We will attempt to give a brief explanation of the following concepts:

- N is a monoid
- Z is an integral domain
- Q is a field
- in the field R the order is complete
- the field C is algebraically complete

If you have been asked by a child to give them arithmetic problems,
so they could show off their newly learned skills in addition and
subtraction I'm sure that after a few problems such as: *2 + 3*,
*9 - 5*, *10 + 2* and *6 - 4*, you tried tossing them
something a little more difficult: *4 - 7*
only to be told ``* That's not allowed.*''

What you may not have realized is that you and the child
did not just have different objects in mind (negative numbers)
but entirely different * algebraic systems*. In other
words a set of objects (they could be natural numbers, integers or
reals)
and a set of operations, or rules regarding how the numbers can be
combined.

We will take a very informal tour of some algebraic systems, but before we define some of the terms, let us build a structure which will have some necessary properties for examples and counterexamples that will help us clarify some of the definitions.

We know that any number that is divided by six will either leave a
remainder,
or will be divided exactly (which is after all the remainder 0). Let us
write any number by the
remainder *n* it leaves after division by six, denoting it as
*[ n ]*. This means that, *7, 55* and *1* will all
be written *[1]*, which we call the * class* to which they all
belong:
i.e. *7 in [1]*, *55 in [1]*, or,
a bit more technically, they are all equivalent to 1 modulo 6.
The complete set of class will contain six elements, and this is called
partitioning numbers into equivalent classes because it separates
(or partitions) all of our numbers into these classes, and any one
number
in a class is equivalent to any other in the same class.

One interesting thing we can do with these classes is to try to add or
to
multiply them. What can *[1] + [3]* mean? We can, rather naively try
out
what they mean in ``normal'' arithmetic: *[1] + [3] = [1 + 3] = [4]*.
So far so good, let us try a second example *25 in [1]* and *45 in [3]*,
their sum is *70* which certainly belongs to *[4]*. Here we see
what we meant above by equivalence, 25 is equivalent to 1 as far
as this addition is concerned. Of course
this is just one example, but fortunately it can be proven that
the sum of two classes is always the class of the sums.

Now this is the kind of thing we all do when we add hours for example, 7 (o' clock) plus 6 hours is 1 (o' clock), and all we are really doing is adding hours (modulo 12).

The neat part comes with multiplication, as we will see later on. But
for now just remember, it can be proven that
something like *[4] x [5] = [2]* will work: the product of two
classes is the class of the product.

Now for some of the necessary terminology.

We need to define a * group*.

Let us take a set of objects and a rule (called a binary operation) which allows us to combine any two elements of this set. Addition is an example from math, or ANDing in some computer language.

The set must be closed under the operation. That means that when two elements are combined the result must also be in the set. For example the set containing even numbers will always give us an even number when two elements are added together. But if we restrict ourselves to odd numbers, their sum is not an odd number and so we know right off the bat that the set of odd numbers and addition cannot constitute a group. Some books will consider closure in the definition of binary operation, and others add it as one of the requirements for a group along with the ones that follow below.

The set and the operation is called a group if the binary operation satisfies the following criteria:

- the operation is associative, which means it doesn't matter how
you
group the elements you are operating on, for example in our set of
remainders:
*[1] + ([3] + [4]) = ([1] + [3]) + [4]* - there is an identity element, meaning: one of the elements combined with the others in the set doesn't change them in the least. For example the zero in addition, or the one in multiplication.
- every element has an inverse with respect to that operation. If you combine an element and its inverse you get the identity (of that operation) back.

(Be careful with this last one, *-3* is the inverse of *3* in addition,
since they give us 0 when added, but *1/3* is the inverse of 3
with respect to multiplication, since *3 x 1/3 = 1* the identity
under
multiplication.)

So we can see that the set of natural numbers N
(with the operation of
addition) is not even a group, since there is no inverse for 5, for
example.
(In other words there is no natural
number which added to 5 will give us zero.) And so the third rule for
our
operation is violated.
But it still has * some* structure, even if it is not as rich as the
ones we'll see later on.

Sets with an associative operation (the first condition above) are called semigroups, and if they also have an identity element (the second condition) then they are called monoids.

Our set of natural numbers under addition is then
an example of a monoid, a structure that is not quite a group because
it
is missing the requirement that every element have an inverse under the
operation (Which is why in elementary school *4 - 7* is not allowed.)

What about the set of integers, is it a group?

By itself this question is nonsensical. Why? Well, we have not mentioned under what operation. OK, let us say: the set of integers with addition.

Now, addition is
associative, the zero does not change any number when added to it, and
for
every number *n* we can add *-n* and get zero. So it's a group all
right.

In fact it is a special kind of group. When we can perform the
operation
on the two elements in any order (e.g *a + b = b + a*) then the group
is called commutative, or * Abelian* in honor of Abel. Not
every operation is commutative, for example three minus two is
certainly
* not* the same as two minus three. Our set of integers under
addition
is then an Abelian group.

If we take an Abelian group (remember: a set with a binary operation) and we define a second operation on it we get a bit more of a structure than we had with just a group.

If the second operation is associative, and
it is distributive over the first then we have a ring. Note that the
second
operation may not have an identity element, nor do we need to find an
inverse
for every element with respect to this second operation. As for
what distributive means, intuitively it is what we do in math when
perform
the following change: *a x (b + c) = (a x b) + (a xc)*.

If the second operation is also commutative then we have what is called a commutative ring. The set of integers (with addition and multiplication) is a commutative ring (with even an identity - called unit element - for multiplication).

