Algebra through Examples

Lesson 1

General Details E-mail: Josie.Shamash@weizmann.ac.il

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.

Comment [R1 :](*)

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)

- 𝑎1𝑏 + 𝑎2𝑏 = 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 ∈𝐼

− 𝑞 ∙ 𝑛 ∈𝐼

= 𝑟 ∈ 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

𝑘 = 𝑎 𝑏𝑐 𝑑

𝑎, 𝑏, 𝑐 ∈ ℝ is a subring but not a left or right Ideal.

e.g.

1 11 1

∙ 𝑎 𝑏0 𝑐

= 𝑎 𝑏 + 𝑐𝑎 𝑏 + 𝑐

∈ 𝑘 𝑜𝑛𝑙𝑦 𝑖𝑓 𝑎 ≠ 0

𝑎 𝑏0 𝑐

∙ 1 11 1

= 𝑎 + 𝑏 𝑎 + 𝑏

𝑐 𝑐 ∈ 𝑘 𝑜𝑛𝑙𝑦 𝑖𝑓 𝑐 ≠ 0

However, 𝐼 = 𝑎 𝑏0 0

𝑎, 𝑣 ∈ ℝ is a right Ideal!

e.g.

𝑎 𝑏0 0

∙ 𝑥 𝑦𝑢 𝑣 =

∗ ∗0 0 ∈ 𝐼

It is not, however, a left Ideal:

𝑥 𝑦𝑢 𝑣 ∙

𝑎 𝑏0 0

= 𝑎𝑥 ∗𝑢𝑎 ∗

𝑖𝑓 𝑢𝑎 ≠ 0 → ∉ 𝐼

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.

𝑅𝐼 = 𝐼 + 𝑎 𝑎 ∈ 𝑅 (equality sets)

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 ∈𝐼

+ 𝑎 + 𝑏 = 𝐼 + (𝑎 + 𝑏)

Note: 𝐼 + 𝑧 = 𝐼, ∀𝑧 ∈ 𝐼

Lets look at 𝐼 + 𝑎′ ∙ 𝑏′

𝐼 + 𝑎′ ∙ 𝑏′ = 𝐼 + 𝑥 + 𝑎 𝑦 + 𝑏 = 𝐼 + 𝑥𝑦 ∈𝐼

+ 𝑎𝑦 ∈𝐼

+ 𝑥𝑏 ∈𝐼

+ 𝑎𝑏 = 𝐼 + 𝑎 ∙ 𝑏

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. ℤ𝑛ℤ

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 0 , 1 , … , 𝑛 − 1 wrt +,∙ 𝑚𝑜𝑑 𝑛

2. 𝐹 𝑥 𝑓 𝑥 𝐹 𝑋 𝑤𝑕𝑒𝑟𝑒 𝐹 𝑖𝑠 𝑎 𝑓𝑖𝑒𝑙𝑑

for instance, when 𝑓 𝑥 = 𝑥2 − 3𝑥 + 2, 𝐹 = ℤ

So in fact:

ℝ 𝑥 𝑥2 − 3𝑥 + 2

= 𝐼 + 𝑎𝑥 + 𝑏 𝑎, 𝑏 ∈ ℝ

Since addition and multiplication are in polynomials mod 𝑥2 − 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 𝑥2 − 3𝑥 + 2 + 𝑟 𝑥

where 𝑞 𝑥 , 𝑟 𝑥 ∈ ℝ 𝑥 ∧ 𝑑𝑒𝑔𝑟𝑒𝑒 𝑟 𝑥 < 2 ∨ 𝑟 𝑥 = 0

Also, since 𝑥2 − 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 + 6𝑥2 − 2𝑥 − 5 = 𝐼 + 6 𝑥2 − 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 = 𝑥2 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 + 𝑝𝑕𝑖 0 → 𝑝𝑕𝑖 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

𝑥2 + 1 ≅ ℂ

𝑎𝑗𝑥𝑗

𝑘

𝑗 =0

Look at homomorphism: 𝑓 𝑥 → 𝑓 𝒾 from ℝ 𝑥 𝜙→ℂ

What is the kernel?

