Download Rings and fields

Rings and fields
Sergei Silvestrov
Spring term 2011, Lecture 3
Contents of the lecture
+ Divisors of zero. Integral domains.
+ Homomorphisms and Isomorphisms of rings.
+ the field of quotients
+ Rings of polynomials.
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Algebra course FMA190/FMA190F
2011, Spring term 2011, Sergei Silvestrov lectures
Divisors of 0 and cancelation
OBS! Repetition from the last part of previous Lecture 2
Definition 1. (p. 62)
Two elements a and b of a ring R are called divisors of 0 if a ∕= 0, b ∕= 0, but ab = 0.
Example 1. The rings (ℤ, +, ⋅), (ℚ, +, ⋅), (ℝ, +, ⋅), and (ℂ, +, ⋅) do not have the divisors of 0. In
the ring ℤn , the divisors of 0 are precisely those nonzero elements that are not relatively prime to n.
Theorem 1. A ring R has no zero divisors if and only if both the right cancellation law
ab = ac ⇒ b = c
and the left cancellation law
ba = ca ⇒ b = c
hold in R.
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Algebra course FMA190/FMA190F
2011, Spring term 2011, Sergei Silvestrov lectures
Integral domains
OBS! Repetition from the last part of previous Lecture 2
Definition 2. (p. 46) A commutative ring R with unity 1 ∕= 0 and no zero divisors is called an integral
domain.
Every field F is an integral domain since ab = 0 and a ∕= 0 imply that b = 1b = (a−1 a)b =
a−1 (ab) = a−1 0 = 0. The converse is not true. For example, the integral domain ℤ is not a field.
However, for any prime p the integral domain ℤ p is a field. It follows from the following result.
Theorem 2. (Th. 3.11, p. 61) Every finite integral domain D is a field.
Proof. Let 0, 1, a1 , . . . , an be all the elements in D. For any a ∈ D ∖ {0}, consider the elements
a1, aa1 , . . . , aan . If aai = aa j , then ai = a j by the left cancellation law, therefore all these elements
are distinct. None of these elements is 0, because D has no 0 divisors. It follows that a1, aa1 , . . . ,
aan are elements 1, a1 , . . . , an in some order, so that either a1 = 1, that is, a = 1, or aai = 1 for
some i. Thus, a has a multiplicative inverse.
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Algebra course FMA190/FMA190F
2011, Spring term 2011, Sergei Silvestrov lectures
The characteristic of a ring
OBS! Repetition from the last part of previous Lecture 2
Definition 3. Let R be a ring. If there is a least positive integer n such that n ⋅ a = 0 for all a ∈ R,
then R is said to have characteristic n. If no such n exists R is said to have characteristic 0.
The next theorem claims that if the ring has unity, it suffices to examine only a = 1.
Theorem 3. Let R be a ring with unity and characteristic n. If n > 0, then n is the least positive
integer such that n ⋅ 1 = 0.
Proof. If k is the least positive integer such that k ⋅ 1 = 0, then for all a ∈ R: k ⋅ a = k ⋅ (1a) =
(k ⋅ 1)a = 0a = 0.
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Algebra course FMA190/FMA190F
2011, Spring term 2011, Sergei Silvestrov lectures
Homomorphisms and isomorphisms (sec. 3.3)
OBS! Repetition from the last part of previous Lecture 2
Definition 4. (p. 71)
Let R and S be rings. A function ϕ : R 7→ S is a homomorphism of rings provided that for all a,
b ∈ R,
ϕ(a + b) = ϕ(a) + ϕ(b)
and
ϕ(ab) = ϕ(a)ϕ(b).
Definition 5. (p. 69)
Invertible (=one-to-one=bijective = injective and surjective) homomorphism is called an isomorphism of rings.
When the context is clear then we shall frequently write “homomorphism" in place of “homomorphism of rings". A homomorphism of rings is, in particular, a homomorphism of the underlying
abelian groups. Consequently the same terminology is used: an isomorphism of rings is a homomorphism of rings which is a one-to-one correspondence.
The kernel of a homomorphism of rings ϕ : R 7→ S is its kernel as a map of abelian groups, that is,
Ker(ϕ) = ϕ −1 (0).
The image of a homomorphism of rings ϕ : R 7→ S is the subset of S
Imϕ = {r ∈ S : s = ϕ(r) for some r ∈ R} = {ϕ(r) : r ∈ R}
Theorem 4. (Cor. 3.13, p.73)
If ϕ : R 7→ S is a homomorphism of rings, then the image Imϕ of ϕ is a subring of S.
Proof. See p. 73 in the book.
Example 2. The canonical map ℤ 7→ ℤm that maps a ∈ ℤ to its remainder modulo m is a homomorphism of rings.
Check this as an exercise!