Now let us go back to our set of remainders. What happens if we
multiply
*[5] x [1]*? We see that we get *[5]*, in fact we can see a number
of things according to our definitions above, *[5]* is its own inverse,
and
*[1]* is the multiplicative element. We can also show easily enough (by
creating a complete multiplication table) that it is commutative. But
notice
that if we take *[3]* and *[2]*, neither of which are equal to the
class
that the zero belongs to *[0]*, and we multiply them, we get *[3]x[2] = [0]*. This bring us to the next definition. In a commutative
ring,
let us take an element which is not equal to zero and call
it *a*. If we can find a non-zero element, say *b* that combined with
*a* equals zero ( *a x b = 0*) then *a* is called a * zero
divisor*.

A commutative ring is called an integral domain if it has no zero divisors. Well the set Z with addition and multiplication fullfills all the necessary requirements, and so it is an integral domain. Notice that our set of remainders is not an integral domain, but we can build a similar set with remainders of division by five, for example, and voilà, we have an integral domain.

Let us take, for example, the set Q of rational numbers with addition and multiplication - I'll leave out the proof that it is a ring, but I think you should be able to verify it easily enough with the above definitions. But to give you a head start, notice the addition of rationals follow all the requirements for an abelian group. If we remove the zero we will have another abelian group, and that implies that we have something more than a ring, in fact, as we will see in the next section.

Now we can make one step further. If the elements of a ring, excluding
the
zero, form an abelian group (with the second operation) then it is a
field.
For example, write the multiplication table of the remainders of
division
by 5, and you will see that it satisfies all the requirements for a
group:
(You will probably have noticed that the group does not contain the
number
five itself since *[5] = [0]*.)
*(tabular)(c | c c c c) 1 2 3 4 ; 1 2 3 4 ; 2 2 4 1 3 ; 3 3 1 4 2 ; 4 4 3 2 1(tabular)*

(Why isn't the set of divisors of six - excluding the zero and under
multiplication - a group? That's easy enough, since we have excluded
the
zero we do not have the result of *[2] x [3]* in our set, so it
isn't
closed.)

Given a ring, we can say that it is ordered when you have a special subset of that ring behaves in a very special way. If any two elements of that special subset are added or multiplied their sum and their product are again in the special subset. Take the negative numbers in R , can they be that special subset? Well the sum seems to be allright, it is also a negative number. But things don't work with the product: it is positive. What about the positive numbers? Yep, and in fact we call that special subset, the set of positive elements. Now, we gave the definition for an ordered ring, we can also define an ordered field the same way.

But what does a complete ordered field mean? Well the definition looks rather nasty: it is complete if every non-empty subset which posesses an upper bound has a least upper bound.

Let's translate some of that, trying to lose as little information on the way. A bound is something that guarantees that all of the elements of your set are on one side of it (reasonably enough). For example, certainly all negative reals are less than 100, so 100 is a bound (it is in fact an upper bound 'cause all negatives are ``below'' it). But there are lots of other bounds, 1, 5, 26 will all do nicely. The question now is, of all of these (upper bounds) which is the smallest, that is which one is ``the border'' so to speak? Does it always exists?

Let's take the rationals, and look at the following numbers:

*1.4, \; 1.41, \; 1.414, \; 1.4142, \; 1.41421, ... *

Now each of these is a rational number (it can be written as a
fraction),
and they are getting closer and closer to a number we've probably seen
before (just take out your calculator and find the square root of two).
So we can write the shorthand for this series as *sqrt(2)*.
Certainly we can find an upper bound for this series, *3* will do
nicely,
but so can *1.5*, or *1.42*. But what is the smallest. Well there isn't
any. Not among the rational at least, because no matter what fraction
you give I can give you one closer to the square root of two. What
about the square root of two itself? Well it's not a rational number
(I'll skip the proof, but it is really rather easy) so you can't use
it.
If you want another series which is really neat look at the section on
``Euler's formula'' in the FAQ.

And that is where the reals come in. Any set or reals that is bounded
you can certainly find the smallest of these bounds. (By the way
this ``least upper bound'' is abbreviated ``l.u.b.'', or ``sup'' for
* supremum*.) We can also turn things around and talk of lower
bounds, and of the largest of these etc. but most of that will be
just a mirror image of what we have dealt with so far.

So that should be it. And for years that did seem to be it, we seemed to have all the numbers we'd ever care to have.

There was just one small stick in the works, but most people just sort
of pretended not to notice, and that was that not all polynomials had
solutions. One simple polynomial of this kind is *x^2 + 1 = 0*.
It's so simple, yet there's no self respecting number that would
solve this polynomial. There were these funny
answers which seemed like they should be solutions but no one could
make any sense out of them, so they were considered imaginary
solutions.
Which was really too bad because they were given the name of imaginary
numbers
and now that the name stuck we realize that they are numbers just as
good as
any of the ones we have been using for centuries. And in fact that
takes us to
the last great pinnacle in this short excursion. The field of complex
numbers.

We can define an algebraically closed field as a field where every
nonconstant
polynomial (i.e. one with an *x* in it from high school days) has a
zero
in the field. Whew! This in short means that as long as the polynomial
is not a constant number (which is no fun anyways) but something which
looks like it wants a solution, like *5 x^3 - 2 x^2 + 6 = 0* it will
always have one, if you are working with complex numbers and not just
reals.

There is another definition which is probably just as good, but may or
may not be easier: A field is algebraically closed if every polynomial
splits
into linear factors.
Linear factors are briefly factors not containing *x* to any power of
two or higher, in other words in the form: *ax + b*. For example
*x^2 + x - 6* can be factored as *(x + 3)(x - 2)*, but if we are in the
field of reals we cannot factor *x^2 + 1*, but we can in the field of
complex numbers: *x^2 + 1 = (x - i)(x + i)*, where, you may recall,
*i^2 = -1*.

Fri Feb 20 21:45:30 EST 1998