Algebra through Examples
Lesson 1
General Details
E-mail: [email protected]
Recommended reading:
- Basic Algebra 1/2 by Jacobs
- TODO: Fill from others
Administrative Details:
- There will be 5 assignments. Each around 5%
- 1 home exam – usually around 80% (best 4 assignments out of the 5 are chosen)
The Axiums of a Field
A field F has two binary operations: +, ∙ such that ∀𝑎, 𝑏, 𝑐, 𝑑 ∈ 𝐹: 𝐹 is closed under them
Addition
(1a) Commutativity: 𝑎 + 𝑏 = 𝑏 + 𝑎
(1b) Associativity: (𝑎 + 𝑏) + 𝑐 = 𝑎 + (𝑏 + 𝑐)
(1c) Neutral element: 𝑎 + 0𝐹 = 𝑎
(1d) Inverses ∀𝑎∃-𝑎, 𝑎 + (-𝑎) = 0𝐹
Multiplication
(1m) Commutativity: 𝑎 ∙ 𝑏 = 𝑏 ∙ 𝑎
(2m) Associativity: (𝑎 ∙ 𝑏) ∙ 𝑐 = 𝑎 ∙ (𝑏 ∙ 𝑐)
(3m) Identity: 𝑎 ∙ 1𝐹 = 𝑎
(4m) Inverses: ∀𝑎 ≠ 0𝐹 ∃𝑎-1 . 𝑎 ∙ (𝑎-1 ) = 1𝐹
We also demand that 0𝐹 ≠ 1𝐹
Distributivity
To connect the two definitions (as they can be independent according to the current
definition) we add distributivity, which states that:
𝑎 ∙ (𝑏 + 𝑐) = 𝑎 ∙ 𝑏 + 𝑎 ∙ 𝑐
Naming
Any set satisfying (∗) is called a group (an additive group)
If also commutatibity is satisfied, we denote it as a commutative (abelian) group.
If the operation is denoted by multiplication, we call it a multiplication group.
(2m, 3m, 4m is satisfied).
Usually denote operation by + only for abelian groups.
A Ring
A ring is any structure that satisfies (1-4a), (2m), (3m) & Distribution.
If the multiplication is commutative, it is called a commutative ring.
If (4m) holds (not necessarily with(1m)), then it is called a division ring.
A ring without (3m) is sometimes referred to as a rng. (a ring without the i).
Examples
Fields
-
ℚ
ℝ
ℂ
ℤp = {0,1, … , p − 1} with respect to addition and multiplication 𝑚𝑜𝑑 𝑝.
For instance, in ℤ5 – 2 ∙ 3 = 1(𝑚𝑜𝑑 𝑝)
Rings
Since fields support additional properties than ring, any field is a ring.
For instance - ℤ
And in addition, here are a few "pure" rings:
- ℝ[𝑥] = Ring of polynomials with real coefficients
- 𝑀𝑛 (ℝ) = Ring of 𝑛 × 𝑛 matrices over ℝ - Not commutative!
- 𝑀𝑛 (𝔽) = Ring of 𝑛 × 𝑛 matrices over some field 𝔽 - Not commutative!
- 𝔽[𝑥] = Ring of polynomials over some field 𝔽
- ℤ[𝑥] = Ring of polynomials over ℤ
- ℤ × ℤ = {(𝑎, 𝑏)|𝑎, 𝑏 ∈ ℤ} with coordinate-wise addition and multiplication:
(𝑎1 , 𝑏1 ) + (𝑎1 + 𝑏1 ) = (𝑎1 + 𝑎2 , 𝑏1 + 𝑏2 )
- If 𝑅, 𝑆 are Rings → 𝑅 × 𝑆 is a Ring.
- ℤ[𝑥, 𝑦] = polynomials in 𝑥 & 𝑦 with coefficients in ℤ.
Commutative Rings
- A sub-Ring if 𝑅 is a Ring.
𝑆 is a sub-Ring if 1𝐹 , 0𝐹 ∈ 𝑆 and 𝑆 is a Ring in respect of operations in R
for instance, 𝑀𝑛 (ℝ) is a sub-Ring of 𝑀𝑛 (ℚ)
Ideals
If 𝑅 is a Ring, 𝐼 ⊆ 𝑅 is an Ideal if and only if:
- 𝐼 is an additive subgroup of 𝑅
- ∀𝑎 ∈ 𝑅, 𝑏 ∈ 𝐼. 𝑎 ∙ 𝑏, 𝑏 ∙ 𝑎 ∈ 𝐼
(𝑅 ∙ 𝐼 ⊆ 𝐼 & 𝐼 ∙ 𝑅 ⊆ 𝐼)
Note that if 1𝐹 ∈ 𝐼 → 𝑅 = 𝐼
Examples
In any Ring 𝑅:
- {0}, 𝑅 are Ideals (Trivial)
In a commutative Ring, if 𝑏 ∈ 𝑅 → 𝑅 ∙ 𝑏 is an Ideal. Is also called principal Ideal and is
denoted by (𝑏)
-
𝑎1 𝑏 + 𝑎2 𝑏 = (𝑎1 + 𝑎2 )𝑏 + 𝑅 ∙ 𝑏
𝑎′ (𝑏 ∙ 𝑎) = (𝑎 ∙ 𝑏)𝑎′ = (𝑎′ ∙ 𝑎)𝑏 ∈ 𝑅 ∙ 𝑏
In case of a non commutative Ring, a left Ideal is an additive subgroup satisfying
multiplication on the left. In the same way, a Right Ideal satisfies multiplications on the
right.
Ideals in ℤ
- 2ℤ
- 7ℤ
- 𝑛ℤ (∀𝑛 ∈ ℤ)
In fact, every Ideal in ℤ is a principal Ideal!
Proof
Let 𝐼 be an Ideal in ℤ (notation: 𝐼 ⊲ 𝑅)
If 𝐼 = {0𝐹 } it is a principal!
So assume 𝐼 ≠ {0𝐹 }. Let 𝑛 be the smallest positive integer in 𝐼.
(𝐼 is closed under addition inverse so must have one!).
Let 𝑚 ∈ 𝐼.
We can find 𝑞, 𝑟 ∈ ℤ s.t. 𝑚 = 𝑞 ∙ 𝑛 + 𝑟 , 0 ≤ 𝑟 < 𝑛
𝑚
⏟ − 𝑞⏟
∙𝑛=𝑟 ∈𝐼
∈𝐼
∈𝐼
But we know 𝑟 < 𝑛 → Contradiction by minimality in choice of 𝑛. So 𝑟 must be 0!
Therefore:
𝑚 = 𝑞 ∙ 𝑛 ∈ 𝑛ℤ
So we proved that ∀𝑚 ∈ 𝐼. 𝑚 ∈ 𝑛ℤ → 𝐼 ⊆ 𝑛ℤ
But also 𝑛ℤ ⊆ 𝐼 since 𝑛 ∈ 𝐼!
Therefore 𝑛ℤ = 𝐼 .
More Ideal Examples
𝑀2 (ℝ) is a non-commutative Ring
𝑎 𝑏
𝑘 = {[
] |𝑎, 𝑏, 𝑐 ∈ ℝ} is a subring but not a left or right Ideal.
𝑐 𝑑
e.g.
1 1 𝑎 𝑏
𝑎 𝑏+𝑐
[
]∙[
]=[
] ∈ 𝑘 𝑜𝑛𝑙𝑦 𝑖𝑓 𝑎 ≠ 0
1 1 0 𝑐
𝑎 𝑏+𝑐
𝑎 𝑏 1 1
𝑎+𝑏 𝑎+𝑏
[
]∙[
]=[
] ∈ 𝑘 𝑜𝑛𝑙𝑦 𝑖𝑓 𝑐 ≠ 0
0 𝑐 1 1
𝑐
𝑐
However, 𝐼 = {[
𝑎
0
𝑏
] |𝑎, 𝑣 ∈ ℝ} is a right Ideal!
0
e.g.
∗ ∗
𝑎 𝑏 𝑥 𝑦
[
]∙[
]=[
]∈𝐼
𝑢
𝑣
0
0
0 0
It is not, however, a left Ideal:
𝑥 𝑦 𝑎 𝑏
𝑎𝑥 ∗
[
]∙[
]=[
] 𝑖𝑓 𝑢𝑎 ≠ 0 → ∉ 𝐼
𝑢 𝑣 0 0
𝑢𝑎 ∗
Fields have no non-trivial ideals.
Quotients of Rings
Let 𝑅 be a Ring and 𝐼 an Ideal.
∀𝑎 ∈ 𝑅 define:
𝐼 + 𝑎 = {𝑥 + 𝑎|𝑥 ∈ 𝐼} − co-set or 𝐼 determined by 𝑎.
𝑅⁄ = {𝐼 + 𝑎|𝑎 ∈ 𝑅} (equality sets)
𝐼
Quotient Ring – we define operations +,∙ to get a ring
(Note: co-sets are disjoint or equal. Proving it would be an assignment).
Define (𝐼 + 𝑎) + (𝐼 + 𝑏) = 𝐼 + (𝑎 + 𝑏)
Define (𝐼 + 𝑎) ∙ (𝐼 + 𝑏) = 𝐼 + (𝑎 ∙ 𝑏)
Must show the definition does not depend on co-sets representatives:
Suppose 𝐼 + 𝑎 = 𝐼 + 𝑎′ and 𝐼 + 𝑏 = 𝐼 + 𝑏′
Need to show: 𝐼 + (𝑎′ + 𝑏 ′ ) = 𝐼 + (𝑎 + 𝑏) and 𝐼 + 𝑎′ ∙ 𝑏 ′ = 𝐼 + 𝑎 ∙ 𝑏
∃𝑥 ∈ 𝐼 𝑎′ = 𝑥 + 𝑎
∃𝑦 ∈ 𝐼 𝑏 ′ = 𝑥 + 𝑏
So - 𝐼 + (𝑎′ + 𝑏 ′ ) = 𝐼 + (𝑥 + 𝑎 + 𝑦 + 𝑏) = 𝐼 + (𝑥
⏟ + 𝑦) + (𝑎 + 𝑏) = 𝐼 + (𝑎 + 𝑏)
∈𝐼
Note: 𝐼 + 𝑧 = 𝐼, ∀𝑧 ∈ 𝐼
Lets look at 𝐼 + 𝑎′ ∙ 𝑏′
𝐼 + 𝑎′ ∙ 𝑏 ′ = 𝐼 + (𝑥 + 𝑎)(𝑦 + 𝑏) = 𝐼 + 𝑥𝑦
⏟ + 𝑎𝑏 = 𝐼 + 𝑎 ∙ 𝑏
⏟ + 𝑎𝑦
⏟ + 𝑥𝑏
∈𝐼
∈𝐼
∈𝐼
In the 𝑅⁄𝐼 quotient ring, the 0𝐹 element is 𝐼.
Since 𝐼 + (𝐼 + 𝑎) = 𝐼 + 𝑎
The 1𝐹 element is 𝐼 + 1 etc…
Examples
1. ℤ⁄𝑛ℤ
For instance, when n=6
(6ℤ + 2) + (6ℤ + 3) = 6ℤ + 5
(6ℤ + 3) + (6ℤ + 4) = 6ℤ + 7 = 6ℤ + 1
TODO: Had a multiplication I did not have time to copy
We can actually think of ℤ⁄𝑛ℤ as {0̅, 1̅, … , ̅̅̅̅̅̅̅
𝑛 − 1} wrt +,∙ 𝑚𝑜𝑑 𝑛
2. 𝐹[𝑥]⁄𝑓(𝑥)𝐹[𝑋] 𝑤ℎ𝑒𝑟𝑒 𝐹 𝑖𝑠 𝑎 𝑓𝑖𝑒𝑙𝑑
for instance, when 𝑓(𝑥) = 𝑥 2 − 3𝑥 + 2, 𝐹 = ℤ
So in fact:
ℝ[𝑥]⁄
= {𝐼 + 𝑎𝑥 + 𝑏|𝑎, 𝑏 ∈ ℝ}
𝑥 2 − 3𝑥 + 2
Since addition and multiplication are in polynomials mod (𝑥 2 − 3𝑥 + 2)
Same as before (with numbers) - ∀𝑓, 𝑔 ∈ ℝ[𝑥]. (𝐼 + 𝑓(𝑥)) + (𝐼 + 𝑔(𝑥)) = 𝐼 +
𝑓(𝑥) + 𝑔(𝑥).
Any polynomial 𝑓(𝑥) can be written in the form:
𝑓(𝑥) = 𝑞(𝑥)(𝑥 2 − 3𝑥 + 2) + 𝑟(𝑥)
where 𝑞(𝑥), 𝑟(𝑥) ∈ ℝ[𝑥] ∧ [𝑑𝑒𝑔𝑟𝑒𝑒(𝑟(𝑥)) < 2 ∨ 𝑟(𝑥) = 0]
Also, since 𝑥 2 − 3𝑥 + 2 = (𝑥 − 1)(𝑥 − 2) →
(𝐼 + (𝑥 − 1)) ∙ (𝐼 + (𝑥 − 2)) = 𝐼
(𝐼 + (2𝑥 + 1)) + (𝐼 + (3𝑥 − 5)) = 𝐼 + (5𝑥 − 4)
(𝐼 + (2𝑥 + 1)) ∙ (𝐼 + (3𝑥 − 5)) = 𝐼 + (2𝑥 + 1)(3𝑥 − 5) =
𝐼 + 6𝑥 2 − 2𝑥 − 5 = 𝐼 + 6(𝑥 2 − 3𝑥 + 2) + (−16𝑥 − 17) =
𝐼 − 16𝑥 − 17
(2𝑥 + 1)(3𝑥 − 5) ≡ −16𝑥 − 17(𝑚𝑜𝑑 𝐼)
𝑎 ≡ 𝑏(𝑚𝑜𝑑 𝐼) ↔ 𝐼 + 𝑎 = 𝐼 + 𝑏
------End of lesson 1
Homo-morphisms of rings
If 𝑅, 𝑆 are Rings, then the function 𝜙: 𝑅 → 𝑆 is a ring homomorphism if
1) ∀𝑎, 𝑏 ∈ 𝑅 𝜙(𝑎 + 𝑏) = 𝜙(𝑎) + 𝜙(𝑏)
2) ∀𝑎, 𝑏 ∈ 𝑅 𝜙(𝑎 ∙ 𝑏) = 𝜙(𝑎) ∙ 𝜙(𝑏)
3) 𝜙(1𝑅 ) = 1𝑅
If 𝜙 satisfies (1) and (2) then: if 𝜙(1) = 𝑥 → 𝜙(1) = 𝜙(1 ∙ 1) = 𝜙(1)2
𝑥 = 𝑥 2 so (𝑥 − 1)𝑥 = 0
If 𝑅 is a domain (𝑎𝑏 = 0 → 𝑎 = 0 𝑜𝑟 𝑏 = 0) then it follows that either 𝑥 = 0 or 𝑥 − 1 = 0.
If 𝑥 = 0 then:
𝜙(𝑎) = 𝜙(𝑎 ∙ 1) = 𝜙(𝑎) ∙ 𝜙(1) = 𝜙 (𝑎) ∙ 𝑥 = 0
Otherwise, get 𝜙(1) = 1
If 𝑅 is not a domain, (1)&(2) 𝜙 ≠ 0 do not in general imply 𝜙(1) = 1.
Claim: If 𝜙: 𝑅 → 𝑆 homomorphism, then 𝑘𝑒𝑟𝜙{𝑎 ∈ 𝑅|𝜙(𝑎) = 0} is an ideal in 𝑅.
Proof – in assignment 1.
𝐼𝑚𝜙{𝜙(𝑎)|𝑎 ∈ 𝑅}
Homomorphism theorem for Rings
1) If 𝜙: 𝑅 → 𝑆 is onto 𝑆 then 𝑅⁄𝑘𝑒𝑟𝜙 ≅ 𝑆 (≅ is isomorphic!)
& isomorphism (homomorphism which is 1-1 & onto) is given by:
𝑘𝑒𝑟𝜙 + 𝑎 → 𝜙(𝑎)
2) If 𝐼 ⊲ 𝑅 ideal then the map 𝑎 → 𝐼 + 𝑎 is a homomorphism from 𝑅 to 𝑅⁄𝐼 & its
kernel is 𝐼.
