Different algebraic systems are used in linear algebra. The most
important are * commutative rings with identity*
and * fields*. I'll begin by stating the axioms
for a * ring*. They will look abstract, because
they are! But don't worry --- lots of examples will follow.

* Definition.* A * ring* is a
set R with two binary operations * addition*
(denoted +) and * multiplication* (denoted ).
These operations satisfy the following axioms:

1. Addition is associative: If , then

2. There is an * identity* for addition, denoted
0. It satisfies

3. Every every of R has an * additive inverse*.
That is, if , there is an element which satisfies

4. Addition is commutative: If , then

5. Multiplication is associative: If , then

6. Multiplication distributes over addition: If , hen

It's common to drop the " " in " " and just write " ". I'll do this except where the " " is needed for clarity.

As a convenience, we can define * subtraction*
using additive inverses. If R is a ring and , then is
defined to be . That is, subtraction is defined as adding
the additive inverse.

You might notice that we now have three of the usual four arithmetic operations: Addition, subtraction, and multiplication. We don't necessarily have a "division operation" in a ring; we'll discuss this later.

If you've never seen axioms for a mathematical structure laid out like this, you might wonder: What am I supposed to do? Do I memorize these? Actually, if you look at the axioms, they say things that are "obvious" from your experience. For example, Axiom 4 says addition is commutative. So as an example for real numbers,

You can see that, as abstract as they look, these axioms are not that
big a deal. But when you do mathematics carefully, you have to be
precise about what the rules are. You will not have much to do in
this course with writing proofs from these axioms, since that belongs
in an abstract algebra course. A good rule of thumb might be to try
to understand by example what an axiom says. And if it seems
"obvious" or "familiar" based on your experience,
don't worry about it. *Where you should pay special attention is
when things don't work in the way you expect.*

If you look at the axioms carefully, you might notice that some familiar properties of multiplication are missing. We will single them out next.

* Definition.* A ring R is *
commutative* if the multiplication is commutative. That is, for
all ,

Note: The word "commutative" in the phrase
"commutative ring" *always* refers to
*multiplication* --- since addition is always assumed to be
commutative, by Axiom 4.

* Definition.* A ring R is a *
ring with identity* if there is an identity for multiplication.
That is, there is an element such that

Note: The word "identity" in the phrase "ring with
identity" *always* refers to an identity for
*multiplication* --- since there is always an identity for
addition (called "0"), by Axiom 2.

A commutative ring which has an identity element is called a * commutative ring with identity*.

In a ring with identity, you usually also assume that . (Nothing stated so far requires this, so you have to take it as an axiom.) In fact, you can show that if in a ring R, then R consists of 0 alone --- which means that it's not a very interesting ring!

Here are some number systems you're familiar with:

(a) The integers .

(b) The rational numbers .

(c) The real numbers .

(d) The complex numbers .

Each of these is a commutative ring with identity. In fact, all of
them except are * fields*. I'll
discuss fields below.

By the way, it's conventional to use a capital letter with the
vertical or diagonal stroke "doubled" (as in
or ) to stand for number systems. It is how you would
write them by hand. If you're typing them, you usually use a special
font; a common one is called *Blackboard Bold*.

You might wonder why I singled out the commutativity and identity axioms, and didn't just make them part of the definition of a ring. (Actually, many people add the identity axiom to the definition of a ring automatically.) In fact, there are situations in mathematics where you deal with rings which aren't commutative, or (less often) lack an identity element. We'll see, for instance, that matrix multiplication is usually not commutative.

The idea is to write proofs using *exactly* the properties you
need. In that way, the things that you prove can be used in a wider
variety of situations. Suppose I had included commutativity of
multiplication in the definition of a ring. Then if I proved
something about rings, you would not know whether it applied to
noncommutative rings without carefully checking the proof to tell
whether commutativity was used or not. If you *really* need a
ring to be commutative in order to prove something, it is better to
state that assumption explicitly, so everyone knows not to assume
your result holds for noncommutative rings.

The next example (or collection of examples) of rings may not be
familiar to you. These rings are * the integers mod
n*. For these rings, n will denote an integer. Actually, n can be
*any* integer if I modify the discussion a little, but to keep
things simple, I'll take .

The * integers mod n* is the set

n is called the * modulus*.

For example,

becomes a commutative ring with identity under the
operations of * addition mod n* and * multiplication mod n*. I won't prove this; I'll
just show you how to work with these operations, which is sufficient
for a linear algebra course. You'll see a rigorous treatment of in abstract algebra.

(a) To add x and y mod n, add them as integers to get .
Then *divide* by n and take the *remainder* ---
call it r. Then .

(b) To multiply x and y mod n, multiply them as integers to get .
Then *divide* by n and take the *remainder* ---
call it r. Then .

Since modular arithmetic may be unfamiliar to you, let's do an extended example. Suppose , so the ring is .

Hence, in .

