Download Section 18 -- Rings and fields

Section 18 – Rings and fields
Instructor: Yifan Yang
Spring 2007
Motivation
• Many sets in mathematics have two binary operations (and
thus two algebraic structures)
• For example, the sets Z, Q, R, Mn (R) (the set of n × n
matrices) all have addition and multiplication.
• Note that the multiplication in Z, Q, and R are
commutative, while that of Mn (R) is not.
• Also, every non-zero element of Q and R has an
multiplicative inverse, but this is not the case for Z and
Mn (R).
• We wish to give an axiomatic study on these subjects, and
determine how various properties (such as commutativity
of multiplication, the existence of multiplicative inverse and
so on) affect the overall algebraic structures of the sets.
Rings and fields
Definition
A ring hR, +, ·i is a set R together with two binary operations +
and ·, called addition and multiplication, such that the following
axioms are satisfied:
1. hR, +i is an abelian group.
2. Multiplication is associative, i.e., a(bc) = (ab)c for all
a, b, c ∈ R.
3. For all a, b, c ∈ R, the left distributive law
a · (b + c) = (a · b) + (a · c) and the right distributive law
(a + b) · c = (a · c) + (b · c) hold.
Examples
• The set Z, Q, R, and C are all rings with the usual addition
and multiplication.
• The set Zn of all residue classes modulo n is a ring with
¯
ā +n b̄ = a + b and ā ·n b̄ = ab.
• The set Mn (R) of all n × n matrices is a ring.
• Let R1 and R2 be rings. Define + and · on R1 × R2 by
(a1 , b1 ) + (a2 , b2 ) = (a1 + a2 , b1 + b2 )
and
(a1 , b1 ) · (a2 , b2 ) = (a1 · a2 , b1 · b2 ).
Then R1 × R2 is a ring, called the direct product of R1 and
R2 .
Examples
• Let F be the set of all continuous functions f : R → R.
Define f + g and f · g to be
f + g : x 7→ f (x) + g(x),
f · g : x 7→ f (x)g(x)
Then hF , +, ·i is ring.
• Let F be the set of all linear transformations from Rn to Rn .
Let addition and multiplication be defined by
f + g : v 7→ f (v ) + g(v ),
f ◦ g : v 7→ f (g(v ))
Then hF , +, ◦i is a ring. (Actually, this is just a different way
to say that Mn (R) is a ring.)
Notation
• For simplicity, we write ab in place of a · b.
• The additive identity of a ring R is denoted by 0. In case
some confusion may arise, we also write 0R .
• For an element a of R, we let −a denote the additive
inverse of a.
• For a positive integer n and an element a of R, the notation
n · a refers to the sum a + · · · + a having n summands.
• By convention we set 0 · a = 0R . Here the 0 on the
left-hand side is the integer 0, while 0R on the right is the
additive identity element of R.
Basic properties
Theorem (18.8)
Let R be a ring. For any a, b ∈ R, we have
1. 0R a = a0R = 0R .
2. a(−b) = (−a)b = −ab.
3. (−a)(−b) = ab.
Proof of (1). Since 0R + 0R = 0R , we have (0R + 0R )a = 0R a.
Then by the distributive law, we have
0R a + 0R a = 0R a.
Then using the cancellation law for groups, we obtain
0R a = 0R .
The proof of a0R = 0R is similar.
Proof of Theorem 18.8, continued
Proof of a(−b) = −ab = (−a)b.
We have, by definition of −b,
b + (−b) = 0R .
By (1),
ab + a(−b) = a0R = 0R .
This means that a(−b) is the additive inverse of ab. That is,
a(−b) = −ab. The proof of (−a)b = −ab is similar.
Proof of Theorem 18.8, continued
Proof of (−a)(−b) = ab.
By (2)
(−a)(−b) = −(a(−b)).
By (2) again,
−(a(−b)) = −(−ab).
This means that (−a)(−b) is the additive inverse of −ab. But
we know that the additive inverse of −ab is ab. Thus, by the
uniqueness of additive inverse, we conclude(−a)(−b) = ab. In-class exercises
Determine whether the following algebraic structures are rings.
1. The set Z[x] of all polynomials over Z.
2. The set GLn (R) of all n × n invertible matrices under the
usual matrix addition and multiplication.
a b
3. The set
: a, b ∈ R under the usual matrix
−b a
addition and multiplication.
4. The set 12 Z = {n/2 : n ∈ Z} under the usual addition and
multiplication.