Proofs: Verification
In (1) you need to check that the map is well-defined
i.e. if 𝑘𝑒𝑟𝜙 + 𝑎 = 𝑘𝑒𝑟𝜙 + 𝑎′ then 𝜙(𝑎) = 𝜙(𝑎′ )
If this holds, then 𝑎 − 𝑎′ ∈ 𝑘𝑒𝑟𝜙
As 𝑎′ = 𝑎′ ∈ 𝑘𝑒𝑟𝜙 + 𝑎′ = 𝑘𝑒𝑟𝜙 + 𝑎
Proof:
∃𝑥 ∈ 𝑘𝑒𝑟𝜙: 𝑎′ = 𝑥 + 𝑎
𝜙(𝑎 = 𝜙(𝑥 + 𝑎) = 𝜙(𝑥) + 𝜙(𝑎) = 𝜙(𝑎)
Note: 𝑘𝑒𝑟𝜙 = {0} ↔ 𝜙 𝑖𝑠 1 − 1.
′)
Our note:
Lets prove the note!
→
Suppose we have 𝑠1 ∈ 𝑆 s.t. ∃𝑥1 , 𝑥2 ∈ 𝑅 𝜙(𝑥1 ) = 𝜙(𝑥2 ) = 𝑠1 .
However: 𝜙(𝑥1 − 𝑥2 ) = 𝜙(𝑥1 ) − 𝜙(𝑥2 ) = 0 → 𝑥1 − 𝑥2 ∈ 𝑘𝑒𝑟𝜙 → 𝑥1 − 𝑥2 = 0 → 𝑥1 =
𝑥2 → Contradiction!
←
First lets prove that 0 is in the 𝑘𝑒𝑟𝜙:
𝑎 = 𝑎 + 0 → 𝜙(𝑎) = 𝜙(𝑎 + 0) → 𝜙(𝑎) = 𝜙(𝑎) + 𝑝ℎ𝑖 (0) → 𝑝ℎ𝑖(0) = 0
Now, since 𝜙 is 1-1, there can only be one element of R going to 0. And we just found it.
So 𝑘𝑒𝑟𝜙 = {0}.
Example
ℝ[𝑥]⁄
(𝑥 2 + 1) ≅ ℂ
𝑘
∑ 𝑎𝑗 𝑥 𝑗
𝑗=0
𝜙
Look at homomorphism: 𝑓(𝑥) → 𝑓(𝒾) from ℝ[𝑥] → ℂ
What is the kernel?
𝑘𝑒𝑟𝜙 = {𝑓(𝑥) ∈ ℝ[𝑥]|𝑓(𝒾) = 0}
= {𝑓(𝑥) ∈ ℝ[𝑥]|𝑓(𝑥)𝑖𝑠 𝑎 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑜𝑓 𝑥 2 + 1 𝑏𝑦 𝑎𝑛𝑜𝑡ℎ𝑒𝑟 𝑝𝑜𝑙𝑦𝑛𝑜𝑚}
(we shall see that later)
Example2
𝜙: ℤ → {0̅, 1̅, … , ̅̅̅̅̅̅̅
𝑛 − 1} that sends 𝑥 ∈ ℤ to 𝑥̅ (𝑚𝑜𝑑 𝑛)= remainder of 𝑥 (𝑚𝑜𝑑 𝑛).
𝑘𝑒𝑟𝜙 = 𝑛ℤ so ℤ⁄𝑛ℤ = ~ℤ𝑛
From now on we’re going to look at commutative Rings!
Commutative Rings
Definition: 𝑅 is a domain if 𝑎𝑏 = 0 → 𝑎 = 0 𝑜𝑟 𝑏 = 0 for all 𝑎, 𝑏 ∈ 𝑅.
Domain – ‫תחום שלמות‬
Examples
ℝ[𝑋], 𝔽[𝑥] (𝔽 𝑠𝑜𝑚𝑒 𝑓𝑖𝑒𝑙𝑑)
ℤ
ℤ[𝑥]
ℤ𝑋ℤ (not a domain!)
ℤ5 𝑋ℤ5 (not a domain!)
− 𝑟𝑖𝑛𝑔 𝑜𝑓 𝑛𝑥𝑛 𝑚𝑎𝑡𝑟𝑖𝑐𝑒𝑠 𝑜𝑣𝑒𝑟 𝑎 𝑓𝑖𝑒𝑙𝑑 (not a domain!)
PID
Definition: R is a principal ideal domain (‫)תחום ראשי‬
If it is a domain & every ideal in it is a principal
(i.e. of the form (𝑎) = 𝑅𝑎, 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑎 ∈ 𝑅 )
Examples
𝔽[𝑋] ← 𝐴𝑠𝑠𝑖𝑔𝑛𝑚𝑒𝑛𝑡 1
Counter example?
ℤ[𝑥] is not a PID! But it’s a domain…
Look at the ideal generated by 𝑥 and 2 (the set of polynomials over ℤ with an even constant
term)
𝑥 ∙ ℤ[𝑥] + 2 ∙ ℤ[𝑥]
For the sake of contradiction, suppose it were a principal ideal. Then there would exist some
polynomial 𝑔(𝑥) which generated the ideal. But since 2 is in the ideal, it must be a multiple
of 𝑔(𝑥), so 𝑔(𝑥) must be a constant, say 𝑛. But 𝑥 is also in the ideal, so it must be the
product of 𝑛 with some 𝑓(𝑥) in ℤ[𝑥]: 𝑥 = 𝑛𝑓(𝑥). Since the coefficient of 𝑥 on the left hand
side is 1, the coefficient of 𝑥 on the right hand side must also be 1. On the other hand, the
coefficient of 𝑥 on the right hand side is a multiple of 𝑛. So 𝑛 = ±1. But this means that our
ideal is actually generated by 1 or -1, which means it is all of ℤ[𝑥]. But this is not true, since
there are elements of ℤ[𝑥] which are not in our ideal – 𝑥 + 1 for instance. Thus, our ideal
must not be a principal ideal!
3 More properties of ℤ
(1) Euclidean property
If 𝑎, 𝑏 ∈ ℤ non-zero, then ∃𝑔, 𝑟 ∈ ℤ s.t. 0 ≤ 𝑟 < |𝑏| and 𝑎 = 𝑏𝑞 + 𝑟.
(2) Every 2 non-zero elements have a greatest common divisor
if 𝑎, 𝑏 ∈ ℤ. gcd(𝑎, 𝑏) = 𝑑, is a number in ℤ s.t. 𝑑|𝑎, 𝑑|𝑏 and if 𝑑′ is also a common
divisor then 𝑑′ |𝑑. (unique up o a sign).
(3) Unique Factorization into primes
Proof of (2):
In ℤ. If 𝑎, 𝑏 ∈ ℤ
Look at the ideal ℤ𝑎 + ℤ𝑏 = principal ideal!
So ∃𝑑 ∈ ℤ. ℤ𝑎 + ℤ𝑏 = ℤ𝑑
𝑎 = 1 ∙ 𝑎 + 0 ∙ 𝑏 ∈ ℤ𝑑 so a multiple of d, 𝑑|𝑎.
Similarily, 𝑏 ∈ ℤ𝑎 + ℤ𝑏 so 𝑑|𝑏.
Now let 𝑑′ ∈ ℤ. 𝑑′ |𝑎 & 𝑑′ |𝑏.
𝑑′ |𝑎 → 𝑎 ∈ ℤ𝑑′ so ℤ𝑎 ⊆ ℤ𝑑′
𝑎|𝑏 → ℤ𝑏 ∈ ℤ𝑑′
And so also ℤ𝑎 + ℤ𝑏 ⊆ ℤ𝑑′
So 𝑑 ∈ ℤ𝑑′ → 𝑑′ |𝑑.
Note: Suppose 𝑑 & 𝑑′ are both gcd’s of 𝑎 & 𝑏 in ℤ.
𝑑|𝑑′ so ∃𝑥 ∈ ℤ. 𝑑𝑥 = 𝑑′
𝑑′|𝑑 so ∃𝑦 ∈ ℤ. 𝑑′𝑦 = 𝑑
𝑑′ 𝑦𝑥 = 𝑑′
𝑑′ (𝑦𝑥 − 1) = 0
𝑑′ ≠ 0, 𝑠𝑜 𝑦𝑥 − 1 = 0
𝑦𝑥 = 1 → 𝑦, 𝑥 ∈ {±1}
So the GCD in ℤ is unique up o a sign.
In general: in any domain, we get uniqueness of the GCD up o an invertible element.
In Rings – invertible elements are referred to as units.
Bezout’s Theorem(In ℤ)
Let 𝑎, 𝑏 ≠ 0 in ℤ & let 𝑑 = gcd(𝑎, 𝑏).
Then, ∃𝑢, 𝑣 ∈ ℤ. 𝑎𝑢 + 𝑏𝑣 = 𝑑
This follows trivially from the fact that ℤ𝑎 + ℤ𝑏 = ℤ𝑑.
Theorem:
Let R be a PID, then if 𝑎, 𝑏 ≠ 0 then 𝑎, 𝑏 have a gcd (unique up to multiplication by a unit)
And Bezout’s theorem holds in R.
Bezout’s theorem holds – if 𝑑 = gcd(𝑎, 𝑏) then ∃𝑢, 𝑣 ∈ 𝑅. 𝑎𝑢 + 𝑏𝑣 = 𝑑.
Definition:
1) If 𝑅 is a Ring and 𝑝 ≠ 0 ∈ 𝑅 is a prime element, whenever 𝑝|𝑎 ∙ 𝑏 (𝑎, 𝑏 ∈ 𝑅) then
𝑝|𝑎 𝑜𝑟 𝑝|𝑏.
2) If 𝑅 is a Ring and 𝑥 ≠ 0 ∈ 𝑅 is an irreducible element then if 𝑥 = 𝑎 ∙ 𝑏 for some
𝑎, 𝑏 ∈ 𝑅 then a or b must be a unit.
In ℤ: prime=irreducible.
Claim: If 𝑅 is a domain then 𝑝 prime→ 𝑝 irreducible.
Proof: Suppose 𝑝 is prime and that 𝑝 = 𝑎 ∙ 𝑏 so also 𝑝|𝑎 ∙ 𝑏 so 𝑝|𝑎 or 𝑝|𝑏. Wlog, We might
as well assume that 𝑝|𝑎. So ∃𝑢 ∈ 𝑅 such that 𝑝𝑢 = 𝑎. So 𝑎𝑏𝑢 = 𝑎 → 𝑎(𝑏𝑢 − 1) = 0 & 𝑎 ≠
0.
So
𝑏𝑢 − 1 = 0 → 𝑏𝑢 = 1 and 𝑏 is a unit.
However, irreducible 𝑛𝑜𝑡 → prime in general.
Example:
ℤ[√−5] = {𝑎 + 𝑏√−5|𝑎, 𝑏 ∈ ℤ} subring of ℂ
This contains irreducible elements that are not prime.
It does contain prime elements!
First, recall that if 𝑥 + 𝑖𝑦 ∈ ℂ → ‖𝑥 + 𝑖𝑦‖2 = 𝑥 2 + 𝑦 2
And if 𝑧1 , 𝑧2 ∈ ℂ, then ‖𝑧1 ‖2 ∙ ‖𝑧2 ‖2 = ‖𝑧1 ∙ 𝑧2 ‖2 .
Use this to show √−5 is a prime element in the ring.
Assume √−5 | 𝑟 ∙ 𝑠 ∈ ℤ[−5]
2
We then got ‖√−5‖ |(‖𝑟‖2 ∙ ‖𝑠‖2 ) so 5|‖𝑟‖2 ‖𝑠‖2 and ‖𝑟‖2 , ‖𝑠‖2 are integers
And so 5|‖𝑟‖2 or 5|‖𝑠‖2
Wlog, 5|‖𝑟‖2
And write 𝑟 = 𝑎 + 𝑏√−5, 𝑎, 𝑏 ∈ ℤ
5|𝑎2 + 5𝑏 2 → 𝑎2 (𝑎𝑛𝑑 ℎ𝑒𝑛𝑐𝑒 𝑎𝑙𝑠𝑜 𝑎) are integer multiples of 5.
So write 𝑎 = 5𝑎′ , 𝑎′ ∈ ℤ.
And 𝑟 = 5𝑎′ + 𝑏√−5 = √−5
⏟⏟
(−√−5𝑎′ + 𝑏)
∈𝑅𝑖𝑛𝑔
∈ℤ[√−5]
So √−5|𝑟 in the ring.
We now show that ℤ[√−5] contains irreducible elements that are not prime.
Look at:
2 ∙ 3 = 6 = (1 + √−5)(1 − √−5)
First note that 2 is irreducible.
Suppose 2 = 𝑟 ∙ 𝑠
4 = ‖2‖2 = ‖𝑟‖2 ∙ ‖𝑠‖2
Case 1:
‖𝑟‖2 = 2 = ‖𝑠‖2
But on the other hand, if 𝑟 = 𝑎 + 𝑏√−5 then we get: 𝑎2 + 5𝑏 2 = 2 which has no solutions
with 𝑎, 𝑏 ∈ ℤ.
Case 2: wlog, ‖𝑟‖ = 1 and ‖𝑠‖2 = 4 then get 𝑎2 + 5𝑏 2 = 1 → 𝑎2 = 1 𝑎𝑛𝑑 𝑏 = 0 → 𝑎 =
±1 and 𝑟 = ±1 and so is a unit.
Note: Can show in a similar way that units of ℤ[√−5] are ±1.
We now show that 2 is not prime in ℤ[√−5].
By (*) we have that 2|(1 + √−5)(1 − √−5)
Suppose 2|1 + √−5.
Then we have 𝑎 + 𝑏√−5, 𝑎, 𝑏 ∈ ℤ: 2(𝑎 + 𝑏√−5) = 1 ± √−5 → 2𝑎 = 1 - impossible.
So 2 divides neither of the factors and so is not prime.
We shall show that In a PID, all irreducibility implies primeness.
Conclusion: ℤ[√−5] I not a PID!
------- end of lesson 2
𝑅 = ℤ[−5] not a PID.
Take 𝐼 = 2𝑅 + (1 + √−5)𝑅
6 = 2 ∙ 3 = (1 + √−5)(1 − √−5)
2 irreducible but not prime.
Also 1+√5
If 𝐼 was principal, then we would have 𝑟 such that 𝑅 ∙ 𝑟 = 2𝑅 + (1 + √−5)𝑅
Giving – 𝑟|2, 𝑟|1 + √−5
So ∃𝑠. 𝑟𝑠 = 2
Case 1: 𝑟 is a unit→ 𝑅 ∙ 𝑟 = 𝑅 → 𝐼 = 𝑅. We will show this is impossible.
Suppose ∃𝑎, 𝑏, 𝑐, 𝑑 ∈ ℤ. 1 = 2(𝑎 + 𝑏√−5) + (𝑐 + 𝑑√−5)(1 + √−5)
1 = 2𝑎 + 𝑐 − 5𝑑 + √−5(2𝑏 + 𝑐 + 𝑑)
So that:
2𝑎 + 𝑐 − 5𝑑 = 1, ⇒ 𝑐 + 𝑑 = 1(𝑚𝑜𝑑 2)
2𝑏 + 𝑐 + 𝑑 = 0 ⇒ 𝑐 + 𝑑 = 0(𝑚𝑜𝑑 2)
Contradiction!
Case 2: 𝑠 is a unit.
𝑟𝑠 −1 = 2 and 𝑟𝑠 −1 𝑠|1 + √−5
So 2|1 + √−5 - contradiction!
Future Assignments:
The grader is Niv Sarig. And he will put the assignments in his web page:
http://www.wesdom.weizmann.ac.il/~nivmoss/ate.html
There is a mailbox for the course!
Claim: In a PID all irreducibles are prime.
Proof: Suppose 𝑎 is irreducible and 𝑎|𝑏 ∙ 𝑐 in a ring 𝑅 (Assuming 𝑏 ∙ 𝑐 ≠ 0).
