Download Basic Properties of Rings - Clayton State University

Basic Properties of Rings
A ring is an algebraic structure with an “addition” operation and a
“multiplication” operation. These operations are required to satisfy many
(but not all!) familiar properties. Some examples of rings:
• Each of the number systems Z, Q, R, and C is a ring; each has some
additional properties.
• When viewed in the right context, the integers modulo n form a ring for
any fixed value of n > 0.
• For any fixed n > 0, the set of n × n matrices is a ring, using matrix
addition and matrix multiplication as the two operations.
Ring Axioms for Addition
For any a, b, c ∈ R, Properties 1-3 hold:
1. Closure: a + b ∈ R
2. Associativity: a + (b + c) = (a + b) + c
3. Commutativity: a + b = b + a
4. Identity: There exists 0R ∈ R such that for every a ∈ R, a + 0R = a.
(the same element, 0R, must work for every element in the ring).
5. Inverses: For any a ∈ R, there exists x ∈ R such that a + x = 0R.
(the value of x will depend on the given element a).
Ring Axioms for Multiplication
For any a, b, c ∈ R, Properties 6-8 hold:
6. Closure: ab ∈ R
7. Associativity: a(bc) = (ab)c
8. Distributive Laws: a(b + c) = ab + ac and (a + b)c = ab + ac
Note: ring multiplication still takes precedence over ring addition.
More importantly, ring multiplication is NOT assumed to be Commutative!
Uniqueness of Zero, Cancellation Laws
We did not assume that 0R was unique, but this is easy to show: Suppose
that z is a fixed element of R such that a + z = a for every a ∈ R. Then
z = z + 0R = 0R + z = 0R.
Any ring satisfies the Cancellation Laws for Addition. If a + b = a + c,
choose x to be a ring element such that a + x = 0R. Then:
b+a=c+a
(b + a) + x = (c + a) + x
b + (a + x) = c + (a + x)
b + 0R = c + 0R
b=c
Additive Inverses
We do not need to assume that a + x = 0R has a unique solution, but this
follows easily from the Cancellation Laws. If y is another element of R such
that a + y = 0R, then a + x = 0R = a + y, and thus x = y.
Given a ∈ R, we can thus define −a to be the unique x ∈ R such that
a + x = 0R. This means a + (−a) = (−a) + a = 0R
A general rule: To show that some expression (say x) is equal to −a, you
need to show that a + x = 0R. Since additive inverses are unique, it then
follows that x = −a. Try proving the following:
• 0R = −0R.
• For any b ∈ R, −(−b) = b.
Multiplication by Zero
Before introducing ring subtraction and proving its familiar properties, we’ll
need the following result:
For any a ∈ R, we have a · 0R = 0R = 0R · a.
Note: We do not assume ring multiplication is commutative, so there really
are two things to prove here. We have:
(a · 0R) + (a · 0R) = a · (0R + 0R) = a · 0R = (a · 0R) + 0R.
By Cancellation, a · 0R = 0R. The other equality is proven similarly.
This is a useful proof to know–it comes up surprisingly often! The two basic
ingredients are the Additive Identity and the Distributive Laws.
Additive Inverses and Subtraction
We can now define ring subtraction: a − b = a + (−b). Prove the following
familiar properties:
• a(−b) = −(ab) = (−a)b.
commutative!
Do not assume that multiplication is
• −(a + b) = (−a) + (−b). Do not use the −1 symbol!
• −(a − b) = (−a) + b = b − a.
• (−a)(−b) = ab. [hint: Try using the results above].
Special Types of Rings
The Ring Axioms do not include many familiar rules from ordinary algebra–
this is done with an eye towards generality. In some cases, we will want to
assume some additional properties. Let R be a ring.
• If ab = ba for every a, b ∈ R, we say that R is a commutative ring.
• If there is an element 1R ∈ R such that a · 1R = a = 1R · a for every
a ∈ R, we say that R is a ring with identity/unity. Unless stated
otherwise, we’ll always assume 1R 6= 0R.
These two properties are independent of one another. See the next slide for
examples.
Examples
• The integers (with ordinary addition/multiplication) are a commutative
ring with identity [with some additional algebraic properties].
• The even integers (with ordinary operations) are a commutative ring
without identity.
• The set of 2 × 2 matrices with real entries are a non-commutative
ring with identity (the ring operations are ordinary matrix addition and
multiplication).
• The set of 2×2 matrices with even integer entries are a non-commutative
ring without identity.
Units of a Ring, Multiplicative Inverses
Let R be a ring with identity (with 1R 6= 0R). An element a ∈ R is a unit
if there exist elements x, y ∈ R such that ax = 1R = ya.
Although we do not assume that x = y in our definition, it is easy to show
that this must be true:
x = 1R · x = (ya)x = y(ax) = y · 1R = y.
It is also easy to show that if there exists x ∈ R such that ax = 1R = xa,
then x must be unique. We can thus define the multiplicative inverse of
a to be this unique element x (if it exists), and write x = a−1 as usual.
Question: What are the units in Z? In Q?
