Download Chapter 3, Rings Definitions and examples. We now have several

Chapter 3, Rings
Definitions and examples.
We now have several examples of algebraic systems with addition and multiplication:
Z, Zn , R, Mn (R), 2Z = { 2n | n ∈ Z }. We will write down a system of axioms which
includes them all.
Definition, p. 42. A ring is a nonempty set R with two binary operations (usually
written as addition and multiplication) such that for all a, b, c ∈ R,
(1) R is closed under addition: a + b ∈ R.
(2) Addition is associative: (a + b) + c = a + (b + c).
(3) Addition is commutative: a + b = b + a.
(4) R contains an additive identity element, called zero and usually denoted by 0 or
0R : a + 0 = 0 + a = a.
(5) Every element of R has an additive inverse: for each a, there exists an x ∈ R such
that a + x = 0 = x + a. We write x = −a.
(6) R is closed under multiplication: ab ∈ R.
(7) Multiplication is associative: (ab)c = a(bc).
(8) Multiplication distributes over addition: a(b + c) = ab + ac and (a + b)c = ac + bc.
Other possible properties are captured by special types of rings. We will encounter
many in this book; many more are studied as well by mathematicians.
Theorem 3.3. Additive inverses are unique.
Proof. Assume that x and y are both inverses of a. Then x = x + 0 = x + (a + y) =
(x + a) + y = 0 + y = y. We don’t have to add axioms about subtraction. We just define a − b to be a + (−b).
Definition. A commutative ring is a ring R that satisfies the additional axiom that
ab = ba for all a, b ∈ R.
Examples are Z, R, Zn , 2Z, but not Mn (R) if n ≥ 2.
Definition. A ring with identity is a ring R that contains a multiplicative identity
element 1R : 1R a = a = a1R for all a ∈ R.
Examples: 1 in the first three rings above,
1
1 0
0 1
in M2 (R). The set of even integers
2
is a ring without identity. The set of odd integers is not a ring. We can also work with
matrices whose elements come from any ring we know about, such as Mn (Zr ).
Example. Let R = M2 (Z2 ). This is a finite (16 elements)
noncommutative ring with
1 0
0 0
and zero element 0R =
. We give an example to show it
identity 1R =
0 1
0 0
is noncommutative:
1 0
0 1
0 1
0 1
1 0
0 0
=
but
=
0 0
0 0
0 0
0 0
0 0
0 0
Example: rings of continuous functions. Let X be any topological space; if you don’t
know what that is, let it be R or any interval in R. We consider the set R = C(X, R),
the set of all continuous functions from X to R. R becomes a ring with identity when we
define addition and multiplication as in elementary calculus: (f + g)(x) = f (x) + g(x) and
(f g)(x) = f (x)g(x). The identity element is the constant function 1. R is commutative
because R is, but it does have zero divisors for almost all choices of X.
There are many, many examples of this sort of ring. The functions don’t have to be
continuous. They can be restricted in many other ways, or not restricted at all. For
example, you can look at polynomial functions or differentiable functions (for some choices
of X).
Definition, p. 46. An integral domain is a commutative ring R with identity 1R 6= 0R
with no zero divisors; that is, ab = 0R implies that a = 0R or b = 0R .
Examples: Z, R, Zp for p prime. Nonexamples: C(R, R), Zn for n composite, the zero
ring {0R }, the even integers 2Z.
Definition, p. 47. A field is an integral domain in which every nonzero element a has
a multiplicative inverse, denoted a−1 .
Examples: R, Q, C, Zp for p prime (Theorem 2.8).
If an element of a ring has a multiplicative inverse, it is unique. The proof is the same
as that given above for Theorem 3.3 if we replace addition by multiplication. (Note that
we did not use the commutativity of addition.) This is also the proof from Math 311 that
invertible matrices have unique inverses.
Definition, p. 60. Any element a in a ring R with identity which has an inverse u (i.e.,
au = 1r = ua) is called a unit.
3
Making new rings.
Theorem 3.1 (Product rings). Let R, S be rings and form the Cartesian product R × S.
Define operations by
(r, s) + (r 0 , s0 ) = (r + r 0 , s + s0 )
(r, s)(r 0 , s0 ) = (rr 0 , ss0 ).
Then R × S is a ring. If R and S are both commutative, so is R × S. If R and S both
have an identity, then (1R , 1S ) is the identity in R × S.
Example. Let R be the ring Z × Z = { (n, m) | n, m ∈ Z }. Note that (1, 0)(0, 1) =
(0, 0) = 0R , so R is not an integral domain. For the same reason, no product ring is an
integral domain.
Definition, p. 49. A subset of a ring which is itself a ring (using the same operations)
is called a subring. A subset of a field which is itself a field is called a subfield.