𝑘𝑒𝑟𝜙 = 𝑓 𝑥 ∈ ℝ 𝑥 𝑓 𝒾 = 0

= 𝑓 𝑥 ∈ ℝ 𝑥 𝑓 𝑥 𝑖𝑠 𝑎 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑜𝑓 𝑥2 + 1 𝑏𝑦 𝑎𝑛𝑜𝑡𝑕𝑒𝑟 𝑝𝑜𝑙𝑦𝑛𝑜𝑚

(we shall see that later)

Comment [u2 :]Consider Deleting

Example2 𝜙: ℤ → 0 , 1 , … , 𝑛 − 1 that sends 𝑥 ∈ ℤ to 𝑥 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!)

ℤ5𝑋ℤ5 (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 = 𝑥2 + 𝑦2

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

We then got −5 2

| ‖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 + 5𝑏2 → 𝑎2(𝑎𝑛𝑑 𝑕𝑒𝑛𝑐𝑒 𝑎𝑙𝑠𝑜 𝑎) are integer multiples of 5.

So write 𝑎 = 5𝑎′ , 𝑎′ ∈ ℤ.

And 𝑟 = 5𝑎′ + 𝑏 −5 = −5 ∈𝑅𝑖𝑛𝑔

− −5𝑎′ + 𝑏

∈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 + 5𝑏2 = 2 which has no solutions

with 𝑎, 𝑏 ∈ ℤ.

Case 2: wlog, ‖𝑟‖ = 1 and ‖𝑠‖2 = 4 then get 𝑎2 + 5𝑏2 = 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

Comment [u3 :]*

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−1𝑠|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 = 𝑟1𝑞2 + 𝑟2, 𝛿 𝑟2 < 𝛿 𝑟1 or 𝑟2 = 0

If 𝑟2 = 0 then 𝑔𝑐𝑑 𝑎, 𝑏 = 𝑟1

Otherwise, I can write 𝑟1 = 𝑟2𝑞3 + 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: 𝐼𝑛∞𝑛=1 = 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 = 𝑥2 + 𝑦2 = ‖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 + 𝑠𝑖 −1, , =𝑟 − 𝑖𝑠

𝑟2 + 𝑠2 𝑟 + 𝑠𝑖 ≠ 0

So 𝑎 ∙ 𝑏−1 ∈ ℚ 𝑖 .

So write: 𝑎 ∙ 𝑏−1 = 𝛼 + 𝛽𝑖, 𝛼, 𝛽 ∈ ℚ. ∃𝑢, 𝑣 ∈ ℤ: 𝑢 − 𝛼 ≤1

2, 𝑢 − 𝛽 ≤

1

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

‖ 𝛼 − 𝑢 + 𝑖 𝛽 − 𝑣 ‖2 = 𝛼 − 𝑢 2 + 𝛽 − 𝑣 2 ≤1

4+

1

4=

1

2

So that 𝛿 𝑟 ≤1

2𝛿 𝑏 < 𝛿 𝑏

Euclidean ⇒ PID.

But PID does not imply Euclidean!

Counter Example:

ℤ 1

2+

−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:

𝑅 = ℤ 𝑥,𝑥

2,𝑥

3, … ,

𝑥

𝑛, … = 𝑥 ∙ ℚ 𝑥 + ℤ

5

6𝑥5 +

2

3𝑥4 + 3 = 5𝑥4 ∙

𝑥

6+ 2 ∙

𝑥

3∙ 𝑥3 + 3

𝑅 is a subring of ℚ 𝑥 .

𝑅 ≠ ℚ 𝑥 as 1

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!

4) Ideals generated by 𝑥,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 𝑛𝑖ℤ + 𝑛𝑗ℤ = ℤ so conditions on ideals 𝑛𝑖ℤ hold etc…

Prove for 𝒏 = 𝟐

We have 𝐼1 + 𝐼2 = 𝑅

So we have 𝑏𝑗 ∈ 𝐼𝑗 𝑠. 𝑡. 𝑏1 + 𝑏2 = 1

Let 𝑥 = 𝑎2𝑏1 + 𝑎1𝑏2

𝑥 + 𝐼1 = 𝑎2𝑏1 ∈𝐼1

+ 𝑎1𝑏2 + 𝐼1 = 𝑎1𝑏2 + 𝐼1 = 𝑎1 1 − 𝑏1 + 𝐼1 = 𝑎1 − 𝑎1𝑏1 + 𝐼1 ∈𝐼1

= 𝑎1 + 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

𝑎1𝑏1 + ⋯ + 𝑎𝑛𝑏𝑛 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

Look at the product: 𝑐𝑖 + 𝑏𝑖𝑛𝑖=2 = 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 ∈𝐼1

+ 𝑏1 ∙ … ∙ 𝑏𝑛 ∈𝐽1

= 1

So that 𝐼1 + 𝐽1 = 𝑅

By the CRT for case 𝑛 = 2 have 𝑦1 ∈ 𝑅 𝑠. 𝑡.

