Gracious Living

Rings, Integral Domains, and Fields
February 2, 2011, 19:36
Filed under: Algebra, Math | Tags: , , , , , ,

In which I sort of breeze through a couple of really awesome and really important concepts.  Last time, we classified abelian groups — now we’ll see what happens if we require additional structure on the groups.  In particular, I’m going to construct $\mathbb{Z}$ and $\mathbb{Q}$ similarly to how the Peano axioms constructed $\mathbb{N}$.

How to construct $\mathbb{Z}$?  The simplest way is to begin with $\mathbb{N}$ and make a group out of what was once only a monoid.  Addition in $\mathbb{N}$ is already associative and has an identity, so all we need to do is adjoin an inverse for every element: for every $n\in\mathbb{N}$, we just define $-n$ by $-n+n=0$.  To make this work with the Peano definitions, we’ll have to have $S(-Sn)+n=-Sn+Sn=0$, so that $S(-Sn)=-n$.  Note that addition is already commutative, so the group we end up with is abelian.  What’s more, it’s the free abelian group on a single generator.

Another way to construct $\mathbb{N}$ and $\mathbb{Z}$, due to Linderholm’s Mathematics Made Difficult, is as universal objects in some suitably defined category.  Don’t worry about what the last sentence means, here’s how to do it: define a pointed monoid to be a monoid (remember, this is a set with an associative operation and identity) with one of its elements labeled as the “basepoint.”  A homomorphism of pointed monoids is then a monoid homomorphism $f:M\rightarrow N$ sending the basepoint of $M$ to the basepoint of $N$.  The universal pointed monoid, if it exists, is a pointed monoid $A$ such that given another pointed monoid $M$, there is a unique homomorphism from $A$ to $M$.  The beauty of this is that if the universal pointed monoid does exist, it is necessarily unique: if $A$ and $B$ are both universal, we must have unique homomorphisms $f:A\rightarrow B,g:B\rightarrow A$.  Then $g\circ f:A\rightarrow A$ is a homomorphism, but so is the identity — and since there is only one map from $A$ to any other pointed monoid, including itself, $g\circ f$ is the identity.  Likewise, $f\circ g$ is the identity on $B$.  So $f$ and $g$ are isomorphisms.  Up to isomorphism, we’re guaranteed uniqueness.

Okay, now I claim that $\mathbb{N}$, with $1$ as the basepoint, is the universal pointed monoid.  Let $M$ be any other pointed monoid, written additively.  Any monoid homomorphism must send $0\in\mathbb{N}$ to the identity of $M$, and if $m$ is the basepoint of $M$, then $1\mapsto m$, so $2\mapsto 2m=m+m,3\mapsto 3m$, and so on.  Clearly, this map always exists, and it’s the only possible one from $\mathbb{N}$ to $M$.  Likewise, $\mathbb{Z}$ is the universal pointed abelian group, an argument which I leave to you.  Notice also that if you go back to unpointed homomorphisms, there is a unique monoid homomorphism from $\mathbb{N}$ to $M$ for each element of $M$, and a unique abelian group homomorphism from $\mathbb{Z}$ to $A$ for each element of $A$. So elements of monoids/abelian groups can be treated as maps from $\mathbb{N}$ or $\mathbb{Z}$. Clever, huh?

I mentioned in the last post a piece of additional structure on abelian groups, that seemed to fall out from our notation for them.  Given $a\in A$, where $A$ is an abelian group, we have elements like $-5a,2a,3a,0a,\dotsc$.  We can think of this as a “multiplication” map $\mathbb{Z}\times A\rightarrow A$.  This map is:

• Associative: we have $(nm)a=n(ma)$.
• Distributive: $(n+m)a=na+ma$, and $n(a+b)=na+nb$.
• Unital: $1a=a$ by definition.
• Inverse-preserving: $(-n)a=-(na)$.
• and it has a zero: $0a=0$.

The case we’re interested in is $A=\mathbb{Z}$, so this is a map $\mathbb{Z}^2\rightarrow\mathbb{Z}$.  In addition to the above properties, this map is now also

• Commutative: $nm=mn$.

So by virtue of its structure as an abelian group, $\mathbb{Z}$ has a built-in multiplication!  An abelian group with an associative, distributive multiplication map is a ring.  Right now, we’re only interested in commutative rings with unit, meaning they satisfy the “commutative” and “unital” conditions as well.  (Exercise: why haven’t I said anything about “inverse-preserving” or “zero-having”?)