5. The set
n
o
a1
an
Z[1/2] = a0 +
+ · · · + n : ai ∈ Z
2
2
under the usual addition and multiplication.
Homomorphisms
Definition
For rings R and R 0 , a function φ : R → R 0 is a (ring)
homomorphism if
1. φ(a + b) = φ(a) + φ(b),
2. φ(ab) = φ(a)φ(b),
for all a, b ∈ R.
The kernel of φ is the set
Ker(φ) = {a ∈ R : φ(a) = 0}.
Examples
• Let n be a positive integer. Let R = Z and R 0 = Zn . Then
the function φ : Z → Zn defined by
φ(a) = a
mod n
(or φ(a) = a + nZ in our notation from the last semester) is
a ring homomorphism.
• The function ψ : Z → 2Z defined by
ψ(a) = 2a
is not a ring homomorphism since ψ(ab) = 2ab, but
ψ(a)ψ(b) = 4ab. Note that if we consider Z and 2Z as
additive groups, then ψ is a group homomorphism.
Example
Let R = R[x] be the set of all polynomials over R. Given a ∈ R,
define φa : R[x] → R by
φa (f (x)) = f (a).
Then φa is an evaluation homomorphism.
Isomorphisms
Definition
Let R and R 0 be two rings. A function φ : R → R 0 is an
isomorphism if
1. φ is a ring homomorphism,
2. φ is one-to-one, or equivalently, Ker(φ) = {0},
3. φ is onto.
We then say R and R 0 are isomorphic.
Example
The map φ : C →
a b
−b a
: a, b ∈ R
φ(a + bi) =
a b
−b a
defined by
is a ring isomorphism. The map
ψ(a + bi) =
is another isomorphism.
a −b
b a
Multiplicative definitions
Definition
• A ring in which the multiplication is commutative is a
commutative ring.
• If an element a of R satisfies
ra = ar = r
for all r ∈ R, then a is the multiplicative identity or the unity,
and is denoted by 1 or 1R .
• A ring with a multiplicative identity is a ring with unity.
• A multiplicative inverse of an element a in a ring with unity
is an element a−1 of R such that a−1 a = aa−1 = 1.
Multiplicative definitions
Definition
Let R be a ring with unity 1 6= 0.
• An element u of R is a unit if it has a multiplicative inverse.
• If every nonzero element of R is a unit, then R is a division
ring (or skew field).
• A commutative division ring is a field.
• A noncommutative division ring is a strictly skew field.
Remarks
• If a ring R has a multiplicative identity element, it is unique.
(See the proof of Theorem 3.13.)
• If an element a of a ring with unity has a multiplicative
inverse, the inverse is unique.
• The assumption 1 6= 0 in the definition of a field is to avoid
the trivial case where the ring consists of just one single
element 0. (In this case 0 is both the additive identity and
the multiplicative identity.)
• Conversely, if 1 = 0, then for all a ∈ R, a = 1a = 0a = 0,
and R = {0} is the trivial ring.
Examples
• The sets Z, Q, R, and C are all commutative rings with
unity under usual addition and multiplication.
• The rings Q, R, and C are fields, while Z is not since only
±1 are units in Z.
• The set Mn (R) of all n × n matrices is a noncommutative
ring. The set of units in Mn (R) is GLn (R). The set Mn (R) is
not a division ring since there exist nonzero matrices that
are not invertible.
• The set Z2 , Z3 , and Z5 are fields. (Take Z5 for example, the
multiplicative inverses of 1̄, 2̄, 3̄, 4̄ are 1̄, 3̄, 2̄, 4̄,
respectively.)
• If gcd(a, n) > 1, then ā never has a multiplicative inverse in
Zn , since āb̄ will be a multiple of gcd(a, n) for any b. This
shows that if n is composite, then Zn is never a field.
In-class exercises
1. Find the units in Z12 .
2. Find the units in Z15 .
3. Let F be the ring of all continuous functions f : R → R with
addition and multiplication given by
f + g : x 7→ f (x) + g(x),
What are the units in F ?
f · g : x 7→ f (x)g(x).
Subrings and subfields
Definition
A subset S of a ring R is a subring of R if S is a ring under the
induced addition and multiplication from R. We denote this
relation by S < R. A subfield is similarly defined.
Homework
Problems 8, 10, 12, 18, 19, 24, 25, 37, 40, 48, 50 of Section 18.