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Some Definitions
Ring: A ring is a set R with 2 binary operations, + and *, st.:
1. R is closed under +
2. + is associative
3. + is commutative
4. There is an additive identity (0r); that is, 0R + a = a + 0R = a for all a ∈ R.
5. For all a, there is a solution, b, to the equation a+b=0. (Additive inverses.)
6. R is closed under *.
7. * is associative.
8. the distributive property holds; a(b+c) = ab+ac and (b+c)a = ba+ca
Ring With Identity. There is a 1R such that a1R = 1R a = a, for all a.
Commutative Ring. * commutes.
Integral Domain. A nonzero commutative ring with 1 st. if ab=0, a=0 or b=0 (no zero
divisors)
Division Ring. A ring with identity in which ax = 1 has a solution for all a.
Unit. A member of Un where Un={x∈R | ∃ y st. xy=1} (those which are invertible).
Field. A non-zero commutative ring with 1 in which every non-zero element is a unit.
Also, a commutative division ring.
Zero Divisor. If ab=0 and a≠0 and b≠0, then a and b are zero divisors.
Additive Inverse. X such that A+X=0
Idempotent. An element is idempotent is a2 = a. A Boolean ring is a ring in which every
element is idempotent.
Nilpotent. An element is nilpotent if an = 0 for some n.
Characteristic. Let R be a ring with identity. If there is a smallest positive integer such
that n1R = 0, then R has characteristic n. Otherwise, R has characteristic 0.
Subring. If R is a ring and S⊆R, S is a subring of R if the + and * of R make S a ring.
Some Theorems about rings
Theorem. Let R and S be rings. Define addition and multiplication on R × S by:
(r, s) + (r’, s’) = (r + r’, s + s’)
(r, s)(r’, s’) = (rr’, ss’)
Then, R × S is a ring. If both rings are commutative, so is R × S. If both have an
identity, so does R × S.
Proof. Verify the axioms.
Proposition. If S ⊆ R, S is a subring iff:
(1) S is closed under +
(2) S is closed under *
(3) S contains 0
(4) If a∈S, then -a∈S.
Proposition. If S is a nonempty subset of R, S is a subring iff:
(1) S is closed under subtraction.
(2) S is closed under multiplication.
Proof. Notice that closure under subtraction implies that 0R is included (a – a), that
additive inverses are included (0 – a), and that aums are included (a – (-b)).
Proposition. If F is a field and a,b∈F are nonzero, then ab≠0 (ie. A field is an integral
domain.)
Theorem. Additive inverses are unique.
Abridged Proof. Suppose a + u = a + v = 0. Then, v = (a + u) + v = (a + v) + u = u.
Lemma (Cancellation Law). If x+y=x+z in R, then y=z.
Proof. Added –x to both sides.
Remark. If x+y=x+z=0, then y=z and this is the additive inverse.
Proposition.
(1) 0R*x=0R
(2) -(ab)=(-a)(b)=a(-b)
(3) -(-a)=a
(4) -(a-b)=-a-b
(5) -(a-b)=b-a
(6) (-a)(-b)=ab
(7) If R has an identity, (-1R)a = -a.
Proof. (1): Use the distributive law. For others, show that they solve the same equations.
Theorem. In a ring, the equations a + x = b has a unique solution, x = b – a.
Proof. Note that this solution works. Any other solution, w, fulfills the equation a + w =
a + (b – a), and must be the same by cancellation.
Theorem. If R is a ring, a1a2…an has a unique value, and association does not change it.
Lemma. If e,f∈R st. ea=ae=1 and fa=af=1, then e=f. (Multiplicative identities are
unique.)
Corollary. If there are both left and right identities, then the left identity equals the right
identity.
Theorem. If a is a unit, then the equations ax = b and ya = b have unique solutions. Note
that, if the ring is not commutative, the solutions might not be the same.
Theorem. Every field is an integral domain.
Proof. If ab = 0, then we may multiply by a-1 or b-1, if a or b is non-zero, to show that the
other must be 0.
Theorem. In any integral domain, if a ≠ 0 and ab = ac, then b = c.
Proof. This equation implies that ab – ca = a(b-c) = 0. Since a is non-zero, b – c = 0.
Theorem. Every finite integral domain is a field.
Proof. Recall that an integral domain has an identity. Suppose the elements in the
integral domain are a1, …, an, and at ≠ 0. Then, ata1, …, atan are unique (by
cancellation). Since this is the number of elements originally in R, one of these must be
1R.
Proposition. If S is a subring of R and 1R∈S, then 1R is an identity for S.
Homomorphisms and Isomorphisms
Homomorphism of Rings.. If f is a mapping from R to S, and f(a+b)=f(a)+f(b) and
f(ab)=f(a)f(b), then f is a homomorphism of rings.
Isomorphism. If f is bijective and a homomorphism, it is an isomorphism.
Injective = Monomorphism = One-to-one
Surjective = Epimorphism = Onto
Bijective = Isomorphism = One-to-one and Onto
Proposition. In a homomorphism of rings:
1. f(0R)=0S
2. f(-a)=-f(a)
3. f(a-b)=f(a)-f(b)
4. f(r) is a subring of S
5. If R has an identity, f(1R) is an identity for f(R), but not necessarily for S. If f is
surjective, it is the identity.
6.
Let R be a ring with identity and let f be surjective. If a is a unit in R then f(a) is a
unit is f(R), and f(a-1) = f(a)-1
Image. Im (f) = {s∈S | ∃r st. f( r) = s}
Proposition. In an isomorphism (f: R à S):
1. a=1R ⇔ f(a)=1S
2. a=0R ó f(a)=0S
3. a is a unit in R ó f(a) is a unit in S
4. ai = 1R ó (f(a))i = 1S
5. The number of elements in R and S is the same (#R=#S)
6. The number of units in R equals the number of units in S