𝑦1 ≡ 1 𝑚𝑜𝑑𝐼1

𝑦1 ≡ 0 𝑚𝑜𝑑𝐽1

Since 𝐽1 ⊆ 𝐼2 ∩ 𝐼3 ∩ …∩ 𝐼𝑛 we also get 𝑦1 ≡ 0 𝑚𝑜𝑑𝐼𝑗 𝑗 > 1

Repeat for each 𝑖: 𝐽𝑖 = 𝐼𝑘 𝑘≠𝑖

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 = 𝑎1𝑦1 + 𝑎2𝑦2+. . +𝑎𝑛𝑦𝑛

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 𝐼1 ∩ …∩ 𝐼𝑛 ≅ 𝑅 𝐼1

× 𝑅 𝐼2 × … × 𝑅 𝐼𝑛

(actually equivalent to CRT)

Proof: Define a homomorphism 𝑓: 𝑅 → 𝑅 𝐼1 × … × 𝑅 𝐼𝑛

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 𝐼1 × 𝑅 𝐼2

× … × 𝑅 𝐼𝑛 to get isomorphism

(by homomorphism theorem)

Let 𝑎1 + 𝐼1 , … , 𝑎𝑛 + 𝐼𝑛 ∈ 𝑅 𝐼1 × 𝑅 𝐼2

× … × 𝑅 𝐼𝑛

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 = 𝑝𝑖𝑟𝑖𝑘

𝑖=1 𝑝𝑖 distinct primes. 𝐼𝑖 = 𝑝𝑖𝑟𝑖ℤ

ℤ 𝑚ℤ ≅ ℤ𝑝1

𝑟1ℤ × … × ℤ𝑝𝑘

𝑟𝑘ℤ

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 ∗

= 1 , 5 =group of two elements

Looking at the group of units on both sides we get:

ℤ 6ℤ ∗

≅𝑖𝑠𝑜𝑚𝑜𝑟𝑝 𝑕𝑖𝑠𝑚 𝑜𝑟 𝑢𝑛𝑖𝑡 𝑔𝑟𝑜𝑢𝑝𝑠

ℤ𝑝1

𝑟1ℤ

∗

× … × ℤ𝑝𝑘

𝑟𝑘ℤ

∗

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.

From (*) we get the formula: 𝜑 𝑚 = 𝜑 𝑝1𝑟1 ∙ … ∙ 𝜑 𝑝𝑘

𝑟𝑘

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: 𝑏𝑠 ≡ 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 𝑝1ℤ

∗∙ ℤ 𝑝2ℤ

∗ has 𝜑 𝑑 = 𝑒 elements.

Recall in a group 𝐺 of order 𝑛

𝑥𝑛 = 1 for all 𝑥 ∈ 𝐺.

Follows from Lagraunge’s theorem – shall prove later.

So that 𝑎𝑒 ≡ 1(𝑚𝑜𝑑 𝑑) 𝑎 = 𝑎 + 𝑑ℤ elements of ℤ 𝑑ℤ ∗

𝑟𝑠 ≡ 1 𝑚𝑜𝑑 𝑒

𝑏𝑠 ≡ 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

Comment [AR4 :]*

ℤ 𝑑ℤ ≅

𝜓 ℤ

𝑝1ℤ × ℤ

𝑝2ℤ

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

𝑥𝑝−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:

𝜓 𝑎𝑒 + 𝑑ℤ = 𝜓 +𝑑ℤ 𝑒

= 0 𝑚𝑜𝑑 𝑝1 , 1 𝑚𝑜𝑑 𝑝2

Again, writing: 𝑟𝑠 = 𝑙𝑒 + 1 we get

𝜓 𝑏𝑠 + 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

Since 𝜓 is an isomorphism we get 𝑎 ≡ 𝑏𝑠 𝑚𝑜𝑑 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 = 𝑕𝑎 𝑕 ∈ 𝐻

𝑎𝐻 left coset = 𝑎𝑕 𝑕 ∈ 𝐻

Properties: Cosets are disjoint or equal.

