MATH 57091 - Algebra for High School Teachers
Basic Properties of Rings
Professor Donald L. White
Department of Mathematical Sciences
Kent State University
D.L. White (Kent State University)
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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)
2/9
Additive Cancellation
Proposition 1
Let R be a ring and let a, b, c ∈ R. If a + c = b + c, then a = b.
Proof: Since c ∈ R, there is an additive inverse −c ∈ R for c.
Given a + c = b + c, we have
(a + c) + (−c) = (b + c) + (−c), because addition is well-defined,
a + (c + (−c)) = b + (c + (−c)), by associativity of addition,
a + 0R
= b + 0R , by definition of additive inverse,
a = b, by definition of additive identity.
Therefore, a = b, as claimed.
D.L. White (Kent State University)
3/9
Uniqueness of Identities
Proposition 2
If R is a ring, then the additive identity 0R is unique.
If R is a ring with 1, then the multiplicative identity 1R is unique.
Proof: Suppose 01 and 02 are both additive identities in R. By definition,
01 + a = a + 01 = a for all a ∈ R,
02 + a = a + 02 = a for all a ∈ R.
Substituting 02 for a in the first equation, we have
01 + 02 = 02
and substituting 01 for a in the second equation, we have
01 + 02 = 01 .
Hence
02 = 01 + 02 = 01 ,
and so 01 = 02 . Therefore, the additive identity is unique.
The proof for the multiplicative identity is similar and is an exercise.
D.L. White (Kent State University)
4/9
Additive Objects, Multiplicative Properties
The remaining properties we wish to prove all deal with
the multiplicative behavior of the additive identity and additive inverses.
Both 0R and −a, for a ∈ R, are defined only in terms of addition.
To study their multiplicative behavior, it will therefore be necessary
to use the distributive laws. This is the case even in Z, Q, and R.
Proposition 3
Let R be a ring. If a is any element of R, then 0R · a = a · 0R = 0R .
Proof: By definition of additive identity, 0R = 0R + 0R . Therefore
a · 0R
= a · (0R + 0R ), because multiplication is well-defined,
= a · 0R + a · 0R , by the left distributive law.
Hence a · 0R + a · 0R = a · 0R , and again by definition of additive identity
a · 0R + a · 0R = 0R + a · 0R ,
and by additive cancellation (Proposition 1), we have a · 0R = 0R .
The proof that 0R · a = 0R is similar, using the right distributive law.
D.L. White (Kent State University)
5/9
Additive Objects, Multiplicative Properties
Proposition 4
Let R be a ring. If a, b ∈ R, then (−a) · b == a · (−b) = −(a · b).
Proof: We will prove (−a) · b = −(a · b) and leave the similar proof
that a · (−b) = −(a · b) as an exercise.
The statement that (−a) · b = −(a · b) is equivalent to the statement
that (−a) · b is the additive inverse of a · b.
Hence we need to prove (−a) · b + a · b = 0R . We have
(−a) · b + a · b = ((−a) + a) · b, by right distributivity,
= 0R · b, by definition of additive inverse,
= 0R , by Proposition 3.
Thus (−a) · b is the additive inverse of a · b; i.e., (−a) · b = −(a · b).
D.L. White (Kent State University)
6/9
Additive Objects, Multiplicative Properties
Applied to the real numbers R, the next result answers the question
“Why is a negative times a negative positive?”
Proposition 5
Let R be a ring. If a, b ∈ R, then (−a) · (−b) = a · b.
Proof: For x ∈ R, we have by definition that
y = −x if and only if x + y = y + x = 0R .
It follows that −(−x) = −y = x. We therefore have
(−a) · (−b) = −(a · (−b)), by Proposition 4,
= −(−(a · b)), by Proposition 4,
= a · b.
Hence (−a) · (−b) = a · b.
D.L. White (Kent State University)
7/9
Additive Objects, Multiplicative Properties
If our ring has a multiplicative identity element, then the additive inverse
of any element a is just (−1R ) · a, where −1R is the additive inverse of 1R .
Proposition 6
Let R be a ring with 1. If a ∈ R, then −a = (−1R ) · a.
Proof: If R has a multiplicative identity 1R , then
(−1R ) · a = −(1R · a), by Proposition 4,
= −a, by definition of multiplicative identity.
Therefore, −a = (−1R ) · a, as claimed.
D.L. White (Kent State University)
8/9
Notation: Multiples and Powers
Let R be a ring and let n be a positive integer.
For r ∈ R, we use the following familiar notation:
n
}|
{
z
nr = r + r + · · · + r
0r = 0R , where “0” denotes the integer 0.
n
z
}|
{
(−n)r = n(−r ) = (−r ) + (−r ) + · · · + (−r )= −(nr )
n
z }| {
rn = r · r · r · · · r
If R has a multiplicative identity,
r 0 = 1R , where “0” denotes the integer 0.
If R has a multiplicative identity and r has a multiplicative inverse,
n
z
}|
{
=
= r −1 · r −1 · r −1 · · · r −1 = (r n )−1 .
NOTE: We have r n · r m = r n+m and (r n )m = r nm ,
but in general, r n s n 6= (rs)n unless rs = sr .
r −n
(r −1 )n
D.L. White (Kent State University)
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