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1-22-2017
Commutative Rings and Fields
Different algebraic systems are used in linear algebra. The most important are commutative rings
with identity and fields.
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 a, b, c ∈ R, then
a + (b + c) = (a + b) + c.
2. There is an identity for addition, denoted 0. It satisfies
0+a=a
and a + 0 = a
for all a ∈ R.
3. Every every of R has an additive inverse. That is, if a ∈ R, there is an element −a ∈ R which
satisfies
a + (−a) = 0 and (−a) + a = 0.
4. Addition is commutative: If a, b ∈ R, then
a + b = b + a.
5. Multiplication is associative: If a, b, c ∈ R, then
a · (b · c) = (a · b) · c.
6. Multiplication distributes over addition: If a, b, c ∈ R, hen
a · (b + c) = a · b + a · c.
It’s common to drop the “·” in “a · b” and just write “ab”. I’ll do this except where the “·” is needed
for clarity.
You’ll study general rings in an abstract algebra course. The rings that occur in linear algebra satisfy
some additional axioms.
Definition. A ring R is commutative if the multiplication is commutative. That is, for all a, b ∈ R,
ab = ba.
• The word “commutative” in the phrase “commutative ring” always refers to multiplication — since
addition is always assumed to be commutative, by Axiom 4.
A ring R is a ring with identity if there is an identity for multiplication. That is, there is an element
1 ∈ R such that
1 · a = a and a · 1 = a for all a ∈ R.
• The word “identity” in the phrase “ring with identity” always refers to an identity for multiplication —
since there is always an identity for addition, 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 1 6= 0. (Nothing stated so far requires this, so you
have to take it as an axiom.) In fact, you can show that if 1 = 0 in a ring R, then R consists of 0 alone —
which means that it’s not a very interesting ring!
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Example. Here are some number systems you’re familiar with:
The integers Z.
The rational numbers Q.
The real numbers R.
The complex numbers C.
Each of these is a commutative ring with identity. In fact, all of them except Z are fields. I’ll discuss
fields below.
Example. (The integers mod n) For this collection of examples, 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 n ≥ 2.
The integers mod n is the set
Zn = {0, 1, 2, . . . , n − 1}.
n is called the modulus.
For example,
Z2 = {0, 1} and Z6 = {0, 1, 2, 3, 4, 5}.
Zn becomes a commutative ring with identity under the operations of addition mod n and multiplication mod n. I’m going to describe these operations in a functional way, which is sufficient for a linear
algebra course. You’ll see a rigorous treatment of Zn in abstract algebra.
(a) To add x and y mod n, add them as integers to get x + y. Then divide x + y by n and take the
remainder — call it r. Then x + y = r.
(b) To multiply x and y mod n, multiply them as integers to get xy. Then divide xy by n and take the
remainder — call it r. Then xy = r.
This is actually much easier to do than it is to describe! Here’s an example. Suppose n = 6, so the ring
is Z6 .
4+5 = 9
(Add them as integers . . . )
= 3 (Divide 9 by 6 and take the remainder, which is 3)
Hence, 4 + 5 = 3 in Z6 .
2·5 =
=
10
(Multiply them as integers . . . )
4 (Divide 10 by 6 and take the remainder, which is 4)
Hence, 2 · 5 = 4 in Z6 .
Notice that when you take the remainder after dividing by 6, you’ll always wind up with a number in
{0, 1, 2, 3, 4, 5}.
Other arithmetic operations work as you’d expect. For example,
34
=
=
81
(Take the power as an integer . . . )
3 (Divide 81 by 6 and take the remainder, which is 3)
Hence, 34 = 3 in Z6 .
Negative numbers in Z6 are additive inverses. Thus, −2 = 4 in Z6 , because 4 + 2 = 0. To deal with
negative numbers in general, add a positive multiple of 6 to get a number in the set {0, 1, 2, 3, 4, 5}. For
example,
(−3) · 5 =
−15
(Multiply them as integers . . . )
= −15 + 18
(Add 18, which is 3 · 6)
=
3
2
Hence, (−3) · 5 = 3 in Z6 .
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, “18 = 0” in Z6 , so adding 18 is like adding 0.
Example. (Addition and multiplication mod 5) Construct addition and multiplication tables for Z5 .
+
0
1
2
3
4
·
0
1
2
3
4
0
0
1
2
3
4
0
0
0
0
0
0
1
1
2
3
4
0
1
0
1
2
3
4
2
2
3
4
0
1
2
0
2
4
1
3
3
3
4
0
1
2
3
0
3
1
4
2
4
4
0
1
2
3
4
0
4
3
2
1
Notice in the multiplication table that
2·3=1
and 4 · 4 = 1.
This means that 2−1 = 3, 3−1 = 2, and 4−1 = 4. In particular, to divide by 2 you multiply by 3: in Z5 ,
the elements 2 and 3 are reciprocals.
Commutative rings with identity come up in discussing determinants, but the algebraic system of
greatest importance in linear algebra is the field.
Definition. Let R be a ring with identity, and let x ∈ R. The multiplicative inverse of x is an element
x−1 ∈ R which satisifies
x · x−1 = 1 and x−1 · x = 1.
Definition. A field F is a commutative ring with identity in which 1 6= 0 and every nonzero element has a
multiplicative inverse.
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By convention, you don’t write “ ” for “x−1 ” unless the ring happens to be Q, R, or C.
x
If an element x has a multiplicative inverse, you can divide by x by multiplying by x−1 . 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.)
Example. The rationals Q, the reals R, and the complex numbers C are fields. Many of the examples will
use these number systems.
