## Basic Notions part 2 – Ideals and the like!

Every once in a while mathematicians have a breakthrough, and this completely revolutionizes maths…opening up a completely new way to view things. It is my aim in this post to explain one of these moments in history!

We are all familliar with how factorisation works in $\mathbb{Z}$. Take an integer, we may factorise it into primes…and this factorisation is unique upto a few sign changes and reordering of the primes in the factorisation. This uniqueness is very important, it allows us to prove many things in elementary number theory since the prime numbers behave very nicely.

We would like to have a similar thing in the ring of integers of any number field, because as we have seen in the past, we can use rings of integers to solve many Diophantine equations and to analyse other problems.

In this case the role of the “sign changes” is taken up by the units of the ring (in $\mathbb{Z}$ the units are exactly $\pm 1$). The role of the prime numbers is taken by the prime/irreducible elements in the ring (in the case of $\mathbb{Z}$ it turns out that primes and irreducibles are the same thing!).

So do we always have uniqueness of factorisation into irreducibles in $\mathfrak{O}_K$? The answer is, unfortunately…no.

In some sense the uniqueness of factorisation in $\mathbb{Z}$ is dictated by the fact that it is a Euclidean Domain (i.e. it is a place where we have a well defined Euclidean algorithm). This is a nice class of integral domains, lying inside the class of Principal Ideal Domains (where the ideals are generated by a single element…this will make sense soon), which also lies inside the class of Unique Factorisation Domains (places where uniqueness of factorisation exists upto multiplication by units and reordering of irreducibles).

The most famous example of non-uniqueness of factorisation is the fact that in $\mathbb{Z}[\sqrt{-5}]$ we have that $6 = 3\times 2 = (1+\sqrt{-5})(1-\sqrt{-5})$. These two factorisations are genuinely different since the units here are $\pm 1$ and all $4$ terms ARE irreducibles in the ring.

The problem is that $\sqrt{3}$ doesn’t lie in the ring…if it did then both sides would factorise further into the same things (you can check this).

How might we go about restoring unique factorisation? Well the natural thing to want to do is to extend the ring so that we get enough extra elements to make all factorisations work uniquely. This, as you can imagine, is quite a messy procedure (how will you know when you have accounted for all extra elements?).

The breakthrough came via Dedekind (following on from work by Kummer). The idea is not to work with elements, but with sets of multiples of elements as objects in their own right!

For example we all know that in $\mathbb{Z}$$6 = 3\times 2, 12 = 3\times 4, 18 = 3\times 6$, and so on. Well really we can sum up all of this behaviour by simply saying that “the multiples of $6$” can all be expressed as “a multiple of $2$” times “a multiples of $3$“. We can write this as $6\mathbb{Z}$ $=$ $(2\mathbb{Z})(3\mathbb{Z})$. There is no other way to factorise a multiple of $6$ because intuitively you must use the primes $2$ and $3$!

Dedekind was the first to invent the notion of an ideal of a ring, following on from Kummers ideal numbers. The structures $6\mathbb{Z}, 2\mathbb{Z}$ and $3\mathbb{Z}$ in the above are all examples of ideals of $\mathbb{Z}$.

So what is an ideal? Well it is meant to capture this multiples business. It should also agree with divisibility properties. Let’s think about the multiples of $6$ again. Well this is certainly a subgroup of $\mathbb{Z}$ under addition. Also we may multiply a multiple of $6$ by ANY integer and still get a multiple of $6$.

This motivates the general definition. An ideal of a ring $R$ is an additive subgroup $I$ such that $ra\in I$ for all $r\in R, a\in I$.

The ideals of $\mathbb{Z}$ are the sets $n\mathbb{Z}$ (these are the only additive subgroups and they all share the multiplicativity property). This ties in with the “multiples” business earlier.

However not all ideals of a ring are generated by one element (i.e. not all ideals are principal ideals). Infact it is possible to show that unique factorisation into irreducibles occurs if and only if EVERY ideal is principal (i.e. if we are working in a principal ideal domain (PID)). In fact in any ring where you have a Euclidean Domain, you automatically have a PID and so automatically have uniqueness of factorisation into irreducibles!

This kinda explains the non-uniqueness of factorisation in certain rings (for example in rings of integers)…some of the ideals are representing the multiples of some “hidden element” that is not actually in the ring!

Anyway, I am not going to go into too much detail on the ring theory behind ideals…any decent algebra text will give you that.

