This part of the module is about generalizing what we know about matrices and vectors.
When we talk about vectors, or matrices, there’s an important thing we have to decide: where do the entries come from? For example, we might work with matrices with real numbers as entries, or with complex numbers as entries. We never really discussed this in the first part of the module, because it doesn’t make any difference to the theory we developed. So it’s natural to ask which other kinds of numbers we could use as entries in our vectors and matrices and still have everything work OK.
The answer is that the entries must come from what is called a field. Roughly speaking, a field is a set with multiplication and addition operations that obey the usual rules of algebra, and where you can divide by any non-zero element. Examples are , the set of all real numbers, , the set of all complex numbers, , the set of all rational numbers. Non-examples are : you can’t divide by 2 in , and , the set of all real matrices, again because we know there are non-zero matrices which aren’t invertible.
The usual way to define a field is to write down a list of axioms. Everything that satisfies the axioms is a field. If you do want to know the field axioms, here they are: you can read them here.
The next section is about an important family of fields we have not seen yet.
Let be a prime number. Then , the finite field of order , is the set with addition and multiplication exactly the same as for ordinary whole numbers, except that we regard all multiples of as equal to 0.
This is easier to understand in an example. Let’s work in . Then
because in . Similarly,
,
.
We can deal with subtractions and negative numbers too:
.
There’s an easy way to summarize all the possible additions and multiplications in : draw tables with rows and columns labelled by , and in row and column put the result of or .
+ | 0 | 1 | 2 | 3 | 4 |
---|---|---|---|---|---|
0 | 0 | 1 | 2 | 3 | 4 |
1 | 1 | 2 | 3 | 4 | 0 |
2 | 2 | 3 | 4 | 0 | 1 |
3 | 3 | 4 | 0 | 1 | 2 |
4 | 4 | 0 | 1 | 2 | 3 |
0 | 1 | 2 | 3 | 4 | |
---|---|---|---|---|---|
0 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 1 | 2 | 3 | 4 |
2 | 0 | 2 | 4 | 1 | 3 |
3 | 0 | 3 | 1 | 4 | 2 |
4 | 0 | 4 | 3 | 2 | 1 |
Here are the addition and multiplication tables for .
+ | 0 | 1 | 2 |
---|---|---|---|
0 | 0 | 1 | 2 |
1 | 1 | 2 | 0 |
2 | 2 | 0 | 1 |
0 | 1 | 2 | |
---|---|---|---|
0 | 0 | 0 | 0 |
1 | 0 | 1 | 2 |
2 | 0 | 2 | 1 |
In a field, you have to be able to divide by non-zero elements. Being able to divide by is equivalent to the existence of an element or , called a multiplicative inverse to (because is just ). So to check that we can divide, we need to check that for each nonzero there is another element which you can multiply by to get 1. We will do this for the examples of and .
In , you can see from the table that
, so
, so
and so every non-zero element of has a multiplicative inverse.
In , you can see from the table that
, so
, so and
, so .
and so every non-zero element of has a multiplicative inverse.
We’re going to give a proof that every element of has a multiplicative inverse and give a method for finding these multiplicative inverses.
First, here is a reminder about division with remainder. There is a fundamental fact about the natural numbers: if you have any two positive whole numbers and , you can try to divide by . When you do that you get a quotient and a remainder , which is a nonnegative integer less than :
It’s easy to convince yourself of the truth of this by imagining starting at zero on the number line and taking steps of size to the right, stopping when the next step would take you past . Call the number of steps taken at the point you stop . You’ve stopped at the number , and the distance from this point to , which is the remainder , must be less than since otherwise you could take another full step without passing .
Let be prime and be a whole number. Then there are whole numbers and such that .
Let be the smallest positive number that can be written as an integer multiple of plus an integer multiple of , so for some integers and . Note that since is a positive number that can be written this way. Divide by to get quotient and remainder , so where . Then
Since and is the smallest positive number that can be written as a multiple of plus a multiple of we must have , that is, divides the prime number . Since we have . ∎
We can even calculate numbers and satisfying quickly. To do this divide by , getting a quotient and a remainder
(we start with a subscript 2 here because we will think of as and as ). If , stop. Otherwise divide by getting a quotient and remainder
If , stop. Otherwise divide by getting a quotient and remainder :
If , stop. Otherwise divide by getting quotient and remainder :
and so on. The sequence satisfies
You can’t have a decreasing sequence of positive integers that goes on forever, so there is some such that . The last few divisions were
(4.1) | ||||
(4.2) | ||||
(4.3) | ||||
(4.4) |
We claim that . For divides , because of the last equation. In (4.3) the terms on the right hand side are multiples of , so the left hand side is a multiple of as well. Repeating the same argument we end up with dividing all the left-hand-sides, in particular, divides the prime number . Since we have . So (4.3) is really
or equivalently
(4.5) |
Now we can express in terms of s with a smaller subscript with equation (4.1), and we can express in terms of s with a smaller subscript using (4.2). When we substitute this in (4.5), we get some multiple of plus some multiple of . And we can keep doing that over and over again until we eventually get a multiple of plus a multiple of , that is, for some whole numbers and .
It helps to see an example. Take and . We have
and so
This helps us find multiplicative inverses in because if then in we have, since multiples of are equal to zero, and is the multiplicative inverse of . In our example, , so the multiplicative inverse of in is which is the same as .