Lest other abelian groups feel left out, some of them have similarly built-in multiplication.  In the group $\mathbb{Z}/n\mathbb{Z}$, if $a\in\mathbb{Z}$ and $b\in\mathbb{Z}/n\mathbb{Z}$, we have $(a+n)b=ab+nb=ab+0=ab$.  So multiplication by $\mathbb{Z}$ is periodic as well, with period $n$.  But this means that we might as well consider it to be multiplication by $\mathbb{Z}/n\mathbb{Z}$, since all the equivalence classes modulo $n$ agree on multiplication.  So $\mathbb{Z}/n\mathbb{Z}$ gets an automatic ring structure as well.

A little investigation might reveal a way in which $\mathbb{Z}$ is better than other rings.  It’s not the only way, but it’s awful important.  In $\mathbb{Z}/6\mathbb{Z}$, we have $3\cdot 2=6=0$.  Unlike in the case of the integers, here it’s possible to multiply two nonzero numbers and get zero back.  We call such numbers zero-divisors.  In fact, the situation is even worse: $3\cdot 2=3\cdot 4=3\cdot 0=0$.  This means that we can’t even cancel: the equation $3x=0$ has three solutions, and $3x=3y$ does not imply $x=y$.  (Exercise: a ring has zero-divisors iff it has “cancellation errors” like the one above.)

We call a ring without zero-divisors an integral domain.  “Integral” after the integers, no fucking clue about “domain.”  Integral domains allow you to cancel common multiples, and thus find unique solutions to linear equations like $ax=y$ — or say that they have no solutions.  One of the ways in which they’re useful is that we can pack division into them.  For equations $ax=y$ which have no solutions in the integral domain (and for which $x\ne 0$), we can extend the ring to a larger one which gives a solution.  Here’s how:

Consider the set of ordered pairs $(a,b)$ of ring elements.  Define the fraction $a/b$ to be the equivalence class of such pairs under the relation $(a,b)\sim (c,d)$ iff $ad-bc=0$.

(Here’s why we can only do this to integral domains: to check transitivity, assume $(a,b)\sim (c,d)\sim (e,f)$.  Then $ad-bc=0=cf-de$.  We want $af-be=0$, and we might attempt to get this by cross-multiplying: $adef=bcef=bde^2$.  But to get from this to $af=be$, we have to cancel $de$, which we can only do in an integral domain.  There is a way to introduce a weaker set of fractions into arbitrary rings, which I’ll describe later.)

Okay, so a fraction $a/b$ is an equivalence class of pairs $(a,b)$ with $ad-bc=0$ defining the equivalence relation between $(a,b)$ and $(c,d)$.  We now define $(a/b)(c/d)=ac/bd,(a/b)+(c/d)=(ad+bc)/bd$.  I leave it to you to show that these are well-defined and ring operations.  But in addition, every nonzero element is invertible: $(a/b)(b/a)=1$.

We call a ring in which every nonzero element is invertible a field.  A field has two groups associated to it: the additive group $F$ or $F^+$, and the multiplicative group $F^\times$ or $F^*$ of nonzero elements.  By contrast, in arbitrary integral domains, the multiplicative group is just a monoid.  In a field, division is defined: division by $a$ is just multiplication by $1/a$.

The field we just constructed was the field of fractions of an integral domain.  This is important for the following reason: any field containing a given integral domain $A$ must contain $1/a$ for all $a\in A-\{0\}$.  But then, by multiplication, it also contains $b/a$ for all $b\in A,a\in A-\{0\}$.  That is, it contains the entire field of fractions — you can check that the ring operations must be the same.  The field of fractions is thus the “universal field” among fields containing the given integral domain.  As you may have guessed, the field of fractions of $\mathbb{Z}$ is $\mathbb{Q}$!  So now we have a universal way of constructing $\mathbb{Q}$ as well.

Let’s look at some more interesting fields.  Notice that $\mathbb{Z}/n\mathbb{Z}$ only has zero-divisors if $n$ is composite.  If it is prime, on the other hand, the ring must be an integral domain.  We could try forming the field of fractions of $\mathbb{Z}/p\mathbb{Z}$, but we’d end up surprised — the field of fractions is the ring itself — the ring is already a field!  For every equation $ax=y$ for $x\ne 0$, there is always an $a$ satisfying the equation.  We call this structure the finite field of order $p$, written $\mathbb{F}_p$.  In fact, there are (unique) finite fields of all orders $p^s$ for $p$ a prime and $s\ge 1$.

So to sum up, we started with abelian groups — the building blocks of addition — and found that some of them had natural multiplication structures as well. We invented rings to generalize this. For integral domains, we found that we could introduce fractions in such a way that division made sense, and we called this structure a field. Along the way, we constructed $\mathbb{Z}$ and $\mathbb{Q}$. All of these topics deserve more attention, but hopefully this introduction was interesting.