Q is a subfield of R, and both are subfields of C. Z is a subring of Q. Z3 is not a
subring of Z. Its elements are not integers, but rather are congruence classes of integers.
2Z = { 2n | n ∈ Z } is a subring of Z, but the only subring of Z with identity is Z itself.
The zero ring is a subring of every ring.
As with subspaces of vector spaces, it is not hard to check that a subset is a subring as
most axioms are inherited from the ring.
Theorem 3.2. Let S be a subset of a ring R. S is a subring of R iff the following
conditions all hold:
(1) S is closed under addition and multiplication.
(2) 0R ∈ S.
(3) −s ∈ S for every element s ∈ S.
Proof. Axioms 1, 4, 5 and 6 are in our hypotheses. Axioms 2, 3, 7 and 8 are inherited
from R. As examples, do exercises from book, page 51–55: 7, 8, 9, 19, 34, discuss 35.
4
Basic properties of rings.
Theorem 3.4–3.5. For all a, b, c in a ring R,
(1) a + b = a + c implies b = c.
(2) a · 0 = 0 · a = 0.
(3) a(−b) = −(ab) = (−a)b.
(4) −(−a) = a.
(5) −(a + b) = (−a) + (−b).
(6) (−a)(−b) = ab.
(7) (−1)a = −a if R has an identity element.
Proof. These make use of the definition of subtraction and negatives. For (1), add −a.
(2) a · 0 = a(0 + 0) = a · 0 + a · 0 and use (1).
(3) Show that a(−b) and (−a)b are additive inverses for ab (we know there is a unique
such element).
(4) and (5) Definition of additive inverse.
(6) Use (3) and (4).
(7) Use (3) or add (−1)a to a. A quicker version of Theorem 3.2 can be obtained using subtraction.
Theorem 3.6. Let R be a ring and ∅ 6= S ⊆ R. S is a subring of R if S is closed under
subtraction and multiplication.
Proof. We need to show S is closed under addition, has 0 and has additive inverses. But
S 6= ∅ implies there is some s ∈ S, hence 0 = s − s ∈ S. For any a ∈ S, −a = 0 − a ∈ S.
And finally, for a, b ∈ S, a + b = a − (−b) ∈ S. Notation: For a in a ring R and n ∈ Z, write na for a sum of n copies of a and an
for a product of n copies of a. This includes the cases 0 · a = 0R and a0 = 1R if R has an
identity element.
Be careful: (a+b)2 = a2 +ab+ba+b2 cannot be simplified unless the ring is commutative.
Theorem 3.7. Let R be a ring and a, b ∈ R. The equation a + x = b has the unique
solution x = b − a in R.
Proof. b − a is a solution: check it! If z is another solution, then a + z = b = a + (b − a),
so z = b − a by cancellation (Theorem 3.4). For multiplication, we need multiplicative inverses to get the cancellation in the proof.
5
Theorem 3.8. Let R be a ring with identity and a, b ∈ R. If a is a unit, then the equations
ax = b and ya = b have unique solutions in R.
Proof. x = a−1 b and y = ba−1 are solutions: check! Uniqueness works as in Theorem 3.7,
using the inverse for cancellation: if z is another solution to ax = b, then az = b = a(a−1 b).
Multiply on the left by a−1 to get z = a−1 az = a−1 a(a−1 b) = a−1 b. A similar argument
works for y. The solutions x = a−1 b and y = ba−1 may not be the same. Exercise 4, p. 62 gives an
example with 2 × 2 matrices.
Sometimes multiplicative cancellation works without inverses (recall the integers).
Theorem 3.10. Let R be an integral domain. If a 6= 0 and ab = ac, then b = c.
Proof. ab = ac implies a(b − c) = 0. Since a 6= 0 and R is an integral domain, we must
have b − c = 0, or b = c. If R is not an integral domain, this fails because of zero divisors:
p. 62. Let a 6= 0 in a ring R. a is a zero divisor if there exists an element b 6= 0 in R
with either ab = 0 or ba = 0.
Example: 2 · 3 = 0 = 2 · 0 in Z6 .
Theorem 3.11. Every finite integral domain is a field.
Combinatorial proof. Let a 6= 0 in the integral domain R. The set aR = { ar | r ∈ R } is a
permutation of the elements of R: ax = ay implies x = y by Theorem 3.10 and there are
only finitely many elements. Therefore some ar must be 1 and a has an inverse. Examples. Discuss exercises 8, 11, 17, 25 on pages 62–65.
6
Isomorphisms and homomorphisms.
Recall from linear algebra that a linear transformation is a function between vector
spaces that preserves the operations on the vector space. For rings we only want to
consider the functions that preserve their two operations.
Definition, p. 71. Let R and S be rings. A function f : R → S (which means the domain
is R and f takes values in S) is called a homomorphism if f (a + b) = f (a) + f (b) and
f (ab) = f (a)f (b) for all a, b ∈ R.
