Abstract-Algebra-1-Rings-and-Fields, guaranteed and verified 100% Pass

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Rings and Fields
Def. A ring (𝑅, +, ∙) is a set 𝑅 with two binary operations + and ∙ , called
addition and multiplication, defined on 𝑅 such that the following axioms are
satisfied.

1) (𝑅, +) is an abelian group
2) Multiplication is associative: 𝑎 ∙ (𝑏 ∙ 𝑐) = (𝑎 ∙ 𝑏) ∙ 𝑐 (and 𝑅 is closed
under multiplication).
3) For all 𝑎, 𝑏, 𝑐 ∈ 𝑅, the left and right distributive laws hold:
𝑎 ∙ (𝑏 + 𝑐 ) = (𝑎 ∙ 𝑏) + (𝑎 ∙ 𝑐)
(𝑏 + 𝑐) ∙ 𝑎 = (𝑏 ∙ 𝑎) + (𝑐 ∙ 𝑎)


Ex. Any subset of ℂ that is a group under + is a ring under the usual addition
and multiplication. Thus (ℂ, +, ∙), (ℝ, +, ∙), (ℚ, +, ∙), and (ℤ, +, ∙) are
rings. We will refer to these rings as ℂ, ℝ, ℚ, and ℤ where the usual addition
and multiplication are understood.



Ex. Let 𝑅 be any ring and let 𝑀𝑛 (𝑅) be the set of all 𝑛 × 𝑛 matrices
having elements of 𝑅 as entries. 𝑀𝑛 (𝑅) is a ring with the usual
addition and multiplication of matrices. We can see this because:

1) 𝑀𝑛 (𝑅) is an abelian group under addition
2) Matrix multiplication is associative (and 𝑀𝑛 (𝑅) is closed under multiplication)
3) Matrix addition and multiplication are distributive:
𝐴 ∙ (𝐵 + 𝐶 ) = (𝐴 ∙ 𝐵) + (𝐴 ∙ 𝐶 )
(𝐵 + 𝐶 ) ∙ 𝐴 = (𝐵 ∙ 𝐴) + (𝐶 ∙ 𝐴).
In particular 𝑀𝑛 (ℂ), 𝑀𝑛 (ℝ), 𝑀𝑛 (ℚ), and 𝑀𝑛 (ℤ) are rings.

Notice in this example + is commutative (which is required by the
definition of a ring) but matrix multiplication is not commutative for
𝑛 ≥ 2. A ring where ∙ is commutative is called a commutative ring.

, 2


Ex. Show 𝐹, the set of all functions 𝑓: ℝ → ℝ, is a ring with the usual addition
and multiplication of functions.


Define multiplication on 𝐹 by: (𝑓 ∙ 𝑔)(𝑥 ) = 𝑓 (𝑥 )𝑔(𝑥).

With the definitions of + and ∙ , 𝐹 is a ring since:

1) We know (𝐹, +) is an abelian group under the usual addition of
functions: (𝑓 + 𝑔)(𝑥 ) = 𝑓 (𝑥 ) + 𝑔(𝑥).


2) ((𝑓 ∙ 𝑔) ∙ ℎ)(𝑥 ) = (𝑓 ∙ 𝑔)(𝑥 ) ∙ ℎ(𝑥 ) = 𝑓(𝑥) ∙ 𝑔(𝑥) ∙ ℎ(𝑥)
(𝑓 ∙ (𝑔 ∙ ℎ)) = 𝑓(𝑥 ) ∙ (𝑔 ∙ ℎ)(𝑥 ) = 𝑓(𝑥) ∙ 𝑔(𝑥) ∙ ℎ(𝑥)
and 𝐹 is closed under multiplication.


3) 𝑓 ∙ (𝑔 + ℎ)(𝑥 ) = 𝑓(𝑥 )(𝑔(𝑥 ) + ℎ(𝑥 )) = 𝑓(𝑥 )𝑔(𝑥 ) + 𝑓 (𝑥 )ℎ(𝑥)
=𝑓∙𝑔+𝑓∙ℎ
(𝑔 + ℎ) ∙ 𝑓(𝑥 ) = (𝑔(𝑥 ) + ℎ(𝑥 )) ∙ 𝑓 (𝑥 ) = 𝑔(𝑥 )𝑓(𝑥 ) + ℎ(𝑥 )𝑓(𝑥)
= 𝑔 ∙ 𝑓 + ℎ ∙ 𝑓.


Notice that if we defined multiplication as composition of functions:
i.e. 𝑓 ∙ 𝑔 = (𝑓 ∘ 𝑔)(𝑥), 𝐹 would not be a ring since this multiplication is not
distributive. For example, let 𝑓 (𝑥 ) = 𝑥 2 , 𝑔(𝑥 ) = 𝑥, ℎ(𝑥 ) = 𝑥.

𝑓 ∘ (𝑔 + ℎ) = 𝑓 ∘ (2𝑥 ) = (2𝑥 )2 = 4𝑥 2
𝑓 ∘ 𝑔 + 𝑓 ∘ ℎ = 𝑥 2 + 𝑥 2 = 2𝑥 2


So 𝑓 ∘ (𝑔 + ℎ) ≠ 𝑓 ∘ 𝑔 + 𝑓 ∘ ℎ.