The ring of integers Z is not a field. For example, 2 is a nonzero integer, but it does not have a
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multiplicative inverse which is an integer. ( is not an integer — it’s a rational number.)
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Q, R, and C are all infinite fields — that is, they all have infinitely many elements. For applications,
it’s important to consider finite fields as well. 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 n · 1 = 0. Notation: char R = n.
If there is no positive integer n such that n · 1 = 0, then char R = 0.
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In fact, if char R = n, then n · x = 0 for all x ∈ R.
Z, Q, R, and C are all rings of characteristic 0. On the other hand, char Zn = n.
Definition. An integer n > 1 is prime if its only positive divisors are 1 and n.
The first few prime numbers are
2,
3,
, 5,
7,
11, . . . .
An integer n > 1 which is not prime is composite. The first few composite numbers are
4,
6,
8,
9, . . . .
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 pn elements.
This field is unique up to ring isomorphism, and is denoted GF (pn ) (the Galois field of order pn ).
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 pn elements are “the same”, in that you can get one from the other by just renaming or reordering the
elements.
Since the characteristic of Zn is n, the first theorem implies the following result:
Corollary. Zn is a field if and only if n is prime.
Example. (Fields of prime characteristic) Z2 , Z13 , and Z61 are fields, since 2, 3, and 61 are prime.
On the other hand, Z6 is not a field, since 6 isn’t prime (because 6 = 2 · 3). Z6 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 Z2 = {0, 1}. Form the field of rational functions Z2 (x). Thus, elements
p(x)
where p(x) and q(x) are polynomials with coefficients in Z2 . Here are some
of Z2 (x) have the form
q(x)
examples of elements of Z2 (x):
1
x2 + x + 1
,
, 1, x7 + x3 + 1.
x
x100 + 1
You can find multiplicative inverses of nonzero elements by taking reciprocals; for instance,
x2 + x + 1
x100 + 1
−1
=
x100 + 1
.
x2 + x + 1
I won’t go through and check all the axioms, but in fact, Z2 (x) is a field. Moreover, since 2 · 1 = 0 in
Z2 (x), it’s a field of characteristic 2. It has an infinite number of elements; for example, it contains
1,
x,
x2 ,
x3 ,
4
....
Example. (A field with 4 elements) Here are the addition and multiplication tables for GF (4), the
Galois field of order 4.
+
0
1
a
b
·
0
1
a
b
0
0
1
a
b
0
0
0
0
0
1
1
0
b
a
1
0
1
a
b
a
a
b
0
1
a
0
a
b
1
b
b
a
1
0
b
0
b
1
a
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.
Example. Find the multiplicative inverses of the nonzero elements of Z5 .
Since 5 is prime, Z5 is a field. This means that every nonzero element has a multiplicative inverse.
1 · 1 = 1,
2 · 3 = 1,
so 1−1 = 1.
so 2−1 = 3
4 · 4 = 1,
and 3−1 = 2.
so 4−1 = 4.
Example. Find 8−1 in Z13 .
In Z13 , I have 8 · 5 = 1, so 8−1 = 5. You could do this by trial and error, since Z13 isn’t that big:
8 · 1 = 8,
8 · 2 = 16 = 3,
8 · 3 = 24 = 11,
8 · 4 = 32 = 6,
8 · 5 = 40 = 1.
Alternatively, you might reason this way: I want x so that 8x = 1. Now an integer equals 1 mod 13 if
it is a multiple of 13 plus 1 (because mod 13 multiples of 13 equal 0). So I take multiples of 13 and add 1,
stopping when I get a number divisible by 8:
13 · 1 + 1 = 14
Not divisible by 8
13 · 2 + 1 = 27
13 · 3 + 1 = 40
Not divisible by 8
Divisible by 8
40
Then
= 5, so 8−1 = 5.
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Even this approach is too tedious to use with large numbers. The systematic way to find inverses is to
use the Extended Euclidean Algorithm.
Example. (a) Show that 2 doesn’t have a multiplicative inverse in Z4 .
(b) Show that 14 doesn’t have a multiplicative inverse in Z18 .
(a) Try all possibilities:
2 · 1 = 2,
2 · 2 = 0,
2 · 3 = 2.
Note that 4 isn’t prime, and Z4 is not a field. This example shows it directly: 2 is nonzero, but 2 does
not have a multiplicative inverse.
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(b) I could do this by trial and error, but it would be tedious because 18 is a bit large. Instead, I’ll show
that there is no multiplicative inverse using proof by contradiction.
Suppose 14x = 1 for x ∈ Z18 . Then
14x = 1
9 · 14x = 9 · 1
0=9
(Note that 9 · 14 = 136 = 7 · 18 = 0 in Z18 .) The last line above is a contradiction, so 14 does not have
a multiplicative inverse in Z18 .
In general, the elements in Zn which have multiplicative inverses are the elements which are relatively
prime to n.
Example. Find the roots of x2 + 5x + 6 in Z10 .
Make a table:
x
0
1
2
3
4
5
6
7
8
9
x + 5x + 6
6
2
0
0
2
6
2
0
0
2
2
The roots are x = 2, x = 3, x = 7 and x = 8.
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 Z10 is not a field).
There are systematic ways for finding inverses of elements in Zn — for example, by using the Extended
Euclidean Algorithm. I’ll stick to simple cases where trial and error is sufficient.
Linear algebra deals with structures based on fields, and you’ve now seen most of the fields that will
come up in the examples. To keep things simple, most of the example involving finite fields will use Zp for
p prime, rather than the more general Galois fields, or infinite fields of characteristic p.
c 2017 by Bruce Ikenaga
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