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Commutative Rings

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COMMUTATIVE RINGS


JAMES D. LEWIS




Contents
1. Properties of the Real Numbers 2
2. Some Examples of Fields 6
3. Mathematical induction 13
4. Division and Factoring 15
5. Computing GCD’s via Prime Decomposition 21
6. Least Common Multiples 21
7. Equivalence relations 22
8. Main Example: The ring Zn 24
9. Solutions of Equations in Rings 34
10. Polynomial Rings 39
11. Evaluation of Polynomials 42
12. Euclid’s Division Algorithm for Polynomials over Fields 45
13. Some ring invariants 64
14. Localization 66
15. Appendix: Zorn’s Lemma 86




Date: April 21, 2010.
Cover page computer artistic work courtesy of Dale R. Lewis.
1

,2 JAMES D. LEWIS


1. Properties of the Real Numbers
The real numbers R are represented by the real line:

− +
←− − − − − − ◦ − − − − − −→
subdivided into the subsets
 a a 
R = R<0 {0} R>0 ,
`
where = disjoint union. We are interested in the substructures:

Z = Integers = {0, ±1, ±2, . . .}
 
a a1 a2
Q = Rational numbers = a, b ∈ Z, b 6= 0 & = ⇔ a1 b2 = b1 a2
b b1 b2
N = {1, 2, 3, . . .} = Natural numbers. There are inclusions

N ⊂ Z ⊂ Q ⊂ R,
n
where the inclusion Z ⊂ Q is given by n ∈ Z 7→ 1 ∈ Q.


Description of the Real Numbers


There is the following picture:
`
R = Q {Irrational numbers}

|| ||

repeating
non − repeating
decimals
decimals
e.g. 0.235

. &

algebraic transcendental
irrationals
√ √ irrationals
e.g. 2, 3 5 e.g. e, π
The algebraic irrational numbers are solutions of equations of the form:
1
x2 − 2 = 0 ; 3x3 − = 0,
2
i.e. single variable polynomial equations with Q-coefficients.


Axiomatic Properties of the Real Numbers

, COMMUTATIVE RINGS 3


R (as well as N, Z, Q) has two (binary) operations +, •:
+
R×R → R

(a, b) 7→ a + b

R×R → R

(a, b) 7→ ab
Properties of [R; +, •]

(1) R is closed under +, •, i.e. a, b ∈ R ⇒√a + b, ab ∈ R. [Analogous notion: R is

not closed under ’s. E.g. −1 ∈ R, but −1 6∈ R.]

(2) Associativity.
(a + b) + c = a + (b + c)
(ab)c = a(bc)
[Hence can write a + b + c, abc.] For example, if we denote by f+ : R × R → R
the + map, i.e., f+ (a, b) = a + b, then associativity implies that f+ (f+ (a, b), c) =
f+ (a, f+ (b, c)).

(3) Commutivity.
a+b=b+a
ab = ba
[I.e. f+ (a, b) = f+ (b, a) and similarly f• (a, b) = f• (b, a), where f• (a, b) := ab is the
corresponding multiplication map.]

(4) Zero element. There exists an element labeled 0 ∈ R such that a + 0 = a for all
a ∈ R. [Note: It will be proven that 0 is unique.]

(5) Identity element (or Unity). There exists an element labelled 1 ∈ R such that
1 · a = a for all a ∈ R. [Note: It will be proven that 1 is unique.]

(6) Additive Inverse. For any a ∈ R, there exists an element labelled −a ∈ R such
that a + (−a) = 0. [Note: It will be proven that additive inverses are unique.]

(7) Multiplicative Inverse. For any a ∈ R, a 6= 0, there exists an element labelled
a−1 ∈ R such that a · a−1 = 1. [Note: It will be proven that multiplicative inverses
are unique.]

(8) Distributive. [Interaction of +, •.]
a(b + c) = ab + ac
(9) 1 6= 0.

, 4 JAMES D. LEWIS


Remark 1.1. 1. If a = b & c = d, then a + c = b + d and ac = bd, i.e. the
operations +, • are well-defined.

2. Given a, b ∈ R, we write
a − b := a + (−b)
and if b 6= 0, then
a
:= a · b−1
b
3. [R; +, •] is an example of a field, i.e. satisfies axiomatic properties (1) -(9)
above. [We only require axiomatic properties (1)-(4), (6), (8) to define a ring.
Roughly then, a “ring including division” amounts to a field.]
Some Consequences of the 9 Axiomatic Properties of [R; +, •]

(i) 0 is unique, i.e. there is only one zero.
[Restatement: If 0̃ also satisfies the property that 0̃ + a = a, for all a ∈ R, then
0̃ = 0.]

0̃ = 0̃ + 0 = 0
↑ ↑
Reason :
def 0 n of def 0 n of
zero 0 zero 0̃
(ii) 1 is unique, i.e. there is only one unity. [Reason: Similar to (i) above.]

(iii) a · 0 = 0 for any a ∈ R.
Reason: We refer to the axiomatic properties (1)-(9) above. Then
(4) (8)
a · 0 = a · (0 + 0) = a · 0 + a · 0
Next, add −(a · 0) to both sides. Thus
(6) (2) (6) (4)
0 = (a·0)+(−(a·0)) = (a·0+a·0)+(−(a·0)) = a·0+(a·0+(−a·0)) = a·0+0 = a·0
Therefore a · 0 = 0.

(iv) a · b = 0 ⇒ a = 0 or b = 0. Reason: If a = 0, then we’re done. Therefore
assume a 6= 0, hence a−1 ∈ R exists. We must then show that b = 0. But
(iii) (2) (5)
0 = a−1 · 0 = a−1 (a · b) = (a−1 · a)b = 1 · b = b
Hence b = 0.

(v) [Cancellation Law for Multiplication] If ac = bc and if c 6= 0, then a = b.
Reason: One shows that ac = bc ⇔ (a − b)c = 0, and then apply (iv) above. In
showing that ac = bc ⇔ (a − b)c = 0, one first shows that −(bc) = (−b)c. This is
an exercise left to the reader.

(vi) [Cancellation Law for Addition] If a + c = b + c then a = b. Reason: Add
“−c” to both sides of “a + c = b + c ”.

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