Suppose 𝐻𝑎 ∩ 𝐻𝑏 ≠ ∅

So have 𝑕, 𝑕′ ∈ 𝐻 𝑠. 𝑡. 𝑕𝑎 = 𝑕′𝑏

𝑕′ −1𝑕𝑎 = 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

𝑝 = 1 2 32 1 3

𝐼𝑑, 𝑟 is a subgroup of G. Which is not normal!

𝐻 ∙ 1 2 33 2 1

= 1 2 33 2 1

, 1 2 32 1 3

∙ 1 2 33 2 1

= 1 2 32 3 1

1 2 33 2 1

∙ 𝐻 = 1 2 33 2 1

, 1 2 33 2 1

∙ 1 2 32 1 3

= 1 2 33 2 1

So this is not the same group!

𝐴3 = set of even permutations = normal subgroup of order 3

𝑟 = 𝐼𝑑, 1 2 32 3 1

, 1 2 33 1 2

𝐴3𝜍 = 𝜍𝐴3 = 𝑆3\𝐴3 = 1 2 32 1 3

𝜍

, 1 2 33 2 1

, 1 2 31 3 2

----- End of lesson 4

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 = 𝑎𝑖𝑥𝑖

𝑖=0 , 𝑎𝑖 ∈ 𝐹

Look at the coset 𝑢 + 𝑝 𝑢 = 𝛼 ∈ 𝐾

𝑝 𝛼 ⊂ 𝑎𝑖𝑢𝑖 = 𝑎𝑖 𝑢 + 𝑝 𝑢 = 𝑎𝑖𝑢

𝑖 + 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 ≡ 𝑕 𝑢 𝑚𝑜𝑑 𝑝 𝑢

Then 𝑔 𝑢 = 𝑕)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 = 𝑎𝑖𝑥𝑖 , 𝑎𝑖 ∈ 𝐹𝑘

𝑖=0

𝜑 𝑓 𝛼 = 𝜑 0𝐾 = 0, 𝑎𝑖 ∈ 𝐹

0 = 𝜑 𝑓 𝛼 = 𝜑 𝑎𝑖𝛼𝑖

𝑘

𝑖=0

= 𝜑 𝑎𝑖 𝜑 𝛼 𝑖

𝑘

𝑖=0

=𝑎𝑖∈𝐹

𝑎𝑖𝜑 𝛼 𝑖

𝑘

𝑖=0

Special case:

𝐾 splitting field for 𝑓 𝑥 ∈ 𝐹 𝑥 then 𝐾 = 𝐹 𝛼1 , … , 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

𝛽 = 𝑎𝑖1…𝑖𝑘 ∙ 𝛼1𝑖1𝛼2

𝑖2 …𝛼𝑘𝑖𝑘

Examples:

1) Galois group of the smallest field of 𝑥4 − 2 over ℚ

Roots of 𝑥4 − 2:

± 24

, ±𝒾 24

𝑥4 − 2 = 𝑥 − 24

𝑥 + 24

𝑥 − 𝒾 24

𝑥 + 𝒾 24

and over k: ℚ 24

, 𝒾

𝜑 ∈ 𝐺𝑎𝑙 𝐾 ℚ = 𝐺 will permute 4 roots

So can think of 𝐺 of being a subgroup of 𝑆4

We know that ℚ 24

, 𝒾 : ℚ = (ℚ 24

, 𝒾 : ℚ 24

∙

ℚ 24

4=𝑑𝑒𝑔𝑟𝑒𝑒 𝑜𝑓𝑚𝑖𝑛𝑖𝑚𝑎𝑙 𝑝𝑜𝑙𝑦

𝑜𝑓 24

𝑜𝑣𝑒𝑟

ℚ 𝑋4−2

: ℚ

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

𝜑 24

= 24

𝜑 − 24

= − 24

𝜑 𝒾 24

= −𝒾 24

𝜑 −𝒾 24

= −𝒾 24

Let 𝜓 be the automorphism that permutes roots cyclically:

𝜓 24

= 𝒾 24

and fixes 𝒾

𝜓 is of order 4

𝜓 − 24

= −𝒾 24

𝜓 𝒾 24

= 𝜓 𝒾 𝜓 24

= 𝒾𝒾 24

= − 24

< 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

𝜍 = 1 2 3 4 5 63 1 4 6 5 2

= 13462 5

1 2 3 4 5 63 5 1 2 4 6

= 13 254 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 = 𝑥3 + 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 = 𝑥𝑛 + 𝑎𝑛−1𝑥𝑛−1 + ⋯ + 𝑎1𝑥 + 𝑎0 , 𝑎𝑖 ∈ ℤ