Since 𝑅 is a PID, 𝑎 & 𝑏 have a gcd.
gcd(𝑎, 𝑏) = 𝑑. Assume 𝑎 = 𝑑 ∙ 𝑎′ .
As 𝑎 is irreducible & 𝑑|𝑎 then either 𝑑 is invertible or 𝑎′ is invertible.
Case 1: 𝑑 is a unit. Wlog d=1.
By bezout: ∃𝑢, 𝑣. 𝑎𝑢 + 𝑏𝑣 = 1
𝑎|𝑏 ∙ 𝑐 so ∃𝑟 ∈ 𝑅. 𝑎𝑥 = 𝑏𝑐
𝑎𝑢𝑥 + 𝑏𝑥𝑣 = 𝑥
𝑎𝑢𝑥 = 𝑏𝑢𝑐
So
𝑏𝑥𝑣 + 𝑏𝑢𝑐 = 𝑥
𝑏(𝑥𝑣 + 𝑢𝑐) = 𝑥 ⇒ 𝑏|𝑥
So ∃𝑏 ′ ∈ 𝑅. 𝑏𝑏; = 𝑥
𝑎𝑥 = 𝑏𝑐
𝑎𝑏𝑏 ′ = 𝑏𝑐
𝑏(𝑎𝑏 ′ − 𝑐) = 0
𝑅 is a domain and 𝑏 ≠ 0 so 𝑎𝑏 ′ − 𝑐 = 0 ⇒ 𝑎𝑏 ′ = 𝑐 𝑎𝑛𝑑 𝑎|𝑐
Case 2: 𝑎′ is a unit.
𝑎(𝑎−1 )−1 = 𝑑
So, 𝑎|𝑑 and 𝑑|𝑏 so 𝑎|𝑏.
Unique Factorization
Definition: A domain 𝑅 (a commutative ring) is a unique factorization domain (𝑈𝐹𝐷) if any
non-unit 𝑎, 𝑎 ≠ 0 can be written as a product of irreducible elements uniquely (up to order
of the factors and units).
𝑒. 𝑔. 6 = 2 ∙ 3 = 3 ∙ 2 = (−3) ∙ (−2)
Example: ℤ, 𝔽[𝑥], 𝑎𝑛𝑦 𝑓𝑖𝑒𝑙𝑑,
ℤ[𝑥]- which is not a PID!
𝑈𝐹𝐷 does not imply 𝑃𝐼𝐷!
But 𝑃𝐼𝐷 ⇒ 𝑈𝐹𝐷.
We showed that ℤ[√−5] is NOT a PID.
Euklidian Property
Definition: A domain 𝑅 is Euclidean if we can define a map 𝛿: 𝑅\{0} → ℕ (called the
Euclidean norm) s.t. for 𝑎, 𝑏 ≠ 0 ∈ 𝑅, ∃𝑞, 𝑟 ∈ 𝑅 such that:
𝑎 = 𝑏𝑞 + 𝑟
and 𝛿(𝑟) < 𝛿(𝑏) or 𝑟 = 0.
And ∀𝑥, 𝑦 ∈ 𝑅. 𝛿(𝑥) ≤ 𝛿(𝑥𝑦)
(definition – Herstein, Jacobson does not require 𝛿(𝑥) ≤ 𝛿(𝑥𝑦))
Examples:
1) ℤ. 𝛿 = | |
2) 𝔽[𝑥], 𝔽 is a field, 𝛿 = degree of a polynomial
3) 𝔽 is a field, 𝛿(𝑎) = 0, ∀𝑎 ≠ 0
Theorem: In a Euclidean domain, every 2 non-zero elements have a gcd.
Proof: Uses Euclid’s algorithm.
Write: 𝑎 = 𝑏𝑞1 + 𝑟1 , 𝛿(𝑟1 ) < 𝛿(𝑏)
If 𝑟1 = 0 then 𝑎 = 𝑏𝑞 and 𝑔𝑐𝑑(𝑎, 𝑏) = 𝑏
If not: write 𝑏 = 𝑟1 𝑞2 + 𝑟2 , 𝛿(𝑟2 ) < 𝛿(𝑟1 ) or 𝑟2 = 0
If 𝑟2 = 0 then 𝑔𝑐𝑑(𝑎, 𝑏) = 𝑟1
Otherwise, I can write 𝑟1 = 𝑟2 𝑞3 + 𝑟3 , 𝛿(𝑟3 ) < 𝛿(𝑟2 ) or 𝑐3 = 0
If 𝑟3 = 0 then gcd 𝑎, 𝑏 = 𝑟2 …
Since 𝛿(𝑏) > 𝛿(𝑟1 ) > 𝛿(𝑟2 ) > ⋯
Is a proper decreasing sequence of units we get
For 𝑘, 𝛿(𝑟𝑘 ) = 0, the last non-zero 𝑧𝑘 is the GCD.
Note: ℤ[√−5] is not Euclidean!
And in assignment 2 you show 6 + 2(1 + √−5) have no GCD.
Theorem: If 𝑅 is Euclidean then 𝑅 is a PID.
Proof: If 𝐼 is an ideal in 𝑅, 𝐼 ≠ 0
Pick 𝑎 ∈ 𝐼 and minimal Euclidean norm. And then 𝐼 = 𝑅𝑎.
Theorem(use for PID→UFD!)
In a PID any increasing chain of Ideals stabilizes.
I.e. Given 𝐼1 ⊆ 𝐼2 ⊆ ⋯ ⊆ 𝐼𝑛 ⊆ 𝐼𝑛+1 ⊆ ⋯ ⊆ 𝑅
𝐼𝑗 Ideals ∃𝑘 𝑠. 𝑡. 𝐼𝑘 = 𝐼𝑘+1 … etc…
Proof:
Look at the union of all the Ideals: ⋃∞
𝑛=1 𝐼𝑛 = 𝐽. 𝐽 is an ideal and so principal.
So ∃𝑎 ∈ 𝑅. 𝐽 = 𝑅𝑎.
𝑎 ∈ 𝐽 so ∃𝑘. 𝑎 ∈ 𝐼𝑘
𝐼𝑘 ⊇ 𝑅𝑎 = 𝐽
So ∀𝑡 ≥ 0. 𝐼𝑘+𝑡 ⊂ 𝐼𝑘 etc. But given 𝐼𝑘+𝑡 ⊇ 𝐼𝑘 ∀𝑡 ≥ 0
So we get equality…
Example:
ℤ[𝑖] =ring of Gaussian integers = {𝑎 + 𝑏𝑖|𝑎, 𝑏 ∈ ℤ}
Turns out – this ring is Euclidean.
Proof: Define 𝛿(𝑥 + 𝑖𝑦) = 𝑥 2 + 𝑦 2 = ‖𝑥 + 𝑖𝑦‖2 .
𝛿 is multiplicative. Need to show Euclidean property holds.
Take 𝑎, 𝑏 ∈ ℤ[𝑖] 𝑎, 𝑏 ≠ 0
ℤ[𝑖] ⊆ ℚ[𝑖] = {𝑟 + 𝑠𝑖|𝑟, 𝑠 ∈ ℚ} - which is a field!
𝑟 − 𝑖𝑠
(𝑟 + 𝑠𝑖)−1 , , = 2
𝑟 + 𝑠𝑖 ≠ 0
𝑟 + 𝑠2
So 𝑎 ∙ 𝑏 −1 ∈ ℚ[𝑖].
1
1
So write: 𝑎 ∙ 𝑏 −1 = 𝛼 + 𝛽𝑖, 𝛼, 𝛽 ∈ ℚ. ∃𝑢, 𝑣 ∈ ℤ: |𝑢 − 𝛼| ≤ 2 , |𝑢 − 𝛽| ≤ 2
Let 𝑞 = 𝑢 + 𝑖𝑣 ∈ ℤ[𝑖]
𝑎𝑏 −1 = 𝑢 + 𝑖𝑣 + (𝛼 − 𝑢) + 𝑖(𝛽 − 𝑣) ∈ ℚ
𝑎𝑏 −1 = 𝑞 + (𝛼 − 𝑢) + (𝛽 − 𝑣)
So 𝛼 = 𝑏𝑞 + [(𝛼 − 𝑢) + (𝛽 − 𝑣)]b
𝑟 = 𝑎 − 𝑏𝑞 ∈ ℤ[𝑖]
Remains to show that 𝛿(𝑖) < 𝛿(𝑏).
𝛿(𝑟) = ‖(𝛼 − 𝑢) + 𝑖(𝛽 − 𝑣)‖2 ∙ ‖𝑏‖2
1
1
1
‖(𝛼 − 𝑢) + 𝑖(𝛽 − 𝑣)‖2 = (𝛼 − 𝑢)2 + (𝛽 − 𝑣)2 ≤ + =
4
4
2
1
So that 𝛿(𝑟) ≤ 2 𝛿(𝑏) < 𝛿(𝑏)
Euclidean ⇒ PID.
But PID does not imply Euclidean!
Counter Example:
1
2
ℤ[ +
√−19
]
2
a PID but not Euclidean. Check…
In 2004 it was shown that ℤ[√14] is Euclidean.
It is easy to show that: ℤ[√−𝑛] (0 > 𝑛 ∈ ℕ) is Euclidean ⇔ 𝑛 = 1 𝑜𝑟 2
In Euclidean domains: we used the Euclidean property to construct the GCDs.
In UFD: Use factorization to construct GCD’s.
𝑎 = 𝑝1 , … , 𝑝𝑘
𝑏 = 𝑞1 , … , 𝑞𝑙
Where they are irreducible.
GCD=product of common factors.
It turns out: Irreducible implies prime in a UFD.
Sum up
Euclidean⇒PID⇒UFD
But the arrows don’t go the other way!
Example:
𝑥 𝑥
2 3
4
𝑥
𝑛
𝑅 = ℤ [𝑥, , , … , , … ] = 𝑥 ∙ ℚ[𝑥] + ℤ
5 5
𝑥
6
2
𝑥
𝑥
+ 3 𝑥 + 3 = 5𝑥 4 ∙ 6 + 2 ∙ 3 ∙ 𝑥 3 + 3
𝑅 is a subring of ℚ[𝑥].
1
𝑅 ≠ ℚ[𝑥] as 2 ∉ 𝑅.
There are very interesting properties:
1) 𝑅 is a bezout Ring (and in particular, every 2 elements ≠ 0 have a GCD)
2) Any finitely generated is principal
3) But 𝑅 is not a PID!
𝑥
4) Ideals generated by {𝑥, 2 , … , … } is not principal!
5) 𝑅 not a UFD. 𝑥 is divisable in this ring, by every integer ≠ 0. So 𝑥 cannot be factored
as products of individuals.
--End of lesson 3
Commutative Rings
Chinese Remainder Theorem
𝑥 ≡ 2(𝑚𝑜𝑑3)
𝑥 ≡ 3(𝑚𝑜𝑑5)
𝑥 ≡ (𝑚𝑜𝑑7)
𝑒. 𝑔. 𝑥 = 23
This is 4th century china
Lady with the eggs
𝑥 ≡ (𝑚𝑜𝑑2)
𝑥 ≡ 1(𝑚𝑜𝑑3)
𝑥≡1(𝑚𝑜𝑑4)
⋮
𝑥≡0(𝑚𝑜𝑑7)
𝑥 = 301
CRT in ℤ
Let 𝑛1 , … , 𝑛𝑘 be pair-wise mutually prime integers. (gcd(𝑛𝑖 , 𝑛𝑗 ) = 1∀𝑖, 𝑗)
And let 𝑎1 , … , 𝑎𝑘 be arbitrary integers.
Then there exists an integer 𝑥 𝑠. 𝑡.
𝑥 ≡ 𝑎𝑖 (𝑚𝑜𝑑 𝑛𝑖 )
Note: There will be no solution 𝑥 𝑠. 𝑡. 𝑥 ≡ 1(𝑚𝑜𝑑2) and 𝑥 ≡ 0(𝑚𝑜𝑑6)
CRT in a commutative ring 𝑹
Let 𝐼1 , … , 𝑖𝑘 be pair-wise co-prime ideals in 𝑅.
(The ideal generated by a sum of any two ideals is 𝑅: 𝐼𝑗 + 𝐼𝑘 = 𝑅 ∀𝑗 ≠ 𝑘)
And 𝑎1 , … , 𝑎𝑛 ∈ 𝑅 arbitrary elements.
Then, there exists 𝑥 ∈ 𝑅 such that 𝑥 ≡ 𝑎𝑗 (𝑚𝑜𝑑𝐼𝑗 )
Or in other words 𝑥 + 𝐼𝑗 = 𝑎𝑗 + 𝐼𝑗 ∀𝑗
Derive 𝐶𝑅𝑇 for ℤ from the general theorem:
If gcd(𝑛𝑖 , 𝑛𝑗 ) = 1 then 𝑛𝑖 ℤ + 𝑛𝑗 ℤ = ℤ so conditions on ideals 𝑛𝑖 ℤ hold etc…
Prove for 𝒏 = 𝟐
We have 𝐼1 + 𝐼2 = 𝑅
So we have 𝑏𝑗 ∈ 𝐼𝑗 𝑠. 𝑡. 𝑏1 + 𝑏2 = 1
Let 𝑥 = 𝑎2 𝑏1 + 𝑎1 𝑏2
𝑥 + 𝐼1 = 𝑎⏟
𝑎1 𝑏1 + 𝐼1 = 𝑎1 + 𝐼1
2 𝑏1 + 𝑎1 𝑏2 + 𝐼1 = 𝑎1 𝑏2 + 𝐼1 = 𝑎1 (1 − 𝑏1 ) + 𝐼1 = 𝑎1 − ⏟
∈𝐼1
∈𝐼1
𝑥 ≡ 𝑎1 (𝑚𝑜𝑑𝐼1 )
Similarly
𝑥 ≡ 𝑎2 (𝑚𝑜𝑑𝐼2 )
If 𝐼, 𝐽 ideals in 𝑅
Denote 𝐼 ∙ 𝐽 =the additive subgroup generated by the products {𝑎𝑏|𝑎 ∈ 𝐼, 𝑏 ∈ 𝐽}
{𝑎1 𝑏1 + ⋯ + 𝑎𝑛 𝑏𝑛 |𝑎𝑖 ∈ 𝐼, 𝑏𝑗 ∈ 𝐽 𝑛 ≥ 0}
Note: {𝑎𝑏|𝑎 ∈ 𝐼, 𝑏 ∈ 𝐽} is closed under multiplication by elements of 𝑅.
Not necessarily closed under addition.
And then 𝐼 ∙ 𝐽 will be an ideal. 𝐼 ∙ 𝐽 ⊆ 𝐼, 𝐽 and in fact 𝐼 ∙ 𝐽 ⊆ 𝐼 ∩ 𝐽 ideal
Examples:
In ℤ
3ℤ ∙ 3ℤ = 9ℤ
But 3ℤ ∩ 3ℤ = 3ℤ
Note: If 𝑝, 𝑞 mutually prime then:
𝑝ℤ ∙ 𝑞ℤ = 𝑝𝑞ℤ = 𝑝ℤ ∩ 𝑞ℤ
In general:
𝐼1 ∙ 𝐼2 ∙ … ∙ 𝐼𝑘 - smallest ideal containing set of products.
We start by writing
𝐼1 + 𝐼2 = 𝑅 ⇒ ∃𝑐2 ∈ 𝐼1 , 𝑏2 ∈ 𝐼2 : 𝑐2 + 𝑏2 = 1
⋮
𝐼1 + 𝐼𝑛 = 𝑅 ⇒ ∃𝑐𝑛 ∈ 𝐼1 , 𝑏𝑛 ∈ 𝐼2 : 𝑐𝑛 + 𝑏𝑛 = 1
𝑛
Look at the product: ∏𝑖=2 𝑐𝑖 + 𝑏𝑖 = 1
Let 𝐽1 = 𝐼2 ∙ … ∙ 𝐼𝑛
The product has elements that has a multiplication of some 𝑐, except for the 𝑏’s.
𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒𝑠
𝑜𝑓 𝑠𝑜𝑚𝑒 𝑐 + 𝑏
⏟1 ∙ … ∙ 𝑏𝑛 = 1
⏟
∈𝐼1
∈𝐽1
So that 𝐼1 + 𝐽1 = 𝑅
By the CRT for case 𝑛 = 2 have 𝑦1 ∈ 𝑅 𝑠. 𝑡.