You can picture arithmetic mod 6 this way:

You count around the circle clockwise, but when you get to where "6" would be, you're back to 0. To see how works, start at 0. Count 4 numbers clockwise to get to 4, then from there, count 5 numbers clockwise. You'll find yourself at 3.

Here is multiplication:

Hence, in .

You can see that as you do computations, you might in the middle get numbers outside . But when you divide by 6 and take the remainder, you'll always wind up with a number in .

Try it with a big number:

Using our circle picture, if you start at 0 and do 80 steps clockwise around the circle, you'll find yourself at 2. (Maybe you don't have the patience to actually do this!) When we divide by 6 then "discard" the multiples of 6, that is like the fact that you return to 0 on the circle after 6 steps.

Notice that if you start with a number that is divisible by 6, you get a remainder of 0:

We see that *in doing arithmetic mod 6, multiples of 6 are equal
to 0.* And in general, *in doing arithmetic mod n, multiples
of n are equal to 0.*

Other arithmetic operations work as you'd expect. For example,

Hence, in .

Negative numbers in are * additive
inverses*. Thus, in , because . To deal with negative numbers in general, add a
positive multiple of 6 to get a number in the set . For example,

Hence, in .

The reason you can add 18 (or any multiple of 6) is that 18 divided by 6 leaves a remainder of 0. In other words, " " in , so adding 18 is like adding 0. In a similar way, you can always convert a negative number mod n to a positive number in by adding multiples of n. For instance,

Remember that multiples of 6 (like 18) are 0 mod 6!

Recall that subtraction is defined as adding the additive inverse. Thus, to do in , use the fact that the additive inverse of 2 (that is, -2) is equal to 4:

We haven't discussed division yet, but maybe the last example tells you how to do it. Just as subtraction is defined as adding the additive inverse, division should be defined as multiplying by the multiplicative inverse. Let's give the definition.

* Definition.* Let R be a ring with identity, and
let . The * multiplicative inverse*
of x is an element which satisifies

If we were dealing with real numbers, then ,
for instance. But going back to the example, *we
don't have fractions* in . So what is
(say) in ? By definition, is the element (if
there is one) in which satisfies

(I could say , but multiplication is commutative in , so the order doesn't matter.)

We just check cases. Remember that if I get a product that is 6 or bigger, I have to reduce mod 6 by dividing and taking the remainder.

I got by dividing 25 by the modulus 6 --- it goes in 4 times, with a remainder of 1.

Thus, according to the definition, . In other words, 5 is its own multiplicative inverse. This isn't unheard of: You know that in the real numbers, 1 is its own multiplicative inverse.

This also means that if you want to divide by 5 in , you should multiply by 5.

What about in ? Unfortunately, if you take cases as I did
with 5, you'll see that for every number n in , you do not
have . Here's a * proof by
contradiction* which avoids taking cases. Suppose
. Multiply both sides by 3:

I made the last step using the fact that is a multiple of 6
(since ), and multiples of 6 are equal to 0 mod 6.
Since " " is a contradiction, is
impossible. So is *undefined* in .

It happens to be true that in , the elements 0, 2, 3, and 4 do not have multiplicative inverses; 1 and 5 do.

And in , the elements 0, 2, 4, 5, 6, and 8 do not have multiplicative inverses; 1, 3, 7, and 9 do.

Do you see a pattern?

You probably don't need much practice working with familiar number systems like the real numbers , so we'll give some examples which involve arithmetic in .

* Example.* (a) Reduce 22 to a number in in .

(b) Reduce -21 to a number in in .

(c) Compute in .

(d) Compute in .

(e) Compute in .

(f) Compute in .

(g) Compute

It's understood for a problem your *final* answer
should be a number in . You can simplify as you
do each step, or simplify at the end (divide by n and take the
remainder).

(a) .

(b) .

Notice that 24 is a multiple of 4, so it's equal to 0 in . You can also do this by dividing by 4 if you do it carefully:

(c) .

(d) .

(e) .

Notice that I added a multiple of 10 (since ) to get a positive number.

(f) .

(g) includes all the numbers from 1 to 25; in particular, it includes 23. So the product is a multiple of the modulus 23, and

* Example.* (a) Find in .

(b) Prove that 6 does not have a multiplicative inverse in .

(a) By trial and error, in . Therefore, .

(b) Suppose for some n in . Then

The last step follows from the fact that is a multiple of 10, so it equals 0 mod 10. Since " " is a contradiction, is impossible, and 6 does not have a multiplicative inverse.

* Example.* (a) Show that 2 doesn't have a
multiplicative inverse in .

(b) Show that 14 doesn't have a multiplicative inverse in .

(a) Try all possibilities:

There is no element of whose product with 2 gives 1. Hence, 2 doesn't have a multiplicative inverse in .

(b) Suppose for . Then

(Note that in .) The last line above is a contradiction, so 14 does not have a multiplicative inverse in .

You may have noticed that the elements in which have
multiplicative inverses are the elements which are *
relatively prime* to n.