𝑟, 𝑠 ∈ ℤ

If 𝑟

𝑠 is a root then: 0 = 𝑓

𝑟

𝑠 =

𝑟𝑛

𝑠𝑛 + 𝑎𝑖𝑟𝑖

𝑠𝑖

𝑛−1𝑖=0

Assume 𝑟, 𝑠 = 1

𝑟𝑛 + 𝑎𝑖𝑟𝑖𝑠𝑛−𝑖

𝑛−1

𝑖=0

= 0

𝑟𝑛 = −𝑎0𝑆𝑛 − 𝑎1𝑆

𝑛−1 + ⋯− 𝑎𝑛−1𝑠𝑟𝑛−1

If 𝑝 is a prime divisor of 𝑠, then 𝑝|𝑟𝑛 so 𝑝|𝑟.

But then, 𝑝|𝑠 and 𝑝|𝑟 which contradicts the fact that 𝑠 and 𝑟 are mutually prime.

So 𝑠 has no prime divisors. So 𝑠 = ±1. Therefore, 𝑟

𝑠∈ 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

𝑥3 + 2𝑥 + 1 = 𝑥 − 𝛼 𝑥2 + 2 + 𝛼 𝑥 + 2 + 𝛼 𝛼 =𝑜𝑣𝑒𝑟 𝑆.𝐹.

𝑥 − 𝛼 𝑥 − 𝛽 𝑥 − 𝛽

Where 𝛽 and 𝛽 are nonreal roots.

So the splitting field will be ℚ 𝛼, 𝛽

ℚ 𝛼, 𝛽 : ℚ = ℚ 𝛼, 𝛽 : ℚ 𝛼 =2 (𝑏𝑦 𝑡𝑕𝑒 𝑒𝑥𝑡𝑟𝑎 𝑓𝑎𝑐𝑡)

∙ ℚ 𝛼 : ℚ =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 𝛽 and so is isomorphic to a subtgroup of

𝑆3 of order 6 ⇒ Galois group ≅ 𝑆3

TOPIC:

Cyclotomic fields and their Galois groups over ℚ

Definition Cyclotomic field is one of the form ℚ 1𝑛

1𝑛

= 𝑒2𝜋𝒾

𝑛 positive with root of 1

Note that ℚ 1𝑛

is a splitting field of the polynomial 𝑥𝑛 − 1 over ℚ

As:

𝑥𝑛 − 1 = Πk=0

𝑛−1

𝑥 − 𝑒2𝜋𝒾𝑛

We also want to factor 𝑥𝑛 − 1 into irreducible factors over ℚ.

E.g. 𝑥3 − 1 = 𝑥 − 1 𝑥2 + 𝑥 + 1 𝑖𝑟𝑟𝑒𝑑𝑢𝑐𝑖𝑏𝑙𝑒 𝑜𝑣𝑒𝑟 ℚ=𝑚𝑖𝑛𝑖𝑚𝑎𝑙 𝑝𝑜𝑙𝑦

Definition: Denote by 𝜆𝑛 𝑥 = minimal polynomial of 1𝑛

over ℚ

So 𝜆3 𝑥 = 𝑥2 + 𝑥 + 1

𝜆𝑛 𝑥 = n’th cyclotomic polynomial

𝜆1 𝑥 = 𝑥 − 1

𝜆2 𝑥 = 𝑥 + 1

𝜆3 𝑥 = 𝑥2 + 1

14

= 𝒾

𝑥4 − 1 = 𝑥2 − 1 𝑥2 + 1 = 𝑥 + 1 =𝜆2

𝑥 − 1 =𝜆1

𝑥2 + 1 =𝜆3

Fact: If 𝑓 𝑥 ∙ 𝑔 𝑥 = 𝑥𝑛 − 1 over ℚ, then 𝑓 𝑥 , 𝑔 𝑥 ∈ ℤ 𝑥

(Follows from Gauss’ lemma – Basic algebra 1)

Interesting fact:

If we factor 𝑥𝑛 − 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

ℚ 1𝑛

: ℚ = deg 𝜆𝑛 =?