𝑦 ≡ 1(𝑚𝑜𝑑𝐼1 )
{ 1
𝑦1 ≡ 0(𝑚𝑜𝑑𝐽1 )
Since 𝐽1 ⊆ 𝐼2 ∩ 𝐼3 ∩ … ∩ 𝐼𝑛 we also get 𝑦1 ≡ 0(𝑚𝑜𝑑𝐼𝑗 ) 𝑗 > 1
Repeat for each 𝑖: 𝐽𝑖 = ∏𝑘≠𝑖 𝐼𝑘
Form 𝐼𝑖 + 𝐽𝑖 = 𝑅
And get 𝑦𝑖 ∈ 𝑅 𝑠. 𝑡.
𝑦𝑖 ≡ 1(𝑚𝑜𝑑𝐼𝑖 )
𝑦𝑖 ≡ 0(𝑚𝑜𝑑𝐽𝑖 )
And so also 𝑦𝑖 ≡ 0(𝑚𝑜𝑑𝐼𝑘 ) 𝑘 ≠ 𝑖
Let 𝑥 = 𝑎1 𝑦1 + 𝑎2 𝑦2 +. . +𝑎𝑛 𝑦𝑛
𝑚𝑜𝑑𝐼1 : 𝑥 ≡ 𝑎1 + 0 + similarly for all 𝑗 𝑥 ≡ 𝑎𝑗 (𝑚𝑜𝑑𝐼𝑗 )
In ℤ
Note that 𝑥 ≡ 𝑎𝑖 (𝑚𝑜𝑑𝑛𝑖 ) ∀𝑖 not unique.
𝑥 + ∏ 𝑛𝑖 will solve all the congruences.
Corollaries:
Let 𝑅 be a commutative ring. 𝐼1 , … , 𝐼𝑛 mutually coprime ideals in 𝑅.
Then
𝑅⁄
𝑅
𝑅
𝑅
(𝐼1 ∩ … ∩ 𝐼𝑛 ) ≅ ( ⁄𝐼1 ) × ( ⁄𝐼2 ) × … × ( ⁄𝐼𝑛 )
(actually equivalent to CRT)
Proof: Define a homomorphism 𝑓: 𝑅 → (𝑅⁄𝐼 ) × … × (𝑅⁄𝐼 )
1
𝑛
By 𝑓(𝑎) = (𝑎 + 𝐼1 , … , 𝑎𝐼𝑛 ) = (𝑎(𝑚𝑜𝑑𝐼1 ), … , 𝑎(𝑚𝑜𝑑𝐼𝑛 ))
Clearly this is a homomorphism. (not so clear. TODO go over it)
Clearly 𝑓 is additive and multiplicative.
𝑓(1) = (1(𝑚𝑜𝑑1 ), … ,1(𝑚𝑜𝑑𝐼𝑛 ))
We calculate ker 𝑓:
𝑎 ∈ ker 𝑓 ⇔ 𝑎 ≡ (𝑚𝑜𝑑𝐼𝑗 ) for all 𝑗 ⇔ 𝑎 ∈ 𝐼1 ∩ … ∩ 𝐼𝑛
ker 𝑓 = 𝐼1 ∩ … ∩ 𝐼𝑛
We need to show 𝑓 is onto (𝑅⁄𝐼 ) × (𝑅⁄𝐼 ) × … × (𝑅⁄𝐼 ) to get isomorphism
1
2
𝑛
(by homomorphism theorem)
Let (𝑎1 + 𝐼1 , … , 𝑎𝑛 + 𝐼𝑛 ) ∈ (𝑅⁄𝐼 ) × (𝑅⁄𝐼 ) × … × (𝑅⁄𝐼 )
1
2
𝑛
We want 𝑥 𝑠. 𝑡. 𝑓(𝑥) = (𝑎1 + 𝐼1 , … , 𝑎𝑛 + 𝐼𝑛 )
Or 𝑥 ≡ 𝑎𝑖 (𝑚𝑜𝑑𝐼𝑖 ) for all 𝑖.
Existence of such an 𝑥 is guaranteed by the CRT.
Special case of corollary
1<𝑚∈ℤ
𝑚=
∏𝑘𝑖=1 𝑝𝑖𝑟𝑖
𝑝𝑖 distinct primes. 𝐼𝑖 =
𝑟
𝑝𝑖 𝑖 ℤ
(ℤ⁄𝑚ℤ) ≅ (ℤ⁄ 𝑟1 ) × … × (ℤ⁄ 𝑟𝑘 )
𝑝1 ℤ
𝑝𝑘 ℤ
Isomorphism of rings
For a commutative ring 𝑅, denote by 𝑅 ∗ = set of units (invertible elements) of 𝑅
Then 𝑅 ∗ =multiplicative abelian group.
∗
e.g. (ℤ⁄6ℤ) = {1̅, 5̅} =group of two elements
Looking at the group of units on both sides we get:
∗
∗
(ℤ⁄6ℤ)
≅
(ℤ⁄ 𝑟1 ) × … × (ℤ⁄ 𝑟𝑘 )
𝑖𝑠𝑜𝑚𝑜𝑟𝑝ℎ𝑖𝑠𝑚 𝑜𝑟 𝑢𝑛𝑖𝑡 𝑔𝑟𝑜𝑢𝑝𝑠
𝑝1 ℤ
𝑝𝑘 ℤ
Denote by 𝜑(𝑚) = #{𝑘|0 < 𝑘 < 𝑚 𝑠. 𝑡. gcd(𝑘, 𝑚) = 1}
(euler phi function)
E.g. 𝜑(6) = 2
∗
Clearly (ℤ⁄𝑚ℤ) has 𝜑(𝑚) elements.
𝑟
𝑟
From (*) we get the formula: 𝜑(𝑚) = 𝜑(𝑝11 ) ∙ … ∙ 𝜑(𝑝𝑘𝑘 )
Application to public key encoding RSA (1975)
Encoding – public
Decoding – secret
Let 𝑝1 , 𝑝2 “very large” prime numbers.
Let 𝑑 = 𝑝1 ∙ 𝑝2
Let 𝑒 = 𝜑(𝑑) = 𝜑(𝑝1 ) ∙ 𝜑(𝑝2 ) = (𝑝1 − 1)(𝑝2 − 1)
Let 𝑟 be any large number co-prime to 𝑒.
By Bezout, we have 𝑠, 𝑡 𝑠. 𝑡. 𝑠𝑟 + 𝑡𝑒 = 1
𝑠𝑟 ≡ 1(𝑚𝑜𝑑𝑒)
We publish only 𝑑 and 𝑟 (and not 𝑠, 𝑒, 𝑝1 , 𝑝2 ).
Let 𝑎 be a positive integer smaller than 𝑑.
We encode 𝑎 as 𝑎𝑟 (𝑚𝑜𝑑 𝑑) = 𝑏
Claim: 𝑏 𝑠 ≡ 𝑎(𝑚𝑜𝑑 𝑑) !
Note: This determines 𝑎 uniquely as 𝑎 was chosen to be less than 𝑑.
Proof:
First case: gcd(𝑎, 𝑑) = 1
∗
∗
∗
(ℤ⁄𝑑ℤ) ≅ (ℤ⁄𝑝 ℤ) ∙ (ℤ⁄𝑝 ℤ) has 𝜑(𝑑) = 𝑒 elements.
1
2
Recall in a group 𝐺 of order 𝑛
𝑥 𝑛 = 1 for all 𝑥 ∈ 𝐺.
Follows from Lagraunge’s theorem – shall prove later.
∗
So that 𝑎𝑒 ≡ 1(𝑚𝑜𝑑 𝑑) 𝑎̅ = 𝑎 + 𝑑ℤ elements of (ℤ⁄𝑑ℤ)
𝑟𝑠 ≡ 1(𝑚𝑜𝑑 𝑒)
𝑏 𝑠 ≡ (𝑎𝑟𝑠 )(𝑚𝑜𝑑 𝑑) ≡ 𝑎𝑙𝑒+1 ≡ (𝑎𝑒 )𝑙 ∙ 𝑎 ≡ 𝑎(𝑚𝑜𝑑 𝑑) - as required.
Second case: gcd(𝑎, 𝑑) ≠ 1
Then wlog can assume 𝑞1 |𝑎 and gcd(𝑎, 𝑝2 ) = 1
ℤ ⁄ ≅ (ℤ ⁄ ) × (ℤ ⁄
𝑝1 ℤ
𝑝2 ℤ)
𝑑ℤ 𝜓
𝜓(𝑎 + 𝑑ℤ) = (𝑎(𝑚𝑜𝑑 𝑝1 ), 𝑎(𝑚𝑜𝑑 𝑝2 )) = (0(𝑚𝑜𝑑 𝑝1 ), 𝑎(𝑚𝑜𝑑 𝑝2 ))
∗
Another corollary from Cauchy’s theorem
Ferma’s little theorem: For a prime 𝑝, 𝑥 ≠ 0
𝑥 𝑝−1 ≡ 1(𝑚𝑜𝑑 𝑝)
So we have 𝑎𝑝2 −1 ≡ 1(𝑚𝑜𝑑 𝑝2 )
𝑎𝑒 = 𝑎(𝑝2 −1)(𝑝1 −1) ≡ 1(𝑚𝑜𝑑 𝑝2 )
𝜓 is an isomorphism so we have:
𝑒
𝜓(𝑎𝑒 + 𝑑ℤ) = (𝜓(+𝑑ℤ)) = (0(𝑚𝑜𝑑 𝑝1 ), 1(𝑚𝑜𝑑 𝑝2 ))
Again, writing: 𝑟𝑠 = 𝑙𝑒 + 1 we get
𝜓(𝑏 𝑠 + 𝑑ℤ) = 𝜓(𝑎𝑟𝑠 + 𝑑ℤ) = 𝜓(𝑎𝑙𝑒+1 + 𝑑ℤ) = 𝜓(𝑎𝑙𝑒 + 𝑑ℤ) ∙ 𝜓(𝑎 + 𝑑ℤ) =
𝜓(𝑎𝑒 + 𝑑ℤ) ∙ (0(𝑚𝑜𝑑 𝑝1 ), 𝑎(𝑚𝑜𝑑 𝑝2 )) =
(0(𝑚𝑜𝑑 𝑝1 ), 1(𝑚𝑜𝑑 𝑝2 )) ∙ (0(𝑚𝑜𝑑 𝑝1 ), 𝑎(𝑚𝑜𝑑 𝑝2 )) = (0(𝑚𝑜𝑑 𝑝1 ), 𝑎(𝑚𝑜𝑑 𝑝2 )) =
𝜓(𝑎 + 𝑑ℤ)
𝑠 (𝑚𝑜𝑑
Since 𝜓 is an isomorphism we get 𝑎 ≡ 𝑏
𝑑)
Short introduction to Group Theory
𝐻 subgroup of 𝐺 if ∀𝑎, 𝑏 ∈ 𝐻 𝑎, 𝑏 −1 ∈ 𝐻 (& 𝐻 ≠ 0)
Cosets of subgroup in 𝐺
𝐻𝑎 right coset = {ℎ𝑎|ℎ ∈ 𝐻}
𝑎𝐻 left coset = {𝑎ℎ|ℎ ∈ 𝐻}
Properties: Cosets are disjoint or equal.
Suppose 𝐻𝑎 ∩ 𝐻𝑏 ≠ ∅
So have ℎ, ℎ′ ∈ 𝐻 𝑠. 𝑡. ℎ𝑎 = ℎ′ 𝑏
(ℎ′ )−1 ℎ𝑎 = 𝑏 and 𝑏 ∈ 𝐻𝑎
𝐻𝑏 ⊆ 𝐻𝑎
And similarly 𝐻𝑎 ⊆ 𝐻𝑏.
Definition:
𝑁 is a normal subgroup of 𝐺 if ∀𝑔 ∈ 𝐺 ∶ 𝑁𝑔 = 𝑔𝑁.
(does not imply 𝑛𝑔 = 𝑔𝑛 ∀𝑁!!!)
If 𝐺 is Abelian, all subgroups are normal!
Example: 𝐺 = 𝑆3 : group of permutations on {1,2,3}
1 2 3
𝑝=(
)
2 1 3
{𝐼𝑑, 𝑟} is a subgroup of G. Which is not normal!
1 2 3
1 2 3
1 2 3
1
𝐻∙(
) = {(
),(
)∙(
3 2 1
3 2 1
2 1 3
3
1 2 3
1 2 3
1 2 3
1
(
) ∙ 𝐻 = {(
),(
)∙(
3 2 1
3 2 1
3 2 1
2
2
2
2
1
3
1
)=(
1
2
3
1
)=(
3
3
2
3
2
2
3
)}
1
3
)}
1
So this is not the same group!
𝐴3 = set of even permutations = normal subgroup of order 3
1 2 3
1 2 3
𝑟 = {𝐼𝑑, (
),(
)}
2 3 1
3 1 2
1 2 3
1
𝐴3 𝜎 = 𝜎𝐴3 = 𝑆3 \𝐴3 = {(
),(
⏟2 1 3
3
𝜎
----- End of lesson 4
2 3
1
),(
2 1
1
2 3
)}
3 2
TODO: Write it
----- end of lesson 5
Theorem: Let 𝑝(𝑥) ∈ 𝐹[𝑥] be irreducible.
Proof
Note: 𝑝(𝑢) maximal so 𝐹[𝑢]⁄𝑝(𝑢) has to be a field!
Consider 𝐹 ⊆ 𝐾 by identifying 𝑎 ∈ 𝐹 with 𝑎 + (𝑝(𝑢))
It remains to show that 𝑝(𝑥) has a root in 𝐾
Suppose 𝑝(𝑥) = ∑𝑖=0 𝑎𝑖 𝑥 𝑖 , 𝑎𝑖 ∈ 𝐹
Look at the coset 𝑢 + (𝑝(𝑢)) = 𝛼 ∈ 𝐾
𝑝(𝛼) ⊂ ∑ 𝑎𝑖 𝑢𝑖 = ∑ 𝑎𝑖 (𝑢 + (𝑝(𝑢))) = ∑ 𝑎𝑖 𝑢𝑖 + (𝑝(𝑢)) =
Want to show 𝐾 unique up to isomorphism minimal such that 𝑝 has a root.
Suppose 𝐿 ⊇ 𝐹, 𝛽 is a root of 𝑝 in 𝐿.
Want to show 𝐾 ≅ subfield of 𝐿.
Map: 𝑔(𝑢) + (𝑝(𝑢)) in 𝐾 to 𝑔(𝛽) ∈ 𝐿.