You might wonder whether there is a systematic way to find
multiplicative inverses in . The best way is to use the * Extended Euclidean Algorithm*; you might see it if
you take a course in abstract algebra. In this course, I'll usually
keep the examples small enough that trial and error is okay for
finding multiplicative inverses when you need them. But here's an
approach that you might prefer. Suppose you want to find
in . Consider *multiples of* 11, plus 1. Stop
with the first such number that's divisible by 7:

From this, I get , because

* Example.* Find in .

In , I have , so . You could do this by trial and error, since isn't that big:

Alternatively, take multiples of 13 and add 1, stopping when you get a number divisible by 8:

Then , so .

Even this approach is too tedious to use with large numbers. The systematic way to find inverses is to use the Extended Euclidean Algorithm.

We saw that in a commutative ring with identity, an element x might
not have multiplicative inverse . That in turn
would prevent you from "dividing" by x. From the point of
view of linear algebra, this is inconvenient. Hence, we single out
rings which are "nice" in that *every* nonzero
element has a multiplicative inverse.

* Definition.* A * field* F
is a commutative ring with identity in which and every nonzero
element has a multiplicative inverse.

By convention, you don't write " " instead of " " unless the ring happens to be a ring with "real" fractions (like , , or ). You don't write fractions in (say) .

If an element x has a multiplicative inverse, *you can divide
by* x *by multiplying by* . Thus, in a field,
you can divide by any nonzero element. (You'll learn in abstract
algebra why it doesn't make sense to divide by 0.)

The rationals , the reals , and the complex numbers are fields. Many of the examples will use these number systems.

The ring of integers is not a field. For example, 2 is
a nonzero integer, but it does not have a multiplicative inverse
*which is an integer*. ( is not an
integer --- it's a rational number.)

, , and are all infinite fields --- that is, they all have infinitely many elements. But (for example) is a field.

For applications, it's important to consider *finite* fields
like . Before I give some examples, I need some
definitions.

* Definition.* Let R be a commutative ring with
identity. The * characteristic* of R is the
smallest positive integer n such that .

Notation: .

If there is no positive integer n such that , then .

In fact, if , then for all .

, , , and are all rings of characteristic 0. On the other hand, .

* Definition.* An integer is *
prime* if its only positive divisors are 1 and n.

The first few prime numbers are

An integer which is not prime is *
composite*. The first few composite numbers are

The following important results are proved in abstract algebra courses.

* Theorem.* The characteristic of a field is
either 0 or a prime number.

* Theorem.* If p is prime and n is a positive
integer, there is a field of characteristic p having
elements. This field is unique up to * ring
isomorphism*, and is denoted (the * Galois field* of order ).

The only unfamiliar thing in the last result is the phrase "ring isomorphism". This is another concept whose precise definition you'll see in abstract algebra. The statement means, roughly, that any two fields with elements are "the same", in that you can get one from the other by just renaming or reordering the elements.

Since the characteristic of is n, the first theorem implies the following result:

* Corollary.* is a field if
and only if n is prime.

The Corollary tells us that , , and are fields, since 2, 3, and 61 are prime.

On the other hand, is not a field, since 6 isn't
prime (because ). In fact, we saw it directly when we
showed that 4 does not have a multiplicative inverse in . Note that *is* a
commutative ring with identity.

For simplicity, the fields of prime characteristic that I use in this
course will almost always be finite. But what would an
*infinite* field of prime characteristic look like?

As an example, start with . Form the * field of rational functions* . Thus, elements of have the
form where and are polynomials with
coefficients in . Here are some examples of elements of
:

You can find multiplicative inverses of nonzero elements by taking reciprocals; for instance,

I won't go through and check all the axioms, but in fact, is a field. Moreover, since in , it's a field of characteristic 2. It has an infinite number of elements; for example, it contains

What about fields of characteristic p other than , , and so on? As noted above, these are called Galois fields. For instance, there is a Galois field with elements. To keep the computations simple, we will rarely use them in this course. But here's an example of a Galois field with elements, so you can see what it looks like.

is the Galois field with 4 elements, and here are its addition and multiplication tables:

Notice that

You can check by examining the multiplication table that multiplication is commutative, that 1 is the multiplicative identity, and that the nonzero elements (1, a, and b) all have multiplicative inverses. For instance, , because .

Since we've already seen a lot of weird things with these new number systems, we might as well see another one.

* Example.* Find the roots of in .

Make a table:

For instance, plugging into gives

The roots are , , and .

You would normally not expect a quadratic to have 4 roots! This shows that algebraic facts you may know for real numbers may not hold in arbitrary rings (note that is not a field).

Linear algebra deals with structures based on fields, and you've now seen most of the fields that will come up in the examples. The modular arithmetic involved in working with may be new to you, but it's not that hard with a little practice. And as I noted, most of the examples involving finite fields will use for p prime, rather than the more general Galois fields, or infinite fields of characteristic p.

Copyright 2020 by Bruce Ikenaga