi) = (a + bi)(a - bi) = a^2 + b^2$. The description of the primes in the Gaussian integers is classical: they are, up to multiplication by units, the primes of the form $4k + 3$ and the integers of the form $a + bi$ such that $a^2 + b^2$ is a prime of the form $4k + 1$ or $2$. Prime factorization in these larger rings is a useful tool for solving Diophantine equations, and in fact a classical way to motivate this subject is to repair a certain “proof” of Fermat’s last theorem.

But it’s not hard to see that we can’t always assume that prime factorization will be around. For example, in $\mathbb{Z}[ \sqrt{-5} ]$ the argument used to establish the division algorithm in the Gaussian integers fails. Indeed, $2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5})$ and it is possible to verify that no terms involved can be factored further, hence we do not have unique factorization in this setting.

After grappling with these issues it comes out that there are two notions that need to be distinguished: prime elements and irreducible elements. Irreducible has the obvious definition: an element is irreducible if, up to units, it cannot be factored any further. But it turns out that the “right” notion of primality is that $p | ab \Rightarrow p | a \text{ or } p | b$. Indeed this is the property of primality that actually matters in the proofs of unique factorization, since it is what allows us to cancel factors in two different prime factorizations of an integer.

Prime elements are always irreducible, but as we saw above, the converse doesn’t hold; $2$ and $3$ are irreducible but don’t divide either of $1 + \sqrt{-5}$ or $1 - \sqrt{-5}$. Indeed, if you’re familiar with the proof that $\mathbb{Z}$ is a UFD you should know that the integral domains such that irreducible elements are prime are precisely the UFDs.

Homomorphisms

The general definition of prime is important enough to merit a shift in perspective, so let’s talk about another Euclidean domain, $\mathbb{C}[t]$. We know by the fundamental theorem of algebra that the primes here are the monic polynomials $t - a, a \in \mathbb{C}$. The uncanny thing about this case is that once you’ve proven existence, you seem to get uniqueness for free: of course a polynomial can’t have two sets of roots. After canceling common factors, you could simply plug in the roots of one side.

This strategy is worth characterizing abstractly so we can see how it applies to the integers. What does “plugging in” mean? It means we are considering the homomorphism $e_a : \mathbb{C}[t] \to \mathbb{C}$ given by replacing $t$ by some fixed $a$. The reason these homomorphisms do what we want is as follows: if $(t - a_1)...(t - a_n) = (t - b_1)...(t - b_m)$ represents two factorizations with disjoint roots, then substituting $t = a_1$ (i.e. applying the homomorphism $e_{a_1}$) gives $0 = (a_1 - b_1)...(a_1 - b_m)$, and this is impossible because $\mathbb{C}$ is an integral domain (in fact a field): nonzero elements cannot multiply to zero.

The analogue of this strategy for the integers is to consider the homomorphisms $e_p : \mathbb{Z} \to \mathbb{F}_p$ that send an integer to its residue $\bmod p$, and in fact once you prove existence, uniqueness follows in exactly the same way because $\mathbb{F}_p$ is also an integral domain (in fact a field). This seems like a foolproof strategy, so what goes wrong for $\mathbb{Z}[ \sqrt{-5} ]$?

The problem is that working, say, $\bmod 3$ no longer gives an integral domain. This is just a restatement of the fact that $2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5})$, but it tells us why the “right” definition of primality is right: primes give rise to homomorphisms into integral domains.

Ideals

Having said that, let me describe the following homomorphism $\phi$ from $\mathbb{Z}[ \sqrt{-5} ]$ into an integral domain: send $a + b \sqrt{-5}$ to $a - b \in \mathbb{F}_3$. Now $\phi$ respects addition and the identity, so to show that it respects multiplication it’s enough to check it for multiplying $1, \sqrt{-5}$, and indeed $\phi(\sqrt{-5} \cdot \sqrt{-5}) = 1$ as expected. $\phi$ also has the property that it sends both sides of the equation $2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5})$ to zero; more precisely it sends $3$ and $1 + \sqrt{-5}$ to zero.

The way Kummer explained this situation is that $\phi$ describes modding out by an “ideal number” that divides both $3$ and $1 + \sqrt{-5}$ (although I don’t claim that this is the terminology Kummer used). Of course such a number does not actually exist, but Kummer was able to do a lot of interesting work by pretending that it and other numbers like it did. By adding in ideal numbers and performing “ideal arithmetic,” Kummer was able to recover unique factorization in some cases.

It was Dedekind who finally clarified the situation. The properties of $\phi$ are completely determined by its kernel, the set of elements such that $\phi(a) = 0$. For homomorphisms such as $e_a$ and $e_p$ the kernels are just the multiples of $t - a$ and $p$ respectively, but $\phi$ has the property that $3$ and $1 + \sqrt{-5}$ are both in its kernel but have no common factor. Kernels are abstractly characterized as ideals, and this particular ideal is not principal, i.e. it is not generated by one element.

And this is the source of our troubles!

The passage from numbers to ideals is very similar to the passage from the rational numbers to the reals (which Dedekind accomplished in much the same way) in that both are a special case of thinking about the Yoneda embedding for posets. To think about the reals in terms of the order properties of $\mathbb{Q}$ one embeds $\mathbb{Q}$ in its power set by defining $S_x = \{ a \in \mathbb{Q} | a \le x \}$; this is an order-preserving injection so one can work with these sets instead of the original rational numbers. One then notices that there are “ideal” sets such as $S = \{ a \in \mathbb{Q} | a^2 \le 2 \}$ that are not obtained in this manner but that are nevertheless useful enough that we simply work with them instead. Similarly, starting with, say, $\mathbb{Z}[ \sqrt{-5} ]$ with the divisibility relation one embeds it into its power set by defining $S_x = \{ a \in \mathbb{Z}[ \sqrt{-5} ] | x \text{ divides } a \}$ (it works better this way), and then one notices that there are “ideal” sets such as the kernel of $\phi$ that are not obtained in this manner, etc.

Let’s summarize the situation thus far. The actual proofs of all the things I’m claiming can be found in any good abstract algebra text.

1. Prime factorization in a ring $R$ comes from the ability to construct homomorphisms from $R$ into integral domains.
2. These homomorphisms are governed by subsets of $R$ called ideals which form their kernels. The first isomorphism theorem holds.
3. The image of a homomorphism is an integral domain if and only if its kernel is a prime ideal, which is a better generalization of the notion of primality than irreducibility.
4. Sometimes all prime ideals are of the form “multiples of a prime element,” but sometimes not. In the latter case we should shift our attention from factoring numbers to factoring ideals.

An integral domain in which every ideal is principal is called a principal ideal domain, and all PIDs have unique factorization, although the converse isn’t true: polynomial rings in more than one variable have unique factorization but aren’t PIDs.

In the next post we’ll discuss ideals from a more geometric perspective.

### One Response

1. on December 25, 2011 at 1:30 pm | Reply Jorge

It is possible to completely classify the domains where Unique Factorization doesn’t hold, without study directly the kernel of such homomorphisms?
And in a Domain where prime ideals are not generated by a prime element can we factor it into another prime ideals, generated by primes in the Domain?