H is independent of choice of coset representative, as if 𝑔(𝑢) ≡ ℎ(𝑢) (𝑚𝑜𝑑 (𝑝(𝑢)))
Then 𝑔(𝑢) = ℎ)𝑢
----- end of lesson 6
Claim: If 𝑓(𝑥) ∈ 𝐹[𝑥] and 𝐹 ⊆ 𝐾 field containing a root of 𝑓(𝑥): 𝛼
Then if 𝜑 ∈ 𝐺𝑎𝑙(𝐾⁄𝐹 ) then 𝜑(𝛼) is a root of 𝑓(𝑥)
In other words, elements of the Galois group permute the roots of 𝑓(𝑥)
Proof: Let 𝑓(𝑥) = ∑𝑘𝑖=0 𝑎𝑖 𝑥 𝑖 , 𝑎𝑖 ∈ 𝐹
𝜑(𝑓(𝛼)) = 𝜑(0𝐾 ) = 0,
𝑎𝑖 ∈ 𝐹
𝑘
𝑘
𝑘
𝑎𝑖 ∈𝐹
0 = 𝜑(𝑓(𝛼)) = 𝜑 (∑ 𝑎𝑖 𝛼 𝑖 ) = ∑ 𝜑(𝑎𝑖 )𝜑(𝛼)𝑖 = ∑ 𝑎𝑖 𝜑(𝛼)𝑖
𝑖=0
𝑖=0
𝑖=0
Special case:
𝐾 splitting field for 𝑓(𝑥) ∈ 𝐹[𝑥] then 𝐾 = 𝐹 (𝛼
⏟1 , … , 𝛼𝑘 )
𝑟𝑜𝑜𝑡𝑠 𝑜𝑓 𝐹
So any 𝜑 ∈ 𝐺𝑎𝑙(𝐾⁄𝐹 ) is determined by images of 𝛼1 , … , 𝛼𝑘 under 𝜑
We now know that these are permuted by 𝜑
𝛽 ∈ 𝐾 so can be written as a polynomial in 𝛼11 , … , 𝛼𝑘 over 𝐹
𝑖
𝑖
𝑖
𝛽 = ∑ 𝑎𝑖1 …𝑖𝑘 ∙ 𝛼11 𝛼22 … 𝛼𝑘𝑘
Examples:
1) Galois group of the smallest field of 𝑥 4 − 2 over ℚ
Roots of 𝑥 4 − 2:
4
4
± √2, ±𝒾 √2
4
4
4
4
𝑥 4 − 2 = (𝑥 − √2)(𝑥 + √2)(𝑥 − 𝒾 √2)(𝑥 + 𝒾 √2)
4
and over k: ℚ(√2, 𝒾)
𝜑 ∈ 𝐺𝑎𝑙 (𝐾⁄ℚ) = 𝐺 will permute 4 roots
So can think of 𝐺 of being a subgroup of 𝑆4
|
4
4
|
4
4
We know that |ℚ(√2, 𝒾): ℚ| = |(ℚ(√2, 𝒾): ℚ(√2)| ∙ | ⏟
ℚ(√2)
4=𝑑𝑒𝑔𝑟𝑒𝑒 𝑜𝑓
|𝑚𝑖𝑛𝑖𝑚𝑎𝑙 𝑝𝑜𝑙𝑦
4
𝑜𝑓 √2 𝑜𝑣𝑒𝑟
ℚ(𝑋 4 −2)
[𝑘: 𝐹] = dimension of 𝐾 over 𝐹.
|𝐺𝑎𝑙 (𝐾⁄ℚ) = 𝐺| = 8
So 𝐺 is isomorphic to an 8-element subgroup of 𝑆4
Possibilities (up to isomorphism) are:
𝐶8 , 𝐶4 × 𝐶2 , 𝐶2 × 𝐶2 × 𝐶2 , 𝐷8 , 𝑄8
|𝑆4 | = 24 (a side note)
𝐶8 – is impossible since 𝑆4 contains no elements of order 8
Let 𝜑 be complex conjugation.
obviously 𝜑 is an element of order 2. 𝜑 ∈ 𝐺
: ℚ|
|
4
4
𝜑(√2) = √2
4
4
𝜑(− √2) = − √2
4
4
𝜑(𝒾 √2) = −𝒾 √2
4
4
𝜑(−𝒾 √2) = −𝒾 √2
Let 𝜓 be the automorphism that permutes roots cyclically:
4
4
𝜓(√2) = 𝒾 √2 and fixes 𝒾
𝜓 is of order 4
4
4
𝜓(− √2) = −𝒾 √2
4
4
4
4
𝜓(𝒾 √2) = 𝜓(𝒾)𝜓(√2) = 𝒾𝒾 √2 = − √2
< 𝜑, 𝜓 > is a group permuted by 𝜑 and 𝜓 ≅ 𝐷8
Cycle notation in 𝑆𝑛 (any permutation can be written as a product of disjoint cycles)
Example 𝜎 ∈ 𝑆4
1 2 3 4 5 6
𝜎=(
) = (13462)(5)
3 1 4 6 5 2
1 2 3 4 5 6
(
) = (13)(254)(6)
3 5 1 2 4 6
Can have 𝜎 ∈ 𝑆5 , 𝜎 = (123)(45)
Elements of 𝑆4 can have orders 1,2,3,4 (again, a side note).
The order of the elements is always the least common multiple of the cycles.
Another example:
Galois group of 𝑝(𝑥) = 𝑥 3 + 2𝑥 + 1 over ℚ
Need to find the splitting field of the polynomial over ℚ.
We first of all show that 𝑝(𝑥) has no roots in ℚ and so is irreducible.
Claim: If 𝑓(𝑥) is a monic polynomial over ℤ, then any rational root will be an integer
Proof: 𝑓(𝑥) = 𝑥 𝑛 + 𝑎𝑛−1 𝑥 𝑛−1 + ⋯ + 𝑎1 𝑥 + 𝑎0 , 𝑎𝑖 ∈ ℤ
𝑟, 𝑠 ∈ ℤ
𝑟
𝑠
𝑟
𝑠
If is a root then: 0 = 𝑓 ( ) =
𝑟𝑛
𝑠𝑛
+ ∑𝑛−1
𝑖=0 𝑎𝑖
𝑟𝑖
𝑠𝑖
Assume (𝑟, 𝑠) = 1
𝑛−1
𝑛
𝑟 + ∑ 𝑎𝑖 𝑟 𝑖 𝑠 𝑛−𝑖 = 0
𝑖=0
𝑟 𝑛 = −𝑎0 𝑆 𝑛 − 𝑎1 𝑆 𝑛−1 + ⋯ − 𝑎𝑛−1 𝑠𝑟 𝑛−1
If 𝑝 is a prime divisor of 𝑠, then 𝑝|𝑟 𝑛 so 𝑝|𝑟.
But then, 𝑝|𝑠 and 𝑝|𝑟 which contradicts the fact that 𝑠 and 𝑟 are mutually prime.
𝑟
𝑠
So 𝑠 has no prime divisors. So 𝑠 = ±1. Therefore, ∈ ℤ
We now show that 𝑝(𝑥) have no integer roots.
𝑝(0) = 1
𝑝(−1) = −2
So there exists 𝛼 ∈ ℝ −1 < 𝛼 < 0 and 𝑝(𝛼) = 0 by continuity of 𝑝(𝑥) as a real function.
But it’s the only real root, since the derivative is always positive, therefore it’s constantly
increasing etc etc…
So 𝑝(𝑥) has no rational roots, and remaining 2 roots are non-real.
Over ℚ(𝛼)
𝑥 3 + 2𝑥 + 1 = (𝑥 − 𝛼)(𝑥 2 + (2 + 𝛼)𝑥 + (2 + 𝛼)𝛼)
Where 𝛽 and 𝛽̅ are nonreal roots.
𝑜𝑣𝑒𝑟 𝑆.𝐹.
=
(𝑥 − 𝛼)(𝑥 − 𝛽)(𝑥 − 𝛽̅ )
So the splitting field will be ℚ(𝛼, 𝛽)
|ℚ(𝛼, 𝛽): ℚ| = ⏟
|ℚ(𝛼, 𝛽): ℚ(𝛼)| ∙ ⏟
|ℚ(𝛼): ℚ|
=2 (𝑏𝑦 𝑡ℎ𝑒 𝑒𝑥𝑡𝑟𝑎 𝑓𝑎𝑐𝑡)
=3
Extra fact:
If 𝛼 is a root of some polynomial 𝑔(𝑥) over a field 𝐹.
And 𝑝(𝑥) is the minimal polynomial of 𝛼 over 𝐹, then 𝑝(𝑥)|𝑔(𝑥) in 𝐹[𝑥]
Proof: Divide 𝑔(𝑥 ) by 𝑝(𝑥) with remainder in 𝐹[𝑥]
𝑔(𝑥) = 𝑝(𝑥)𝑞(𝑥) + 𝑟(𝑥)
deg 𝑟 < deg 𝑝 or 𝑟 = 0
Substitute 𝑥 = 𝛼: 0 = 𝑔(𝛼) = 𝑝(𝛼)𝑞(𝛼) + 𝑟(𝛼)
So 𝛼 root of 𝑟(𝑥) of smaller degree than 𝑝(𝑥) - contradiction!
So |𝐺𝑎𝑙(ℚ(𝛼, 𝛽)/ℚ)| = 6.
Elements of Galois group permute the set {𝛼, 𝑏𝑒𝑡𝑎 𝛽̅ } and so is isomorphic to a subtgroup of
𝑆3 of order 6 ⇒ Galois group ≅ 𝑆3
TOPIC:
Cyclotomic fields and their Galois groups over ℚ
𝑛
Definition Cyclotomic field is one of the form ℚ( √1)
𝑛
2𝜋𝒾
√1 = 𝑒 𝑛 positive with root of 1
𝑛
Note that ℚ( √1) is a splitting field of the polynomial 𝑥 𝑛 − 1 over ℚ
As:
𝑛−1
𝑥 𝑛 − 1 = Π (𝑥 − 𝑒
k=0
2𝜋𝒾
𝑛 )
We also want to factor 𝑥 𝑛 − 1 into irreducible factors over ℚ.
(𝑥 2 + 𝑥 + 1)
E.g. 𝑥 3 − 1 = (𝑥 − 1)
⏟
𝑖𝑟𝑟𝑒𝑑𝑢𝑐𝑖𝑏𝑙𝑒 𝑜𝑣𝑒𝑟 ℚ=𝑚𝑖𝑛𝑖𝑚𝑎𝑙 𝑝𝑜𝑙𝑦
𝑛
Definition: Denote by 𝜆𝑛 (𝑥) = minimal polynomial of √1 over ℚ
So 𝜆3 (𝑥) = 𝑥 2 + 𝑥 + 1
𝜆𝑛 (𝑥) = n’th cyclotomic polynomial
𝜆1 (𝑥) = 𝑥 − 1
𝜆2 (𝑥) = 𝑥 + 1
𝜆3 (𝑥) = 𝑥 2 + 1
4
√1 = 𝒾
𝑥 4 − 1 = (𝑥 2 − 1)(𝑥 2 + 1) = (𝑥
⏟+ 1) (𝑥
⏟− 1) (𝑥⏟2 + 1)
=𝜆2
=𝜆1
=𝜆3
Fact: If 𝑓(𝑥) ∙ 𝑔(𝑥) = 𝑥 𝑛 − 1 over ℚ, then 𝑓(𝑥), 𝑔(𝑥) ∈ ℤ[𝑥]
(Follows from Gauss’ lemma – Basic algebra 1)
Interesting fact:
If we factor 𝑥 𝑛 − 1 over ℚ(i.e. over ℤ!)
Turns out up to 𝑛 = 105 all coefficients are ∈ {0, ±1}!
For 𝑛 = 105 get coefficients = 2
105 = 3 ∙ 5 ∙ 7
𝑛
|ℚ( √1): ℚ| = deg 𝜆𝑛 =?
Examples:
1)
4
ℚ(𝑖) = ℚ(√1)
Can be thought of a 2 dimensional vector space over ℚ
𝑎 + 𝒾𝑏
(𝑎 + 𝒾𝑏)(𝑐 + 𝒾𝑑) = 𝑎𝑐 − 𝑏𝑑 + 𝒾(𝑎𝑑 + 𝑏𝑐)
We can think of them as vectors with regular dot multiplication.
3
2) ℚ(𝜔) = ℚ(√1)
|ℚ(𝜔): ℚ| = 2 irreducible polynomial 𝜆3 of 𝜔 is 𝑥 2 + 𝑥 + 1
2 dimensional vector space over ℚ - addition – as usual
(𝑎 + 𝜔𝑏)(𝑐 + 𝜔𝑑) = 𝑎𝑐 + 𝜔2 (𝑏𝑑) + 𝜔(𝑎𝑑 + 𝑏𝑐) = 𝑎𝑐 − 𝑏𝑑 + 𝜔(𝑎𝑑 + 𝑏𝑑 − 𝑏𝑑)
Since:
𝜔2 + 𝜔 + 1 = 0
𝜔2 = −1 − 𝜔
5
3) ℚ(√1)
𝜆5 (𝑥) = 𝑥 4 + 𝑥 3 + 𝑥 2 + 𝑥 + 1
5
|ℚ(√1): ℚ| = 4
5
1, 𝜌, 𝜌2 , 𝜌3 basis for ℚ( √1) over ℚ
In general
𝑝
4) 𝑝 is prime ℚ( √1)
𝑥 𝑝 − 1 = (𝑥 − 1)(𝑥 𝑝−1 + 𝑥 𝑝−2 + ⋯ + 𝑥 + 1)
The second part is irreducible using einsensteins criterion (lang algebra) = 𝜆𝑝 (𝑥)
𝑝
|ℚ( √1): ℚ| = 𝑝 − 1
5) N=6
Let’s factor it over ℚ:
𝑥 6 − 1 = (𝑥 3 − 1)(𝑥 3 + 1) = (𝑥 − 1)(𝑥 2 + 𝑥 + 1)(𝑥 + 1 )(𝑥 2 − 𝑥 + 1)
6
√1 = 𝜌
𝜔 = 𝜌2
𝜔2 = 𝜌4
df
Roots areL
Roots (Accoringly) 1, 𝜔, 𝜔2 −1 𝜌, 𝜌5 = 𝜌̅
What is ℚ(𝜌)??
2 dimensions over ℚ. What is the multiplication rule?
Notice: −𝜔 is a 6th root of (−𝜔)2 = 𝜔
So can take 𝜌 = −𝜔
ℚ(𝜌) = ℚ(𝜔)!!!!
It’s actually the same field! Not isomorphic – same field!
--- end of lesson
𝑛
Theorem: [ℚ( √1): ℚ] = 𝜑(𝑛) =Eular 𝜑-function
Recheck:
𝜑(6) = |{1,5}| = 2
𝜑(5) = 4
𝜑(4) = |{1,3}| = 2
𝜑(3) = 2
𝜑(𝑝) = 𝑝 − 1
𝑝 is prime
𝑛
Denote 𝜉 = √1
Proof: [ℚ(𝜉): ℚ] =degree of the minimal polynomial of 𝜉 over ℚ = deg 𝜆𝑛 (𝑥)
Note: 𝜉 𝑘 is a primitive n’th root of 1 ⇔ gcd(𝑘, 𝑛) = 1
|{𝜉 𝑘 |𝜉 𝑘 𝑝𝑟𝑖𝑚𝑒 𝑛′ 𝑡ℎ𝑟𝑜𝑜𝑡 𝑜𝑓 1}| = 𝜑(𝑛)
So in fact, 𝜆𝑛 (𝑥) = ∏gcd(𝑘,𝑛)=1(𝑥 − 𝜉 𝑘 )
1≤𝑘<𝑛
This is a key fact!
By gauss’ Lemma, 𝑥 𝑛 − 1 factors over ℚ into polynomials in ℤ[𝑥]
So in fact, as 𝜆𝑛 (𝑥)|𝑥 𝑛 − 1 over ℚ (since 𝜉 is a root of 𝑥 𝑛 − 1 and 𝜆𝑛 (𝑥) is its root
polynomnial)
We in fact have that 𝜆𝑛 (𝑥) ∈ ℤ[𝑥]
Suppose 𝑑|𝑛:
Then, any d’th root of 1 is also an n’th root of 1.
So the roots of 𝜆𝑑 (𝑥) satisfy 𝑥 𝑛 − 1 = 0
So 𝜆𝑑 (𝑥)|𝑥 𝑛 − 1 over ℚ
Conclusion: 𝜆𝑑 (𝑥)|𝑥 𝑛 − 1 for all 𝑑|𝑛.
Conversely:
Suppose 𝑝(𝑥) is an irreducible monic factor of 𝑥 𝑛 − 1 (in ℚ[𝑥])
Any root 𝛼 of 𝑝(𝑥) is a root of 𝑥 𝑛 − 1 and so 𝛼 𝑛 = 1
If 𝑑 minimal such that 𝛼 𝑑 = 1 then 𝑑|𝑛.
So 𝛼 is a primitive d’th root of 1. Its minimal polynomial is 𝜆𝑑 (𝑥)
And so 𝜆𝑑 (𝑥)|𝑝(𝑥) but 𝑝(𝑥) is irreducible and monic and so 𝜆𝑑 (𝑥) = 𝑝(𝑥).
So every irreducible factor of 𝑥 𝑛 − 1 over ℤ is of the form 𝜆𝑑 (𝑥) for some 𝑑|𝑛.