The interesting thing we get from turning to ideals is the breakthrough that I alluded to at the beginning of this post…namely that in certain integral domains (namely the Dedekind domains, which the ring of integers always belongs to), we have unique factorisation of ideals into prime ideals $\mathfrak{a} = \mathfrak{p}_1^{e_1} ... \mathfrak{p}_g^{e_g}$. Further this factorisation only depends on the ordering of the prime ideals! There is no trouble with units anymore!

The notion of prime ideal corresponds nicely with the notion of prime in $\mathbb{Z}$. The formal definition is that an ideal $\mathfrak{p}$ is prime if whenever a product $ab\in\mathfrak{p}$ we must have that either $a\in\mathfrak{p}$ or $b\in\mathfrak{p}$.

(If we return to our notion that an ideal represents “multiples” then this notion of prime ideal corresponds to the notion of prime number in $\mathbb{Z}$. Indeed $ab\in\mathfrak{p}$ translates to “if $p|ab$” and $a\in\mathfrak{p}$ or $b\in\mathfrak{p}$ translates to “then $p|a$ or $p|b$“.)

Everything fits in nicely here. But how do we explain the example of non-unique factorisation we had above? Well it turns out that you get the two different factorisations of $6$ by simply reordering the ideals in the prime ideal factorisation! (See the book by Stewart/Tall or equivalently search google to see this done explicitly).

Now it turns out that in proving this uniqueness of factorisation we have to go a little bit further and invent so called fractional ideals. You see the set of ideals of $\mathfrak{O}_K$ do not form a group but form something very close to a group. The inverses are a problem since things in a ring aren’t necessarily invertible. However if we look instead at $\mathfrak{O}_K$-submodules of $K$ (i.e. the posh way of saying “allow fractional generators”) then we do get a group!

In fact in algebraic number theory we use this group quite a lot. The main way we use it is to form the Ideal class group of the number field $K$. This group creates a measure of the non-uniqueness of factorisation into irreducibles in $\mathfrak{O}_K$ in quite a clever way.

The basic idea is as follows. We know that factorisation into irreducibles is unique if and only if every ideal is principal. Thus, if we take the group of all fractional ideals, denoted $I_K$ and mod out by the principal ones, denoted $P_K$, we will get a group $I_K / P_K = Cl_K$ (called the ideal class group). whose non-identity classes each represent an essentially different obstacle to unique factorisation. So we are taking a fractional ideal, factorising it into prime ideals and shedding away the principal bits (since they do not contribute to the non-uniqueness of factorisation into irreducibles).

Alternatively the ideal class group can be made via an equivalence relation placed on $I_K$.

The amazing thing about the ideal class group is that no matter what number field you work over, $Cl_k$ is ALWAYS finite! In other words there are only finitely many obstacles to unique factorisation of elements of $\mathfrak{O}_K$. This is quite a difficult thing to prove without going into an entire realm of number theory where we study the “geometry of numbers”.

We call the order of the ideal class group the class number of $K$, denoted $h_K$.

Using a few bits of group theory we can find out a few facts about how the ideals in $\mathfrak{O}_K$ behave. For example, take any ideal $\mathfrak{a}$ of the ring of integers. Then automatically $\mathfrak{a}^{h_K}$ is a principal ideal. This is none other than the corollary of Lagrange’s theorem (the order of the ideal class group is $h_K$, a finite number).

Also say we find that $\mathfrak{a}^n$ is principal for some $n$ coprime to $h_K$. Then $\mathfrak{a}$ itself is automatically principal by Lagrange’s theorem!

Kummer was the first to notice how the ideal theory ties in with Fermat’s last theorem.  Following on from the work of Lame (which assumed uniqueness of factorisation in cyclotomic fields…a mistake), Kummer decided to turn to ideals. Factorising the equation in $\mathbb{Q}(\zeta_p)$ as $\prod_{i=0}^{p-1} (x - \zeta_p^i y) = z^p$ and taking ideals of both sides tells us (by unique factorisation of ideals) that there are ideals $\mathfrak{a}_i$ such that $\langle x - \zeta_p^i y\rangle$ $=$  $\mathfrak{a}_i^p$ for each $i$.

Kummer immediately realised the significance of the class number here…since to be able to conclude that $\mathfrak{a}_i$ is principal, we would need to know that $p$ is coprime to the class number of $\mathbb{Q}(\zeta_p)$. He defined a regular prime to be a prime satisfying this property. From this Kummer was able to prove FLT for all regular prime exponents! This settled a big amound of FLT, although noone knows whether there are infinitely many regular primes!

This concludes the post, hopefully next time I will talk about how in most cases we can get the factorisation into prime ideals via the minimal polynomial of the generator for the number field and how such factorisations work in extensions of number fields (specifically in Galois extensions).