MATH 57091 - Algebra for High School Teachers
Examples of Rings
Professor Donald L. White
Department of Mathematical Sciences
Kent State University
D.L. White (Kent State University)
1 / 10
Review
Definition
A ring is a set R with two binary operations, denoted + and ·,
that have the following properties.
1
[Associativity of Addition] (a + b) + c = a + (b + c) for all a, b, c ∈ R.
2
[Commutativity of Addition] a + b = b + a for all a, b ∈ R.
3
[Additive Identity Element] There is an element 0R ∈ R
such that 0R + x = x + 0R = x for all x ∈ R.
4
[Additive Inverse] For each element x ∈ R, there is an element y ∈ R
such that x + y = y + x = 0R . We denote y = −x.
5
[Associativity of Multiplication] (a · b) · c = a · (b · c) for all a, b, c ∈ R.
[Distributive Laws] For all a, b, c ∈ R,
6
i
ii
a · (b + c) = a · b + a · c and
(b + c) · a = b · a + c · a.
D.L. White (Kent State University)
2 / 10
Review
Definition
Let R be a ring.
If a · b = b · a for all a, b ∈ R, we say R is a commutative ring.
Definition
Let R be a ring.
If there is an element 1R ∈ R, 1R 6= 0R , such that 1R · a = a · 1R = a
for all a ∈ R, we say R is a ring with 1.
Definition
A commutative ring with 1 in which every non-zero element is a unit
(i.e., has a multiplicative inverse) is called a field.
D.L. White (Kent State University)
3 / 10
Examples
We have seen previously that
Z, Q, R, and Zm are all commutative rings with 1,
Q, R, and Zp (p a prime) are all fields, and
Z and Zm (m not prime) are not fields.
We will introduce two new examples:
1
a commutative ring without a multiplicative identity, and
2
a non-commutative ring.
D.L. White (Kent State University)
4 / 10
Examples
1
We
but
If x
We
The set of even integers, 2Z = {2z | z ∈ Z},
is a commutative ring without a multiplicative identity,
under the usual addition and multiplication of integers.
know that addition and multiplication are binary operations on Z,
we must check that the subset 2Z is closed under both operations.
and y are in 2Z, then x = 2m and y = 2n for some m, n ∈ Z.
then have
x + y = 2m + 2n = 2(m + n).
Since Z is closed under addition, m + n ∈ Z and x + y = 2(m + n) ∈ 2Z.
Thus 2Z is closed under addition; i.e., addition is a binary operation on 2Z.
Similarly,
xy = (2m)(2n) = 2(2mn).
Since Z is closed under multiplication, 2mn ∈ Z and xy = 2(2mn) ∈ 2Z.
Hence 2Z is closed under multiplication;
i.e., multiplication is a binary operation on 2Z.
D.L. White (Kent State University)
5 / 10
Examples
We now have that addition and multiplication are binary operations on 2Z.
Since 2Z is a subset of Z, with the same operations as Z,
associativity and commutativity of both operations and the distributive law
are “inherited” by 2Z from Z. That is, since these properties
hold for all integers, they hold in particular for even integers.
For example, since Z is a ring, a · (b + c) = a · b + a · c for all integers
a, b, and c, and so the property holds if a, b, and c happen to be
even integers, and so the distributive law holds.
Similarly for associativity and commutativity of both operations.
Since 0 = 2 · 0, we have that 0 is in 2Z, and of course,
0 + a = a + 0 = a for all a in 2Z. Hence 0 is the additive identity of 2Z.
If x is in 2Z, so x = 2m, m ∈ Z, then −x = −2m = 2(−m) and −m ∈ Z.
Hence −x = 2(−m) is in 2Z.
Since x + (−x) = (−x) + x = 0, −x is the additive inverse of x.
Therefore, 2Z is a commutative ring.
D.L. White (Kent State University)
6 / 10
Examples
Finally, we need to show that 2Z does not have a multiplicative identity.
That is, we must show that there is no element u in 2Z such that
u · x = x for all x in 2Z.
CAUTION: This is not the same as saying that the integer 1 is not in 2Z.
It is possible for a ring R to have a multiplicative identity 1R ,
and a subset S that is a ring with multiplicative identity 1S
under the same operations, and yet 1R is not in S, hence 1S 6= 1R .
(See Example 3 below.)
Suppose u is an element of 2Z such that u · x = x for all x ∈ 2Z.
Then u = 2k for some integer k, and, in particular, u · 2 = 2;
that is, (2k) · 2 = 2. But then 4k = 2 implies k is not an integer,
a contradiction. Hence no such u can exist.
Therefore, 2Z does not have a multiplicative identity.
D.L. White (Kent State University)
7 / 10
Examples
2
Let R be any ring (for example, R = Z) and define
a b a,
b,
c,
d
∈
R
.
M2 (R) =
c d M2 (R) is a non-commutative ring under the usual matrix operations:
0
a b
a b0
a + a0 b + b 0
+
=
c d
c0 d0
c + c0 d + d0
and
a b
c d
D.L. White (Kent State University)
0
0
aa + bc 0 ab 0 + bd 0
a b0
·
=
.
c0 d0
ca0 + dc 0 cb 0 + dd 0
8 / 10
Examples
Since R is closed under addition and multiplication, M2 (R) is also;
that is, addition and multiplication are binary operations on M2 (R).
Verification of associativity of addition and multiplication,
commutativity ofaddition,
and the distributive laws is straightforward.
0 0
The zero matrix
is the additive identity.
0 0
a b
−a −b
The additive inverse of
is
. Hence M2 (R) is a ring.
c d
−c −d
Matrix multiplication is not commutative in general; e.g., in M2 (Z),
0 1
0 0
1 0
0 0
0 1
0 0
·
=
and
·
=
.
0 0
1 0
0 0
1 0
0 0
0 1
M2 (R) is a ring with 1 if and only
if R is a ring with 1,
1R 0
and in this case, I =
is the multiplicative identity.
0 1R
D.L. White (Kent State University)
9 / 10
Examples
3
Let S be the following subset of M2 (Z):
a 0 a∈Z .
S=
0 0 Verify that S is a ring under the usual matrix operations.
Observe that the multiplicative identity
1 0
I =
0 1
of M2 (Z) is not in S.
Verify that
1 0
J=
0 0
is the multiplicative identity element of S.
D.L. White (Kent State University)
10 / 10