Conclusion: 𝑥 𝑛 − 1 = ∏𝑑|𝑛 𝜆𝑑 (𝑥) over ℚ. And 𝜆𝑑 (𝑥) ∈ ℤ[𝑥]
(𝑥 − 1) ⏟
(𝑥 + 1 ) ⏟
(𝑥 2 + 𝑥 + 1) ⏟
(𝑥 2 − 𝑥 + 1)
Example: 𝑥 6 − 1 = ⏟
=𝜆1 (𝑥)
𝜆2 (𝑥)
𝜆3 (𝑥)
𝜆6 (𝑥)
Corollary from conclusion:
From degree of polynomials we get:
𝑛 = ∑ deg 𝜆𝑑 (𝑥) = ∑ 𝜑(𝑑)
𝑑|𝑛
𝑑|𝑛
Example:
𝑥 12 − 1 = (𝑥 6 + 1 )(𝑥 6 − 1) =
(𝑥 2 + 1) ⏟
(𝑥 4 − 𝑥 2 + 1) ⏟
(𝑥 − 1) ⏟
(𝑥 + 1 ) ⏟
(𝑥 2 + 𝑥 + 1) ⏟
(𝑥 2 − 𝑥 + 1)
⏟
𝜆4
𝜆2 (𝑥)
𝜉,𝜉 11 ,𝜉 5 ,𝜉 7
=𝜆1 (𝑥)
1
𝜆2 (𝑥)
−1
𝜆3 (𝑥)
𝜔,𝜔2
𝜆6 (𝑥)
−𝜔,−𝜔2
12
𝜉 = √1
𝒏
Galois grups of ℚ(𝒙𝒊 ) over ℚ, 𝝃 = √𝟏
ℚ(𝜉)
⁄ℚ) = 𝐺
Let 𝐺𝑎𝑙 (
Elements of 𝐺 permute primitive roots of unity and are determine by the image of 𝜉.
So 𝐺 subroup of group of permutations {𝜉 𝑘 | gcd (𝑘, 𝑛) = 1} i.e. of 𝑆𝜑(𝑛)
1≤k<𝑛
Let gcd(𝑘, 𝑛) = 1:
𝜓𝑘
𝜉 → 𝜉 𝑘 determines an automorphism of ℚ(𝜉)
Conversely, every automorphism must be of this form.
|𝐺| = [ℚ(𝜉): ℚ] = 𝜑(𝑛)
Suppose gcd(𝑙, 𝑘) = 1 = gcd(𝑛, 𝑘)
𝜑𝑘 ∙ 𝜓𝑙 (𝜉) = 𝜓𝑘 (𝜉 𝑘 ) = 𝜉 𝑘𝑙 = 𝜓𝑘𝑙 (𝜉)
𝜓𝑙 𝜓𝑘 (𝜉) = 𝜓𝑙 (𝜉 𝑘 ) = 𝜉 𝑙𝑘
So the group is abelian!
More precisely:
𝜓𝑘 = 𝜓𝑙 = 𝜓𝑚 where 𝑚 ≡ 𝑘𝑙(𝑚𝑜𝑑 𝑛)
In fact: The map 𝑘 → 𝜓𝑘
∗
Is group homomorphism between (ℤ⁄𝑛ℤ) and 𝐺
∗
So 𝐺 ≅ (ℤ⁄𝑛ℤ)
E.g. 𝑛 = 12
∗
(ℤ⁄12ℤ) = {1,5,7,11} multiplication mod 12.
12
𝜉 = √1
Note: 𝜉 → 𝜉11 is complex conjugation
Finite Fields
If 𝐹 is finite then its characteristics must be some prime 𝑝
And its prime field ≅ ℤ⁄𝑝ℤ.
So every finite field can be considered to be an extension of ℤ⁄𝑝ℤ.
In fact, it is an algebraic extension.
(if 𝛼 transcendental then 1, 𝛼, 𝛼 2 , 𝛼 3 , … infinitely linearly independent set so any field
containing 𝛼 will be infinite).
First difference between characteristic 0 case and the characteristic 𝒑
case
We had quadratic extensions of ℚ e.g.
ℚ(√2), ℚ(𝜔), ℚ(𝑖) which are isomorphic as fields!
By contrast, ℤ⁄𝑝ℤ has a unique quadratic extension up to isomorphism.
Example: ℤ⁄2ℤ clearly unique up to isomorphism. Call it 𝔽2 or 𝐺𝐹(2)
Now look at 𝑥 2 + 𝑥 + 1 which is irreducible over ℤ⁄2ℤ
Extend 𝔽2 to get a field in which 𝑥 2 + 𝑥 + 1 has a root.
𝔽 [𝑥]
𝑘= 2 ⁄ 2
𝑥 +𝑥+1
{𝐾: 𝔽} = dim𝔽 𝐾 = 2 ⇒ 𝐾 2 dimensional vector space over 𝔽2 and so has 4 elements.
Elements of 𝐾 can be considered to be remainders of polynomials in 𝑥 over 𝔽2
After division by 𝑥 2 + 𝑥 + 1 i.e. linear polynomials.
0,1, 𝑥, 𝑥 + 1
+
0
0
0
1
1
𝑥
𝑥
𝑥+1 𝑥+1
∙
0
1
𝑥
𝑥+1
1
1
0
𝑥+1
𝑥
𝑥
𝑥+1
𝑥
𝑥+1
𝑥+1
𝑥
0
1
1
0
0
1
𝑥
0
0
0
0
1
𝑥
0
𝑥
𝑥+1
0 𝑥+1
1
𝑥+1
0
𝑥+1
1
𝑥
Very easy to show directly that every field of order 4 is isomorphic to 𝐾.
Note: 𝑥 2 + 𝑥 + 1 is actually the only irreducible quadratic polynomial over 𝔽
Theorem: Let 𝐹 be a finite field then |𝐹| = 𝑝𝑘 elements for some prime 𝑝, 1 ≤ 𝑘 ∈ ℕ.
Conclusion: there is no field of order 6,10,15, etc!
Proof: Let ℤ⁄𝑝ℤ = 𝔽𝑝 to be the prime field of 𝐹 then 𝐹 is a vector space over 𝔽𝑝 .
And as 𝐹 is finite, it is finite dimensional over 𝔽𝑝 . Say dim 𝐹 = 𝑘.
(𝑘)
So 𝐹 ≅ 𝔽𝑝 as a vector space and so |𝐹| = 𝑝𝑘
Example:
Look at 𝑥 4 + 𝑥 3 + 1 over 𝐺𝐹(2)
Claim: 𝑥 4 + 𝑥 3 + 1 is irreducible over 𝐺𝐹(2)
Clearly it has no roots.
If it factored as 2 irreducible quadratics then we would have 𝑥 4 + 𝑥 3 + 1 = (𝑥 2 + 𝑥 + 1)2
But (𝑥 2 + 𝑥 + 1)2 = 𝑥 4 + 𝑥 2 + 1
So 𝐺𝐹(2)[𝑥]⁄(𝑥 4
gives an extension of degree 4 and so a field of order 16!
+ 𝑥 3 + 1)
Its elements can be considered as polynomials of degree less or equal to 3.
Or, vectors of length 4 over 𝔽2 .
Addition is very easy with both notations (mod 2)
(𝑥 3 + 𝑥) + (𝑥 2 + 𝑥 + 1) = 𝑥 3 + 𝑥 2 + 1
𝑎
𝑏
𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑 ↔ ( )
𝑐
𝑑
Multiplication on the other hand, is harder
(𝑥 3 + 𝑥) ∙ (𝑥 2 + 𝑥 + 1) = 𝑥 5 + 𝑥 3 + 𝑥 4 + 𝑥 2 + 𝑥 3 + 1 = 𝑥 5 + 𝑥 4 + 𝑥 2 + 𝑥
≡ 𝑥 2 (𝑚𝑜𝑑 𝑥 4 + 𝑥3 + 1
1
0
0
0
1
1
( )( ) = ( )
1
1
0
0
1
0
Another Notation
Let 𝛼 = 𝑥 + (𝑥 4 + 𝑥 3 + 1) in 𝐹
So 𝛼 root of 𝑥 4 + 𝑥 3 + 1 in 𝐹. 𝛼 4 + 𝛼 3 + 1 = 0
1, 𝛼, 𝛼 2 , 𝛼 3 are linearly independent over ℤ⁄2ℤ and so distinct.
Note that 𝐹 ∗ is a group of order 15.
So 𝛼 has order dividing 15⇒ 𝛼 has order 1,3,5,15
𝛼4 = 𝛼3 + 1
𝛼 5 = 𝛼(𝛼 3 + 1) = 𝛼 4 + 𝛼 = 𝛼 3 + 1 + 𝛼 = 𝛼 3 + 𝛼 + 1 ≠ 1. Otherwise, 𝛼 3 + 𝛼 = 0 and 𝛼
satisfies polynomials of degree 3 – contradiction.
Conclude: 𝛼 has order 15! So 𝐹 ∗ is cyclic and generated by 𝛼.
So 𝐹 = {0,1, 𝛼, … , 𝛼 14 }
This notation is convenient for multiplication:
𝛼 𝑖 ∙ 𝛼 𝑗 = 𝛼 𝑖+𝑗(𝑚𝑜𝑑 15)
(Addition - problematic!)
Note: Over 𝐹 𝑥 4 + 𝑥 3 + 1 factors into linear factors and so is a splitting field for this
polynomial over 𝔽2
Notice that: 𝛼 4 + 𝛼 3 + 1 = 0
(Over ℤ⁄𝑝ℤ: (𝑥 + 𝑦)𝑝 = 𝑥 𝑝 + 𝑦 𝑝 )
So 0 = (𝛼 4 + 𝛼 3 + 1)2 = 𝛼 8 + 𝛼 6 + 1 ⇒ 𝛼 2 is a root of 𝑥 4 + 𝑥 3 + 1
(𝛼 8 + 𝛼 6 + 1)2 = 𝛼 16 + 𝛼 12 + 1 ⇒ 𝛼 4 is a root of 𝑥 4 + 𝑥 3 + 1
Same for (𝛼 16 + 𝛼 12 + 1)2 which leads to 𝛼 8 is a root as well
So 𝑥 4 + 𝑥 3 + 1 = (𝑥 − 𝛼)(𝑥 − 𝛼 2 )(𝑥 − 𝛼 4 )(𝑥 − 𝛼 8 )
Theorem: The multiplicative group of a finite field is cyclic.
Proof: next lesson!
Note: If |𝐹| = 𝑞 then all its nonzero elements will satisfy 𝑥 𝑞−1 = 1
As |𝐹 ∗ | = 𝑞 − 1
Over a field, the polynomial has at most 𝑞 − 1 different roots. So in this case the set of
elements in 𝐹 ∗ is precisely the set of roots of 𝑥 𝑞−1
If we take 𝑥 𝑞 − 𝑥 then every element of 𝐹 (including 0!) is a root and 𝐹 is the splitting field
of 𝑥 𝑞 − 𝑥.
--- end of lesson 8
Fundemental theorem of Abelian groups:
Every Abelian group is a direct product of cyclic groups.
(If the group is finite – get a direct product of a finite number of finite cyclic groups).
Proof: Jacobson Basic Algebra 1.
For the finite case, you can always write:
𝐺 = 𝐻1 × … × 𝐻𝑟
𝐻𝑖 = direct product of cyclic groups of orders that are powers of a fixed prime 𝑝𝑖
𝑝1 , … , 𝑝𝑟 direct primes.
Theorem: If 𝐹 is a finite field, then 𝐹 ∗ is cyclic.
Proof: Assume 𝐹 ∗ = 𝐻1 × … × 𝐻𝑠 as above.
Each 𝐻𝑖 can be written as a direct product:
𝑝𝑖 = 𝑝- 𝐻𝑖 = 𝐶𝑝𝑘𝑖1 × 𝐶𝑝𝑘𝑖2 × … × 𝐶𝑝𝑘𝑖𝑟
Can assume 𝑘1 ≥ ⋯ ≥ 𝑘𝑟
𝐶𝑘 = cyclic of order 𝑘
So every element 𝑎 of 𝐻𝑖 satisfies 𝐴𝑝
𝑘1
=1
So every element of 𝐻𝑖 is a root of the polynomial 𝑥 𝑝
𝑘1
−1=0
𝐻𝑖 ⊂ 𝐹 and in 𝐹 there are at most 𝑝𝑘1 roots of this polynomial. So |𝐻𝑖 | = 𝑝𝑘1 . Meaning, 𝑟 =
1.
So 𝐻𝑖 = 𝐶𝑝𝑘1 and in general we get:
So 𝐹 ∗ = 𝐶𝑝𝑘1 × … × 𝐶𝑝𝑘𝑠
1
𝑠
𝑝1 , … , 𝑝𝑠 are distinct primes!
So 𝐹 ∗ is cyclic generated by the product of the generators of 𝐶𝑝𝑘1 , … , 𝐶𝑝𝑘𝑠 .
1
𝑠
Corollary: If 𝐹 is a finite field of order 𝑞. Then it is the splitting field of 𝑥 𝑞 − 𝑥 (where 𝑞 =
𝑝𝑘 , 𝑝 is prime) over ℤ⁄𝑝ℤ. And so unique up to isomorphism.
Proof: All the elements of 𝐹 ∗ are roots of 𝑥 𝑞−1 − 1 and so together with 0 all the elements
of 𝐹 are roots of 𝑥 𝑞−𝑥 − 𝑥.
So every element is a root and the set of roots = 𝐹.
We shall show that if 𝐹 and 𝐹 ′ are both fields of order 𝑞 = 𝑝𝑘 then they are isomorphic:
Let 𝛼 ∈ 𝐹 ∗ generator.
So 𝑎 is algebraic over ℤ⁄𝑝ℤ so is a root of an irreducible monic polynomial 𝑚(𝑥) ∈ ℤ⁄𝑝ℤ [𝑥]
So 𝑚(𝑥)|𝑥 𝑞 − 𝑥
𝐹 ′ is also a splitting field of 𝑥 𝑞 − 𝑥 over ℤ⁄𝑝ℤ.
So 𝑚(𝑥) has a root 𝛽 in 𝐹 ′ .
We map 𝛼 𝑖 to 𝛽 𝑖 ∀𝑖 and 0 to 0.
We need to show that the map is onto 𝐹 ′ (and so 1-1)
And that it is additive! (it is multiplicative by definition).
Suppose 𝛽 𝑟 = 1 for 𝑟 < 𝑞 − 1.
Then 𝛽 is a root of 𝑥 𝑟 − 1 in 𝐹 ′ .
𝑚(𝑥) is the minimal polynomial of 𝛽 so that 𝑚(𝑥)|𝑥 𝑟 − 1 over ℤ⁄𝑝ℤ
So that 𝛼 𝑟 = 1 in 𝐹.
But 𝛼 is of order 𝑞 − 1 so 𝑞 − 1|𝑟 and 𝑟 ≥ 𝑞 − 1 - contradiction!
We now show the map is additive:
a) If 𝛼 𝑖 + 𝛼 𝑗 = 𝛼 𝑘 then need to show 𝛽 𝑟 + 𝛽 𝑠 = 𝛽 𝑡
b) If 𝛼 𝑖 + 𝛼 𝑗 = 0 then need to show 𝛽 𝑟 + 𝛽 𝑠 = 0
We shall show (a):
𝛼 𝑖 + 𝛼 𝑗 = 𝛼 𝑘 implies 𝛼 is a root of 𝑥 𝑟 + 𝑥 𝑠 − 𝑥 𝑡 so 𝑚(𝑥)|𝑥 𝑟 + 𝑥 𝑠 − 𝑥 𝑡
So then 𝛽 root of 𝑥 𝑟 + 𝑥 𝑠 − 𝑥 𝑡 and so 𝛽 𝑟 + 𝛽 𝑠 = 𝛽 𝑡 .
Note: It also follows that the roots of 𝑥 𝑞 − 𝑥 over ℤ⁄𝑝ℤ are distinct.
Theorem: For any prime 𝑝 and 1 ≤ 𝑘 ≤ ℕ there exists a field of order 𝑝𝑘 .
𝑘
Proof: Take ℤ⁄𝑝ℤ and extend to a splitting field for 𝑥 𝑝 − 𝑥.
This will be a field of order 𝑝𝑘 (and will be unique!).
Corollary: For any 𝑘 ≥ 1 integer and prime 𝑝, there exists an irreducible polynomial of
degree 𝑘 over ℤ⁄𝑝ℤ.