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Algebra course FMA190/FMA190F
2011, Spring term 2011, Sergei Silvestrov lectures
The field of quotients
OBS! From here on starts new material of Lecture 3
Every subring of an integral domain is itself an integral domain. Since fields are integral
domains, it follows that every subring of a field is an integral domain. The converse of this is true,
and it is much more interesting: Every integral domain is a subring of a field.
The proof of the next theorem is a straightforward generalisation of the usual construction of
the field of rational numbers ℚ from the integral domain of integers ℤ.
Theorem 5. (Sec. 9.4, pp.317-322)
If R is an integral domain, then there is a field F (the field of quotients) containing R as a subring.
Moreover, F can be chosen so that, for each f ∈ F , there are a, b ∈ R with b ∕= 0 and f = ab−1 .
Proof. Let X = { (a, b) ∈ R : b ∕= 0 } and define a relation ≡ on X by (a, b) ≡ (c, d) if ad = bc. We
claim that ≡ is an equivalence relation. Verifications of reflexivity and symmetry are straightforward;
here is the proof of transitivity. If (a, b) ≡ (c, d) and (c, d) ≡ (e, f ), then ad = bc and c f = de.
But ad = bc gives ad f = b(c f ) = bde. Cancelling d , which is nonzero, gives a f = be; that is,
(a, b) ≡ (e, f ).
Denote the equivalence class of (a, b) by [a, b], define F as the set of all equivalence classes
[a, b], and equip F with the following addition and multiplication:
[a, b] + [c, d] = [ad + bc, bd]
and
[a, b][c, d] = [ac, bd].
First, since b ∕= 0 and d ∕= 0, we have bd ∕= 0, because R is an integral domain, and so the formulas
make sense. Let us show that addition is well-defined. If [a, b] = [a′ , b′ ] (that is, ab′ = a′ b) and
[c, d] = [c′ , d ′ ] (that is, cd ′ = c′ d ), then we must show that [ad + bc, bd] = [a′ d ′ + b′ c′ , b′ d ′ ]. But this
is true:
(ad + bc)b′ d ′ = ab′ dd ′ + bb′ cd ′ = a′ bdd ′ + bb′ c′ d = (a′ d ′ + b′ c′ )bd.
A similar argument shows that multiplication is well-defined.
The verification that F is a commutative ring is now routine: The zero element is [0, 1], the one
is [1, 1], and the additive inverse of [a, b] is [−a, b]. It is easy to see that the family R′ = { [a, 1] : a ∈
R } is a subring of F , and we identify a ∈ R with [a, 1] ∈ R′ .
To see that F is a field, observe that if [a, b] ∕= [0, 1], then a ∕= 0, and the inverse of [a, b] is
[b, a].
Finally, if b ∕= 0, then [1, b] = [b, 1]−1 , and so [a, b] = [a, 1][b, 1]−1 .
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Algebra course FMA190/FMA190F
2011, Spring term 2011, Sergei Silvestrov lectures
The set R[x]
We will now take somewhat unexpected point of view on polynomials. This new simple and
rigorous point of view however is central for many exciting modern applications such wavelets in signal and image processing, various important computer algorithms, algorithms using pseudo-random
numbers generators, coding and decoding algorithms, quantum computing algorithms, automatic
control, fractals, stochastic processes and statistics methods based on time-series and Markov
chains such as for example Google’s pagerank, various parts of physics for example in quantum
mechanics and nano physics, etc ...
This different way of treating polynomials allows to gain also more fundamental and deeper understanding of polynomials including the meaning of writing them as you did in school and
analysis courses, different forms for writing/treating polynomials and operation on them, various
properties of single polynomials and the set of polynomials as a whole, etc .....
Definition 6. (Apendix G, pp 542-548)
Let R be a ring. A sequence σ in R is a function σ : ℤ+ ∪ {0} 7→ R.
In what follows, we write ai instead of σ (i), and denote a sequence by σ =
(a0 , a1 , . . . , am , . . . ).
Definition 7. A sequence σ belongs to the set R[x] if it has the form
σ = (a0 , a1 , . . . , am , 0, 0, . . . ).
Definition 8. An indeterminate x is the following element of the set R[x]:
x = (0, 1, 0, 0, . . . ).
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Algebra course FMA190/FMA190F
2011, Spring term 2011, Sergei Silvestrov lectures
Ring operations in R[x]
Define addition and multiplication in the set R[x] as follows: if σ = (a0 , a1 , . . . ) and τ =
(b0 , b1 , . . . ), then
σ + τ = (a0 + b0 , a1 + b1 , . . . )
and
σ τ = (c0 , c1 , . . . ),
where ck = ∑ki=0 ai bk−i .
Theorem 6. If R is a ring then R[x] is a ring that contains R as a subring.
Proof. Verification of the axioms in the definition of a ring is routine. The subset { (a, 0, 0, . . . ) : a ∈
R } is a subring of R[x] that we identify with R.
Theorem 7. If σ = (a0 , a1 , . . . , am , 0, 0, . . . ), then
σ = a0 + a1 x + ⋅ ⋅ ⋅ + am xm .
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