Examples. Z → Q defined by n 7→ n is the natural embedding of the integers into the
rational numbers.
Z → Zn defined by n 7→ [n]. This is a homomorphism by the definition of addition and
multiplication in Zn . (Theorem 2.6)
ιR : R → R, the identity map for any ring R.
f : C → C defined by f (a + bi) = a − bi (complex conjugation). Check the definition.
The last two examples are special in that they are one-to-one (injective) and onto
(surjective). In this case, we say the homomorphism is an isomorphism. If f : R → S is
a ring isomorphism, we say R and S are isomorphic. An interesting example in the book
is given on p.
69: the
complex numbers are isomorphic to the ring of real 2 × 2 matrices
a b
of the form
, which corresponds to the complex number a + bi.
−b a
Besides the identity, there is one other rather trivial example: the zero mapping z : R →
S defined by z(r) = 0S for all r ∈ R is a ring homomorphism.
Example. Recall Exer. 7, p. 51: R∗ = { (r, r) | r ∈ R }. Define f : R → R∗ by
f (r) = (r, r), Check that f is an isomorphism.
Theorem 3.12. Let f : R → S be a ring homomorphism.
(1) f (0R ) = 0S .
(2) f (−a) = −f (a) for all a ∈ R.
(3) f (a − b) = f (a) − f (b) for all a, b ∈ R.
If R is a ring with identity and f is surjective, then
(4) S is a ring with identity and f (1R ) = 1S .
(5) Whenever u ∈ R is a unit, then f (u) is a unit in S and f (u)−1 = f (u−1 ).
Proof. (1) f (0) = f (0 + 0) = f (0) + f (0) implies f (0R ) = 0S .
(2) f (−a) + f (a) = f (−a + a) = f (0) = 0S by (1), so f (−a) = −f (a).
(3) f (a − b) = f (a + (−b)) = f (a) + f (−b) = f (a) − f (b) by (2).
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(4) Let s ∈ S. We must show sf (1) = f (1)s = s. Since f is surjective, there is some r ∈ R
with f (r) = s. Then sf (1) = f (r)f (1) = f (r · 1) = f (r) = s and similarly for f (1)s.
(5) f (u−1 )f (u) = f (u−1 u) = f (1) = 1 by (4). Note that we really need surjectivity in (4) and (5). If S = R × R and we define
f : R → S by f (r) = (r, 0), then f (1) = (1, 0) is not the identity in S. Furthermore, 1 ∈ R
is a unit, but f (1) is not a unit.
When the homomorphism f : R → S is not surjective, it is handy to have a name for
the subset of S that it maps onto. We write im(f ) = { f (r) | r ∈ R } and call this set the
image of f . (The book does not have a symbol for the image or for the kernel which we
shall define later.) Now we can say that f : R → S is surjective if S = im(f ).
Corollary 3.13. Given a ring homomorphism f : R → S, the image of f is a subring of
S.
Proof. im(f ) 6= ∅ since f (0) = 0 ∈ im(f ). Closure under subtraction is Theorem 3.12(3).
Closure under multiplication follows from the definition of homomorphism: f (a)f (b) =
f (ab) ∈ im(f ). Students should read pages 74–75 for hints on how to decide (or prove) whether two
rings are isomorphic or not.
Examples. 1. Z6 is isomorphic to Z2 × Z3 . They have the same number of elements.
We must find a bijection which preserves the operations. Define f : Z6 → Z2 × Z3 by
f ([n]) = ([n]2 , [n]3 ). Check that f is a homomorphism. Since both sets have 6 elements, f
will be a bijection if it is either injective OR surjective. Surjectivity is an example of the
Chinese Remainder Theorem (page 408). We show injectivity. Assume f ([n]) = f ([m]).
Then f ([m − n]) = f ([m]) − f ([n]) = 0, so it suffices to prove only [0] maps to ([0], [0]).
(Recall the situation for vector spaces!) If [a] 7→ ([0]2 , [0]3 ), then a is divisible by both 2
and 3, hence by 6 since they are relatively prime. Therefore [a] = 0 in Z6 .
2. Z4 is not isomorphic to Z2 × Z2 . What fails in trying to use the previous proof is that
2 and 2 are not relatively prime. What is really different about them? Any isomorphism
must take 1 ∈ Z4 to (1, 1) ∈ Z2 × Z2 . But in Z4 , we must add 1 to itself four times to get
zero, while (1, 1) + (1, 1) = (0, 0). Thus any homomorphism f : Z4 → Z2 × Z2 will have
f (2) = f (0) = 0 and will not be injective.
These two examples can be generalized. See exercises 39, 40 on page 79.
3. Pages 76–79, exercises 18, 25, 33.