Proof: Take 𝛼 a generator of 𝐹 ∗ where 𝐹 field of order 𝑝𝑘 = 𝑞. (𝐹 = 𝐺𝐹(𝑞))
ℤ⁄ [𝛼] = 𝐹 and ℤ⁄ [𝛼] is a vector space of dimension 𝑙 over ℤ⁄ where 𝑙 is the degree
𝑝ℤ
𝑝ℤ
𝑝ℤ
of the minimal polynomial of 𝛼.
So ℤ⁄𝑝ℤ [𝛼] is of order 𝑝𝑙 so 𝑘 = 𝑙 and minimal polynomial is irreducible of degree 𝑘.
Factorization of 𝑿𝒏 − 𝟏 over finite fields
Example: 𝐺𝐹(16) = 𝐺𝐹(2)[𝛼]
𝛼 root of 𝑥 4 + 𝑥 3 + 1 over 𝐺𝐹(2).
Every element in this field is a root of 𝑥 16 − 𝑥.
So 𝑥 4 + 𝑥 3 + 1|𝑥 16 − 𝑥 over 𝐺𝐹(2).
Roots of 𝑥 4 + 𝑥 3 + 1 in 𝐺𝐹(16) were: 𝛼, 𝛼 2 , 𝛼 4 , 𝛼 16
0 root of 𝑥. (so 𝑥|𝑥 16 − 𝑥)
1 root of 𝑥 + 1 (so 𝑥 + 1|𝑥 16 − 𝑥)
𝑥 16 − 𝑥 = 𝑥(𝑥 + 1)(𝑥 4 + 𝑥 3 + 1) ∙ ℎ(𝑥), ℎ(𝑥) ∈ 𝐺𝐹(2)[𝑥] of degree 10.We want to factor
ℎ(𝑥)
Definition:
Let 𝑓(𝑥) = polynomial of degree 𝑛.
The reciprocal of 𝑓(𝑥) is 𝑔(𝑥) = 𝑥 𝑚 𝑓(𝑥 −1 )
Example:
5
𝑥 𝑓(𝑥
−1 )
=𝑥
5 (𝑥 −5
𝑓(𝑥) = 𝑥 5 − 2𝑥 4 + 3𝑥 2 − 7𝑥 + 19
− 2𝑥 −4 + 3𝑥 −2 − 7𝑥 −1 + 19) = 1 − 2𝑥 + 3𝑥 2 − 7𝑥 4 + 19𝑥 5
Use question 4 in assignment 4 to get the reciprocal of 𝑥 4 + 𝑥 3 + 1:
𝑥4 + 𝑥 + 1
−1
So 𝑥 4 + 𝑥 + 1 is irreducible and 𝛼
⏟
is a root and also 𝛼 −2 = 𝛼 13 , 𝛼 −4 = 𝛼 11 , 𝛼 −8 = 𝛼 7 .
=𝛼14
We conclude that 𝑥 4 + 𝑥 + 1|𝑥 16 − 𝑥
So ℎ(𝑥) has 𝑥 4 + 𝑥 + 1 as an irreducible factor over 𝐺𝐹(2)
Note also: 𝑥 5 − 1|𝑥 15 − 1. Since (𝑥 5 − 1)(𝑥 10 + 𝑥 5 + 1) = 𝑥 15 − 1.
Over 𝐹𝐺(2) we have 𝑥 5 − 1 = (𝑥 + 1)(𝑥 4 + 𝑥 3 + 𝑥 2 + 𝑥 + 1)
So 𝑥 4 + 𝑥 3 + 𝑥 2 + 𝑥 + 1|𝑥 16 − 𝑥 and is irreducible (question 1 in assignment 4).
Note also: 1, 𝛼 5 , 𝛼 10 are roots of 𝑥 3 − 1 in 𝐺𝐹(16): 𝛼 3 , 𝛼 6 , 𝛼 12 , 𝛼 24 = 𝛼 9
𝑥 3 − 1 factors to: (𝑥 − 1)(𝑥 2 + 𝑥 + 1)
So 𝑥 2 + 𝑥 + 1 is the minimal polynomial of 𝛼 5 , 𝛼 10 .
So over 𝐺𝐹(2):
𝑥 16 − 𝑥 = 𝑥(𝑥 − 1)(𝑥 2 + 𝑥 + 1)(𝑥 4 + 𝑥 3 + 1)(𝑥 4 + 𝑥 + 1)(𝑥 4 + 𝑥 3 + 𝑥 2 + 𝑥 + 1)
Roots (in the appropriate order of the factors):
0, 1, 𝛼 5 , 𝛼 10 , 𝛼 , 𝛼 2 , 𝛼 4 , 𝛼 8 , 𝛼 14 , 𝛼 13 , 𝛼 11 , 𝛼 7 , 𝛼 3 , 𝛼 6 , 𝛼 9 , 𝛼 12
Note: 𝛼, 𝛼 −1 = 𝛼 14 are primitives elements (i.e. generators of 𝐺𝐹(16)∗ but the roots of
𝑥 4 + 𝑥 3 + 𝑥 2 + 𝑥 + 1 are not generators for 𝐺𝐹(16)∗
Though we can use this polynomial to construct 𝐺𝐹(16) over 𝐺𝐹(2). And every element of
𝐺𝐹(16) is a polynomial in 𝛼 3 (but not a power of 𝛼 3 !)
Every element of 𝐺𝐹(𝑝𝑘 ) satisfies 𝑥 𝑝
𝑛
𝑘 −1
= 1.
𝑘
If 𝑥 − 1 has a root in 𝐺𝐹(𝑝 ).
Must have 𝑛|𝑝𝑘 − 1
Can see which are the subfields of 𝐺𝐹(16) by looking at the factorization of 𝑥 16 − 𝑥.
Possible subfields (are of order 2𝑚 , 𝑚 ≤ 4):
𝐺𝐹(2) - prime field and so a subfield!
𝐺𝐹(4) – {0,1, 𝛼 5 , 𝛼 10 } as 𝐺𝐹(4) splitting field of 𝑥 2 + 𝑥 + 1
𝐺𝐹(8) - Don’t have any irreducible polynomials of degree 3 dividing 𝑥 16 − 𝑥! 𝐺𝐹(8) is the
splitting field of an irreducible cubic over 𝐺𝐹(2)! So this is not a subfield of 𝑮𝑭(𝟏𝟔).
𝐺𝐹(16) (clearly).
Also: 𝐺𝐹(16) could not be a vector space over 𝐺𝐹(8) otherwise 16 would equal an integral
power of 8.
--- end of lesson
𝑥 𝑛 − 𝑥 over 𝐺𝐹(2)
-
What are the subfields of a given finite field 𝐺𝐹(𝑞), 𝑞 = 𝑝 𝑥 , 𝑝 𝑝𝑟𝑖𝑚𝑒.
Lemma: 𝑥 𝑚 − 1|𝑥 𝑛 − 1 ⇔ 𝑚|𝑛
Proof: Divide = 𝑥 𝑛 − 1 by 𝑥 𝑚 − 1 with remainder (over ℤ):
𝑥 𝑛 − 1 = (𝑥 𝑚 − 1)(𝑥 𝑛−𝑚 + 𝑥 𝑛−2𝑚 + 𝑥 𝑛−3𝑚 + ⋯ + 𝑥 𝑛−𝑘𝑚 ) + ⏟
𝑥 𝑛−𝑘𝑚 − 1
𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟
𝑘 is such that 𝑘𝑚 ≤ 𝑛 but (𝑘 + 1)𝑚 > 𝑛.
So remainder is 0 ⇔ 𝑛 = 𝑘𝑚 ⇔ 𝑚|𝑛
Theorem: 𝐺𝐹(𝑝𝑚 ) ⊆ 𝐺𝐹(𝑝𝑛 ) ⇔ 𝑚|𝑛
Proof:
If 𝑚|𝑛 then by the lemma 𝑥 𝑚 − 1|𝑥 𝑛 − 1
So in particular setting 𝑥 = 𝑝 we get 𝑝𝑚 − 1|𝑝𝑛 − 1
Using the lemma again, we get that 𝑥 𝑝
𝑚 −1
− 1|𝑥 𝑝
𝑝𝑚 −1
𝑛 −1
−1
So all the roots of 𝑥
− 1 are contained in 𝐺𝐹(𝑝
(which is the set of roots of 𝑥 𝑝
1)
Meaning 𝐺𝐹(𝑝𝑚 )∗ ⊆ 𝐺𝐹(𝑝𝑛 )∗ so 𝐺𝐹(𝑝𝑚 ) ⊆ 𝐺𝐹(𝑝𝑛 )
Now assume 𝐺𝐹(𝑝
⏟ 𝑚 ) ⊆ 𝐺𝐹(𝑝
⏟ 𝑛)
𝐿
𝑛 )∗
𝑛 −1
−
𝐾
So 𝐾 is a vector space over 𝐿, finite. So of finite dimension, say 𝑘 over 𝐿.
|𝐿|𝑘 = |𝐾|
So 𝑝𝑚𝑘 = 𝑝𝑛 so 𝑚|𝑛.
Example:
𝑥 16 − 𝑥
𝑛 = 4 subfields are of order 2𝑚 for 𝑚|4
𝑛 = 1, 𝑛 = 2, 𝑛 = 4: 𝐺𝐹(2), 𝐺𝐹(4), 𝐺𝐹(16)
Note: If 𝐺𝐹(𝑝𝑚 ) ⊆ 𝐺𝐹(𝑝𝑛 ), then 𝜑: 𝐺𝐹(𝑝𝑛 ) → 𝐺𝐹(𝑝𝑛 ) is frobenius automorphism 𝑎 → 𝑎𝑝
𝑚
Then 𝜑𝑚 (𝑎) = 𝑎𝑝
So set if fixed points under
𝑚
𝜑𝑚 = {𝑎|𝜑𝑚 (𝑎) = 𝑎, 𝑎 ∈ 𝐺𝐹(𝑝𝑛 )} = {𝑎 ∈ 𝐺𝐹(𝑝𝑛 )|𝑎𝑝 = 𝑎} =
{𝑎 ∈ 𝐺𝐹(𝑝𝑛 )∗ |𝑎𝑝
𝑛 −1
𝑛
1 = 0} ∪ {0} = set of roots of 𝑥 𝑝 − 𝑥 in 𝐺𝐹(𝑝𝑛 )
Note: If 𝐹 finite field |𝐹| = 𝑝𝑛 and we look at roots of 𝑥 𝑘 − 1 in 𝐹.
Then 𝑎 is a root ⇔ 𝑎𝑘 = 1 in 𝐹 meaning either: 𝑘 = 0 and 𝑎 = 1 or 𝑘|𝑝𝑛 − 1.
The nontrivial factorizations of polynomials of type 𝑥 𝑘 − 1 are only for 𝑘|𝑝𝑛 − 1
(as if gcd(𝑘, 𝑝𝑛 − 1) = 1 only roots will be 1: (𝑥 𝑘 − 1) = (𝑥 − 1)(𝑥 𝑘−1 + ⋯ ))
𝑛
In general, we want to factor 𝑥 𝑝 − 𝑥 or 𝑥 𝑝
𝑛 −1
− 1 over 𝐺𝐹(𝑝).
𝑛
Theorem: over ℤ⁄𝑝ℤ = 𝐺𝐹(𝑝) 𝑥 𝑝 − 𝑥 is a product of all monic irreducible polynomials over
𝐺𝐹(𝑝) where degree divides 𝑛 (each one exactly once as roots are distinct!)
Example:
(𝑥 4 + 𝑥 3 + 1)(𝑥 4 + 𝑥 + 1)(𝑥 4 + 𝑥 3 + 𝑥 2 + 𝑥 + 1)
𝑥 16 − 𝑥 = ⏟
𝑥(𝑥 + 1) (𝑥
⏟ 2 + 𝑥 + 1) ⏟
𝑖𝑟𝑟𝑒𝑑𝑢𝑐𝑖𝑏𝑙𝑒
𝑜𝑓 𝑑𝑒𝑔𝑟𝑒𝑒 1
𝑖𝑟𝑟𝑒𝑑𝑢𝑐𝑖𝑏𝑙𝑒
𝑜𝑟 𝑑𝑒𝑔𝑟𝑒𝑒 2
𝑎𝑙𝑙 𝑖𝑟𝑟𝑒𝑑𝑢𝑐𝑖𝑏𝑙𝑒𝑠
𝑜𝑓 𝑑𝑒𝑟𝑒𝑒 4
Proof: Suppose 𝑓(𝑥) ∈ ℤ⁄𝑝ℤ [𝑥] monic, irreducible of degree 𝑚 and 𝑚|𝑛.
Extend 𝐺𝐹(𝑝) to a field containing a root of 𝑓 denoted 𝛼. This field will have 𝑝𝑚 elements.
We know by the last theorem, since 𝑚|𝑛 this field is contained in a field of 𝐺𝐹(𝑝𝑛 ).
𝑛
𝑛
And so satisfies 𝛼 𝑝 = 𝛼. If 𝛼 = 0, 𝑓(𝑥) = 𝑥 and 𝑥|𝑥 𝑝 − 𝑥!
Otherwise 𝛼 ≠ 0, 𝛼 𝑝
𝑛 −1
− 1 = 0 so 𝛼 root of 𝑥 𝑝
And so its minimal polynomial 𝑓(𝑥) divides 𝑥
𝑛 −1
𝑝𝑛 −1
−1
𝑛
and so 𝑥 𝑝 − 𝑥.
𝑛
Conversely: Suppose now 𝑓(𝑥)|𝑥 𝑝 − 𝑥 ,monic irreducible and its degree is 𝑚.
If 𝛼 is a root of 𝑓(𝑥), then extending 𝐺𝐹(𝑝) to a field containing 𝛼 we get an extension of
dimension 𝑚 over 𝐺𝐹(𝑝) i.e. a field of order 𝑝𝑛 .
𝑛
So 𝛼 is also a root of 𝑥 𝑝 − 𝑥.
And so 𝐺𝐹(𝑝𝑚 ) = 𝐺𝐹(𝑝)(𝛼)
In other words, every element of 𝐺𝐹(𝑝𝑚 ) is a polynomial in 𝛼.
𝑛
𝑛
𝛼 is also a root of 𝑋 𝑝 − 𝑥 as 𝑓(𝑥)|𝑥 𝑝 − 𝑥
So 𝛼 ∈ 𝐺𝐹(𝑝𝑛 ). Giving that 𝐺𝐹(𝑝𝑚 ) = 𝐺𝐹(𝑝)(𝛼) ⊆ 𝐺𝐹(𝑝𝑛 )
But then by the lemma – 𝑚|𝑛.
Error-Correcting Codes
e.g. spellcheck: eleqhant
bed bod
With binary information – location of an error means we can correct it! (0 ↔ 1)
Naïve way:
Transmit the same message 3 times and take a majority check.
The probability of having an error in exactly the same position twice is very low.
Very waistul! We might have a more sophisticated way of doing it…
Parity-Check Digit
Transmit an extra digit at the end of the message.
Send 1 if the message has an odd number of ones.
Send 0 if the message has an even number of ones.
e.g. message = 10101 ⏟
0
𝑝𝑎𝑟𝑖𝑡𝑦
If we get a message with an odd number of ones we know there’s an error, but we don’t
know where it is.
If we get an even number we could have had a double error. But this happens with a
relatively low probability.
Example:
ID with a Sifrat Bikoret
03569657
12121212
0+6+5+3+9+3+5+5=26
10-last digit = 4!
Hamming Code (7,4)
Locates (and so corrects) single errors.
Code words will be of length 7. There will be 4 “information digits” + 3 “redundancy digits”.
We call them also parity check digits even though they do not check parity.
Assumption: very low probability of double errors.
𝑝 = probability of error in transmitting a digit.
Probability of a correctly transmitted message is (1 − 𝑝)7
Probability of transmitting exactly one error: 7𝑝(1 − 𝑝)6
So if you add them together you get: (1 − 𝑝)7 + 7𝑝(1 − 𝑝)6
If 𝑝 = 0.1 get 0.853 of a message with ≤ 1 errors.
