# The Unapologetic Mathematician

## Different kinds of rings

There are a number of different kinds of rings differentiated (sorry) by properties of their multiplications. Most of them lead into their own specialized areas of study. I mentioned that a ring may or may not be commutative, and it may or may not have an identity, but there are a few more that will be useful.

One initially counterintuitive idea is that it’s entirely possible that a ring has “zero divisors”: two nonzero elements that multiply to give zero. Imagine starting with two copies of the integers, $\mathbb{Z}$ and $\bar{\mathbb{Z}}$, writing elements of the second copy as integers with a bar over them. Now consider pairs of elements, one from each copy, $(a,\bar{b})$. Add pairs by adding the two components, but multiply them like this: $(a,\bar{b})(c,\bar{d})=(ac,\bar{ad+bc})$
Notice that the product of any two elements of $\bar{\mathbb{Z}}$ is zero! Weird. Eerie.

To be explicit: an element of this ring coming from $\bar{\mathbb{Z}}$ is $(0,\bar{a})$. We calculate the product: $(0,\bar{a})(0,\bar{b})=(00,\bar{0b+a0})=(0,\bar{0})$

So, any element $a$ for which there is a $b$ so that $ab=0$ is called a left zero divisor. Right zero divisors are defined similarly. If a ring has no zero divisors, so the product of two nonzero elements is always nonzero, we call it an “integral domain”. The integers are just such an integral domain, fittingly enough.

Now if a ring has a multiplicative identity we can start talking about multiplicative inverses. We say an element $a$ has a left inverse $b$ if $ba=1$, or a right inverse $c$ if $ac=1$. If a ring has both a left and a right inverse they’re the same, since $b=b1=b(ac)=(ba)c=1c=c$
In this case we call $a$ a unit and write its inverse as $a^{-1}$. We can also see that an element having a left (right) inverse cannot be a left (right) zero divisor: $ax=0\Rightarrow x=1x=(ba)x=b(ax)=b0=0$
If every nonzero element of a ring is a unit, we call it a division ring.

In the case of commutative rings, all these distinctions between “left” and “right” (zero divisors, inverses, etc.) disappear, since multiplication doesn’t care about the order of the factors. We actually have a special name for a commutative division ring: we call it a “field”, though everyone else in the world except the Belgians seems to call it a “(dead) body” (körper, corps, поле, test, lichaam, …).

[EDIT: added explicit calculation verifying that elements from $\bar{\mathbb{Z}}$ in the example are zero-divisors.]

March 29, 2007 Posted by | Ring theory | 3 Comments

## The ring of integers

As I mentioned before, the primal example of a ring is the integers $\mathbb{Z}$. So far we’ve got an ordered abelian group structure on the set of (equivalence classes of) pairs of natural numbers. Now we need to add a multiplication that distributes over the addition.

First we’ll figure out how to multiply natural numbers. This is pretty much as we expect. Remember that a natural number is either ${}0$ or $S(b)$ for some number $b$. We define $a\cdot0=0$ $a\cdot S(b)=(a\cdot b)+a$

Firstly, this is commutative. This takes a few inductions. First show by induction that ${}0$ commutes with everything, then show by another induction that if $a$ commutes with everything then so does $S(a)$. Then by induction, every number commutes with every other. I’ll leave the details to you.

Similarly, we can use a number of inductions to show that this multiplication is associative — $(a\cdot b)\cdot c=a\cdot(b\cdot c)$ — and distributes over addition of natural numbers — $a\cdot(b+c)=a\cdot b+a\cdot c$. This is extremely tedious and would vastly increase the length of this post without really adding anything to the exposition, so I’ll again leave you the details. I’m reminded of something Jeff Adams said (honest, I’m not trying to throw these references in gratuitously) in his class on the classical groups. He told us to verify that the commutator in an associative algebra satisfies the Jacobi identity because, “It’s long and tedious and doesn’t add much, but I had to do it when I was a grad student, so now you’re grad students and it’s your turn.”

So now these operations — addition and multiplication — of natural numbers make $\mathbb{N}$ into what some call a “semiring”. I prefer (following John Baez) to call it a “rig”, though: a “ring without negatives”. We use this to build up the ring structure on the integers.

Recall that the integers are (for us) pairs of natural numbers considered as “differences”. We thus define the product $(a,b)\cdot(c,d)=(a\cdot c+b\cdot d,a\cdot d+b\cdot c)$

Our life now is vastly easier than it was above: since we know addition and multiplication of natural numbers is commutative, the above expression is manifestly commutative. No work needs to be done! Associativity is also easy: just set up both triple products and expand out, checking that each term is the same by the rig structure of the natural numbers. Similarly, we can check distributivity, that $(1,0)$ acts as an identity, and that the product of two integers is independent of the representing pair of natural numbers.

Lastly, multiplication by a positive integer preserves order. If $a and $0 then $ac. Together all these properties make the integers as we’ve defined them into a commutative ordered ring with unit. The proofs of all these things have been incredibly dull (I actually did them all today just to be sure how they worked), but it’s going to get a lot easier soon.

March 29, 2007