Division Rings and Fields
Let R be a ring with identity. Note that 0R cannot be a unit.
• Suppose that for each a 6= 0R, there exist elements x, y ∈ R such that
ax = 1R = ya. Then R is called a division ring.
As before, the condition for a division ring implies that x = y and that
x is unique.
• A commutative division ring is called a field. Examples: Q, R, C.
Zero Divisors
Let R be any ring. A zero divisor is an element a 6= 0R such that for some
b ∈ R, ab = 0R and/or ba = 0R.
Example: In the ring of 2 × 2 matrices, we have:
1 0
0 0
0 0
·
=
0 0
0 1
0 0
Example: In the ring of integers (mod 6), 2 and 3 are zero divisors, since
2 · 3 ≡ 0 (mod 6).
Integral Domains
The familiar rings (Z, Q, R, C) have no zero divisors. This is usually stated
in the following equivalent form:
• If ab = 0R, then a = 0R or b = 0R (or possibly both).
Using induction and associativity, this extends to:
• If a product (of finitely many factors) is equal to 0R, then at least one
factor equals 0R.
A commutative ring with identity (with 1R 6= 0R) having no zero divisors is
called an integral domain.
Cancellation Laws for Multiplication
Suppose that R is an integral domain. If a 6= 0R and ab = ac, then
ab − ac = ac − ac
ab − ac = 0R
a(b − c) = 0R
b − c = 0R
b=c
Cancellation on the right can be proven similarly, and also follows from that
fact than multiplication is commutative in an integral domain.
Cancellation Laws for Multiplication
Conversely, suppose that R is a commutative ring with identity, 1R 6= 0R,
and the multiplicative Cancellation Laws hold in R.
• If ab = 0R and a 6= 0R, then ab = 0R = a · 0R.
• By Cancellation, b = 0R.
• Thus R is an integral domain.
In particular, note that any field is an integral domain. If ab = 0R and
a 6= 0R, then we can multiply both sides by a−1 and simplify to get b = 0R.
Z is an integral domain that is not a field (2x = 1 has no solution in Z).
Repeated Ring Addition
Let R be a ring, a ∈ R, and n ∈ Z. Define na (in that order!) as follows:
• For n = 0, 0a = 0R [note that 0 ∈ Z, but 0R ∈ R].
• For n > 0, na = (n − 1)a + a [note that − is done in Z, + is done in
R, and the “multiplication” is what we’re defining with this notation].
• For n < 0, na = (−n)(−a) [−n > 0; −a is the additive inverse in R].
Practically speaking, if n is a positive integer, then we have:
• na = a + a + ... + a [this is n copies of a].
• (−n)a = (−a) + (−a) + ... + (−a) [this is n copies of −a].
Be VERY CAREFUL with this notation! Although na is written as
multiplication, it is not the same thing as ring multiplication. In particular,
we always have 1a = a (1 ∈ Z), even when R does not have an identity 1R.
It follows immediately from the definition that (−1)a = −a for any a ∈ R.
We also have the following distributive/associative-like laws:
• For any n, m ∈ Z and a ∈ R, (n ± m)a = na ± ma.
• For any n ∈ Z and a, b ∈ R, n(a ± b) = na ± nb.
• For any n ∈ Z and a, b ∈ R, n(ab) = (na)b = a(nb).
• For any n, m ∈ Z and a, b ∈ R, (nm)(ab) = (na)(mb).
Note that the same notation is used for different operations above. Note
also that multiplication is commutative in Z, but might not be in R.
Repeated Ring Multiplication
We use the ordinary power notation to denote repeated multiplication. For
any a ∈ R, we define an, where n is a positive integer, recursively:
• a1 = a.
• For n > 1, an = a(an−1).
If R is a ring with identity, we define a0 = 1R for all a 6= 0R.
If a has a multiplicative inverse, we define a−n = (a−1)n for n > 0.
Question: Can you simplify (ab)2, or, more generally, (ab)n? What about
(a + b)2?
Since ring multiplication might not be commutative...
Unless R is a Commutative Ring, you cannot assume (ab)n = anbn.
In particular (ab)−1 does not always equal a−1b−1.
However, it is easy to check that if a and b have multiplicative inverses,
then ab has the multiplicative inverse b−1a−1; note the the order has been
reversed!
Unless R is a Commutative Ring, you cannot assume (a+b)2 = a2 +2ab+b2.
Similar considerations hold for higher powers of binomials.
In general, (a + b)2 = a2 + ab + ba + b2. This is a good exercise to try on
your own; use the distributive laws for rings very carefully!
Fortunately, an+m = (an)(am), with appropriate restrictions on zero or
negative exponents.
Exercises
Let R be a ring, a, b ∈ R.
• If R is a ring with identity, show that (−1R)a = −a.
• Is a2 − b2 = (a − b)(a + b)? If not, what conditions are required?
• Does a2 = 0R imply a = 0R? If not, what conditions are required?
• If n, m ∈ Z, does na = ma imply n = m? For an easier case, take
m = 0.
• If a and b are units, is a + b a unit? Is ab a unit?