Sending 4 digits (with no redundancy) correctly has probability (1 − 𝑝)4
If 𝑝 = 0.1 get 0.6561.
So 0.853 is a big improvement of sending only 4 digits and no errors!
This is a linear code, 𝑖. 𝑒. our code words are elements of a vector space over 𝐺𝐹(2):
elements of 𝐺𝐹(2)7
Subspace of dimension 4. i.e. there are going to be 16 possible code words.
(same number of code words in 𝐺𝐹(2)4 )
We define our code by giving a basis: 4 vectors of length 7.
(in a 4 × 7 matrix).
𝑣1 1 0
𝑣2 0 1
𝑣3 0 0
𝑣4 0 0
Suppose we want to transmit 1101?
Send instead 𝑣1 + 𝑣2 + 𝑣4 = 1101001
0
0
1
0
0
0
0
1
0
1
1
1
1
0
1
1
1
1
0
1
Big advantage: Efficient decoding and locates ≤ 1 errors.
Use an analog to inner product/scalar multiplication. Induced by matrix multiplication over
𝐺𝐹(2).
[𝑥1
7
𝑦1
… 𝑥𝑛 ] [ ⋮ ] = ∑ 𝑥𝑖 𝑦𝑖 (𝑚𝑜𝑑 2)
𝑦𝑛
𝑖=1
It is a bilinear form on 𝐺𝐹(2)4 .
Decoding:
Suppose we receive 𝑦 ∗ = [1 1 0 1 1 1
𝑣1 + 𝑣2 = 𝑦 = [1 1 0 0 1 1 0]
0]
We compute:
𝑦∗ ∙ 𝑎 = 1 + 1 + 1 = 1
𝑦∗ ∙ 𝑏 = 1 + 1 = 0
𝑦∗ ∙ 𝑐 = 1 + 1 = 0
The result is sequence 100
Which happens to be the binary representation of 4. And the error is in the fourth digit!
If there’s no error, we get 0
𝑎 = [0001111]
𝑏 = [0110011]
𝑐 = [1010101]
Hamming matrix:
1
0
[
0
0
0
1
0
0
0
0
1
0
0
0
0
1
0
1
1
1
1
0
1
1
1
1
]
0
1
The trick is in fact - Orthogonal complements:
Recall: 𝑉 is a vector space over𝐹.
𝐵: 𝑉 × 𝑉 → 𝐹 Is a bilinear form if it is linear in both variables:
𝐵(𝑎1 𝑣1 + 𝑎2 𝑣2 , 𝑤) = 𝑎1 𝐵(𝑣1 , 𝑤) + 𝑎2 + 𝐵(𝑣2 , 𝑤)
𝐵(𝑣, 𝑎1 𝑤2 + 𝑎2 𝑤2 ) = 𝑎1 𝐵(𝑣, 𝑤) + 𝑎2 (𝑣, 𝑤2 )
And for any subspace 𝑊 of 𝑉 we can define
𝑊
⏟⊥
= {𝑤 ∈ 𝑉|𝐵(𝑢, 𝑤) = 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑤 ∈ 𝑊}
𝑂𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙
𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡
𝑜𝑓 𝑊 𝑤𝑟𝑡 𝐵
𝑊 ⊥ is a subspace of 𝑉.
If 𝐹 has charactaristics 0 and 𝐵 is non-degenerate bilinear form.
e.g. If 𝐹 = ℝ and 𝐵 is dot product.
If 𝐹 = ℂ and 𝐵 is inner product (𝑣, 𝑤) = 𝑣 𝑇 ∙ 𝑤
̅
Then we have that:
𝑊 ⊕ 𝑊⊥ = 𝑉
For 𝑉 finite dimension.
Proof: uses fact that 𝑊 ∩ 𝑊 ⊥ = {0} so that the union of base for 𝑊 and a base for 𝑊 ⊥ is a
base for 𝑉.
In general, for 𝐹 or characteristic 𝑝 and arbitrary bilinear form this is not true!
e.g. Taking product defined in 𝐺𝐹(27 ) can see that [1
to itself!
E.g.
If 𝑊 = 𝑠𝑝𝑎𝑛{[1
e.g.
1 0 0
0 0
[0
1 0
0 0 0
0]} then 𝑊 ⊊ 𝑊 ⊥
0 1 1
0 0
0] ∈ 𝑊 ⊥ \𝑊
And 𝑊 ⊥ ≠ 𝐺𝐹(2)7
But: dim 𝑊 + dim 𝑊 ⊥ = dim 𝑉 ← proof in Basic Algebra 1 (Jacobson)
E.g. dim 𝑊 ⊥ above will be 6!
Take as a basis for 𝑊 ⊥ :
0 0 1 0 0 0 0
0 0 0 1 0 0 0
0 0 0 0 1 0 0
0 0 0 0 0 1 0
0 0 0 1 0 0 0
[1 1 0 0 0 0 0]
--- end of lesson
0] is orthogonal
The parity check matrix is defined to be a matrix whose columns are a basis for the
orthogonal complement of the code.
Correcting Errors in linear codes over GF(2)
Given a vector which contains errors, we want to correct it to the code word that differs
from it in the fewest digits.
Define -Hamming distance: 𝑑(𝑣, 𝑤) = # of digits which 𝑣 and 𝑤 differ.
e.g.
𝑣 = (1 0 1 1 0 0 1 1),
𝑤 = (0 1 1 1 1 0 1 0)
𝑑(𝑣, 𝑤) = 4
Turns out, that in the hamming code, every 2 words/vectors are at distance ≥ 3.
TODO: Draw words in the code in a schematic way
Circle of radius 1 around 𝑤 = all vectors 𝑣 such that 𝑑(𝑤, 𝑣) = 1.
So any vector with one error can only be corrected in one way o a codeword.
General: We can correct 𝑟 errors if the minimal distance between two code words ≥ 2𝑟 + 1
Note: In the hamming code we have 16 elements. In the whole space, we have 27 = 128
elements. The elements at distance exactly 1 from a codeword = 7 ∙ 16.
So in fact, every element in the space is either in the code or at distance 1 from a codeword
as 7 ∙ 16 + 16 = 128.
BCH Code
Bose-Chandhuri-Hocquenghem
Double error correcting code that uses 𝐺𝐹(16) and has a nice decoding algorithm similar to
that of the hamming code.
Construct by starting with the parity check matrix 𝐻 (and then the code will be orthogonal
complement of its rows).
The elements will be vectors in 𝐺𝐹(2)15
(need minimal hamming distance to be at least 5!)
𝐺𝐹(16)∗ = {1, 𝛼, … , 𝛼 14 } where 𝛼 is the root of 𝑥 4 + 𝑥 3 + 1 over 𝐺𝐹(2).
Use: representation of 𝐺𝐹(16) as vectors over 𝐺𝐹(2) of length 4.
Form of 𝐻 is going to be as follows:
8 × 15 matrix over 𝐺𝐹(2)
𝑏
𝐻=[ 1
𝑐1
4
Where 𝑏𝑖 , 𝑐𝑖 ∈ 𝐺𝐹(2) row vectors.
We think of also as elements of 𝐺𝐹(16).
𝑏2
𝑐2
… 𝑏15
]
… 𝑐15
Take 𝑏𝑖 = vector of length 4 corresponding to 𝛼 𝑖−1 in the table.
So we have 1, 𝛼, … , 𝛼 14 in the top half of the matrix.
𝑐𝑖 ’s will be defined later…
We want: If 𝑥 = (𝑥1 … 𝑥 15 ) codeword, we want:
(1) 𝐻 ∙ 𝑥 𝑇 = 0 ⇔ 𝑥 in code
(2) If 𝑥 has at most 2 errors, want it to detect by multiplication by 𝐻.
Suppose 𝑥 has exactly 2 errors in positions 𝑖 and 𝑗. Then we can write:
𝑥 = 𝑥𝑐 + 𝑒𝑖 + 𝑒𝑗
And then:
𝑏𝑖 + 𝑏𝑗
𝐻 ∙ 𝑥 = 𝐻𝑥 + 𝐻𝑒𝑖 + 𝐻𝑒𝑗 = 𝐻𝑒𝑖 + 𝐻𝑒𝑗 = (
)
𝑐𝑖 + 𝑐𝑗
So we want to choose the 𝑐𝑖 ’s so we can recover from this vector.
𝑏
Bad choice: 𝑐𝑖 = 𝑏𝑖 . Get 𝐻𝑥 = ( ) - in this case we cannot recover 𝑖 and 𝑗.
𝑏
1
1
1
0
0
0
0
0
1
1
If 𝑏 = ( ). We could have had: ( ) + ( ) But also: ( ) + ( ) And a lot of other
0
0
0
1
1
⏟
⏟
⏟
⏟
1
1
0
1
0
𝑏1
𝑏4
𝑏8
𝑏0
possibilities.
Another bad choice: define 𝑐𝑖 = (𝑏𝑖 )2 (thinking of 𝑏𝑖 as an element of 𝐺𝐹(16) so that 𝑐𝑖
corresponding to 𝛼 2𝑖−2
So we should then get:
𝑏𝑖 + 𝑏𝑗
𝑏𝑖 + 𝑏𝑗
𝑏
𝐻𝑥 = ( 2
=
)
(
2) = ( 2)
𝑏𝑖 + 𝑏𝑗2
𝑏
(𝑏𝑖 + 𝑏𝑗 )
If you square you get the same thing….
Definition: Take 𝑐𝑖 = 𝑏𝑖3.
(
𝑏𝑖 + 𝑏𝑗
𝑏𝑖3 + 𝑏𝑗3
𝑏
) = ( ) want to show 𝑖 and 𝑗 determined uniquely and how to find them.
𝑐
𝑐 = 𝑏𝑖3 + 𝑏𝑗3 = (𝑏𝑖 + 𝑏𝑗 )(𝑏𝑖2 + 𝑏𝑖 𝑏𝑗 + 𝑏𝑗2 ) = 𝑏(𝑏𝑖2 + 𝑏𝑖 𝑏𝑗 + 𝑏𝑗2 ) = 𝑏(𝑏 2 + 𝑏𝑖 𝑏𝑗 )
(regarding the elements of 𝐺𝐹(16))
We first assume we have exactly 2 errors. So 𝑖 ≠ 𝑗 and 𝑏 ≠ 0.
Get 𝑐𝑏 −1 + 𝑏 2 = 𝑏𝑖 𝑏𝑗
So 𝑏𝑖 and 𝑏𝑗 are roots in 𝐺𝐹(16) of the quadratic equation:
(𝑥 − 𝑏𝑖 )(𝑥 − 𝑏𝑗 ) = 𝑥 2 − (𝑏𝑖 + 𝑏𝑗 )𝑥 + 𝑏𝑖 𝑏𝑗 = 𝑥 2 − 𝑏𝑥 + 𝑐𝑏 −1 + 𝑏 2
So given 𝑏 and 𝑐, construct this polynomial.
𝑏𝑖 and 𝑏𝑗 are its unique solutions (in the field 𝐺𝐹(16)).
For convenience write: 𝐻 ′ = 𝐻 with 𝛼 notation.
2
14
𝐻 ′ = [1 𝛼3 𝛼 6 … 𝛼 12 ]
1 𝛼 𝛼
… 𝛼
Suppose 𝑦 is a received message with errors in positions 𝑖 and 𝑗.
𝑖−1
𝑗−1
5
And suppose 𝐻 ′ 𝑦 = ( 𝛼3𝑖−3 + 𝛼 3𝑗−3 ) = (𝛼 7 )
𝛼
𝛼
+𝛼
1
0
1
1
Equivalently: 𝐻 ∙ 𝑦 =
polynomial will be: 𝑥 2 + 𝛼 5 𝑥 + 𝛼 8
0
1
1
(1)
Since: 𝑐𝑏 −1 + 𝑏 2 = 𝛼 7 ∙ 𝛼 −5 + 𝛼 10 = 𝛼 2 + 𝛼 10 = 𝛼 3
Need 𝑖 and 𝑗 such that: 𝛼 𝑖−1 + 𝛼 𝑗−1 = 𝛼 5 and 𝛼 𝑖−1 ∙ 𝛼 𝑗−1 = 𝛼 8
𝑖 + 𝑗 − 2 ≡ 8(𝑚𝑜𝑑 15)
𝑖 + 𝑗 ≡ 10 (𝑚𝑜𝑑 15)
Checking possibilities: Get only 𝑖 = 3, 𝑗 = 7 satisfies 𝛼 𝑖−1 + 𝛼 𝑗−1 = 𝛼 5 as well.
Note: If the quadratic polynomial has no roots, then it cannot result from a double error.
Meaning in fact that some triple errors are detectable but not correctable.
Single errors are also correctable using 𝐻:
𝑏
It is the only case where we get a vector of the form: ( 3 ) and then determine 𝑏𝑖 = 𝑏 by
𝑏
checking.
So the polynomial will be 𝑥(𝑥 − 𝑏).
We want to determine the dimension of the code and how to calculate a matrix for the
code.
Claim: 𝑟𝑎𝑛𝑘𝐻 = 8
Conclusion: dim 𝑐𝑜𝑑𝑒 = 7
We shall show, that the first eight columns are linearly independent.
𝑏𝑖
0
Suppose ∑8𝑖=1 𝑎𝑖 ( 3 ) = ( ) and 𝑎𝑖 ∈ 𝐺𝐹(2)
𝑏𝑖
0
𝑖−1
𝑖
Then we also get ∑8𝑖=1 𝑎𝑖 ( 𝛼3𝑖−3 ) = 0 ⇒ ∑7𝑖=0 𝑎𝑖+1 ( 𝛼3𝑖 ) = 0 ⇔
𝛼
𝛼
7
7
𝑖
3𝑖
∑𝑖=0 𝑎𝑖+1 𝛼 = 0 and ∑𝑖=0 𝑎𝑖+1 𝛼 = 0
Look at the polynomial ∑7𝑖=0 𝑎𝑖+1 𝑥 𝑖 = 0 over 𝐺𝐹(2) And 𝛼 and 𝛼 3 are both roots.
So their minimal polynomials both divide ∑7𝑖=0 𝑎𝑖+1 𝑥 𝑖
7
4
3
𝑥 + 𝑥 + 1,
4
3
2
𝑥 + 𝑥 + 𝑥 + 𝑥 + 1| ∑ 𝑎𝑖+1 𝑥 𝑖
𝑖=0
The product (𝑥 4 + 𝑥 3 + 1)(𝑥 4 + 𝑥 3 + 𝑥 2 + 𝑥 + 1) which is a polynomial of degree 8
divides ∑7𝑖=0 𝑎𝑖+1 𝑥 𝑖 which is of degree less or equal to 7! So ∑7𝑖=0 𝑎𝑖+1 𝑥 𝑖 is the zero
polynomial! Therefore all coefficients are zero and therefore linearly independent.
Thus are also a basis for our vector space.
We construct 𝐶 = matrix for the code.
𝐻 will be of the form: 7 × 15
Where the first 8 columns are are the redundancy digits and the last 7 columns are the
information digits.
7
𝛼
1
Take ( ) , ( 3 ) , … , ( 𝛼21 ) first 8 columns of 𝐻 ′ .
𝛼
1
𝛼
8
𝑖
𝛼
The 9’th column ( 24 ) is a linear combination of the first 8 columns: ∑7𝑖=0 𝑠𝑖 ( 𝛼3𝑖 )
𝛼
𝛼
So the row vector (𝑠0 𝑠1 … 𝑠7 1 0 … 0) orthogonal to all rows of 𝐻 ′ and 𝐻!
Take as the first row of 𝑐.
9
Similarly, column 10: ( 𝛼27 ) = linear combination of 8 columns of 𝐻 ′ .
𝛼
7
9
1
𝛼
𝑡0 ( ) + ⋯ + 𝑡7 ( 21 ) = ( 𝛼27 )
1
𝛼
𝛼
So
7
9
1
0
𝑡0 ( ) + ⋯ + 𝑡7 ( 𝛼21 ) + ( 𝛼27 ) = ( )
1
0
𝛼
𝛼
So take the vector (𝑡0 … 𝑡7 0 1 0 … 0) orthogonal to rows of 𝐻 ′ take to be row
2 of 𝐶 etc.