Examples:

1)

ℚ 𝑖 = ℚ 14

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.

2) ℚ 𝜔 = ℚ 13

ℚ 𝜔 : ℚ = 2 irreducible polynomial 𝜆3 of 𝜔 is 𝑥2 + 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

3) ℚ 15

𝜆5 𝑥 = 𝑥4 + 𝑥3 + 𝑥2 + 𝑥 + 1

ℚ 15

: ℚ = 4

1, 𝜌, 𝜌2 , 𝜌3 basis for ℚ 15

over ℚ

In general

4) 𝑝 is prime ℚ 1𝑝

𝑥𝑝 − 1 = 𝑥 − 1 𝑥𝑝−1 + 𝑥𝑝−2 + ⋯ + 𝑥 + 1

The second part is irreducible using einsensteins criterion (lang algebra) = 𝜆𝑝 𝑥

ℚ 1𝑝

: ℚ = 𝑝 − 1

5) N=6

Let’s factor it over ℚ:

𝑥6 − 1 = 𝑥3 − 1 𝑥3 + 1 = 𝑥 − 1 𝑥2 + 𝑥 + 1 𝑥 + 1 𝑥2 − 𝑥 + 1

16

= 𝜌

𝜔 = 𝜌2

𝜔2 = 𝜌4

df

Roots areL

Roots (Accoringly) 1, 𝜔, 𝜔2 −1 𝜌, 𝜌5 = 𝜌

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

Theorem: ℚ 1𝑛

: 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

Denote 𝜉 = 1𝑛

Proof: ℚ 𝜉 : ℚ =degree of the minimal polynomial of 𝜉 over ℚ = deg 𝜆𝑛 𝑥

Note: 𝜉𝑘 is a primitive n’th root of 1 ⇔ gcd 𝑘, 𝑛 = 1

𝜉𝑘 𝜉𝑘 𝑝𝑟𝑖𝑚𝑒 𝑛′𝑡𝑕𝑟𝑜𝑜𝑡 𝑜𝑓 1 = 𝜑 𝑛

So in fact, 𝜆𝑛 𝑥 = 𝑥 − 𝜉𝑘 gcd 𝑘 ,𝑛 =11≤𝑘<𝑛

This is a key fact!

By gauss’ Lemma, 𝑥𝑛 − 1 factors over ℚ into polynomials in ℤ 𝑥

So in fact, as 𝜆𝑛 𝑥 |𝑥𝑛 − 1 over ℚ (since 𝜉 is a root of 𝑥𝑛 − 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 𝑥𝑛 − 1 = 0

So 𝜆𝑑 𝑥 |𝑥𝑛 − 1 over ℚ

Conclusion: 𝜆𝑑 𝑥 |𝑥𝑛 − 1 for all 𝑑|𝑛.

Conversely:

Suppose 𝑝 𝑥 is an irreducible monic factor of 𝑥𝑛 − 1 (in ℚ 𝑥 )

Any root 𝛼 of 𝑝 𝑥 is a root of 𝑥𝑛 − 1 and so 𝛼𝑛 = 1

If 𝑑 minimal such that 𝛼𝑑 = 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 𝑥𝑛 − 1 over ℤ is of the form 𝜆𝑑 𝑥 for some 𝑑|𝑛.

Conclusion: 𝑥𝑛 − 1 = 𝜆𝑑 𝑥 𝑑|𝑛 over ℚ. And 𝜆𝑑 𝑥 ∈ ℤ 𝑥

Example: 𝑥6 − 1 = 𝑥 − 1 =𝜆1 𝑥

𝑥 + 1 𝜆2 𝑥

𝑥2 + 𝑥 + 1 𝜆3 𝑥

𝑥2 − 𝑥 + 1 𝜆6 𝑥

Corollary from conclusion:

From degree of polynomials we get:

𝑛 = deg 𝜆𝑑 𝑥

𝑑|𝑛

= 𝜑 𝑑

𝑑|𝑛

Example:

𝑥12 − 1 = 𝑥6 + 1 𝑥6 − 1 =

𝑥2 + 1 𝜆4

𝑥4 − 𝑥2 + 1 𝜆2 𝑥

𝜉 ,𝜉11 ,𝜉5 ,𝜉7

𝑥 − 1 =𝜆1 𝑥

1

𝑥 + 1 𝜆2 𝑥 −1

𝑥2 + 𝑥 + 1 𝜆3 𝑥

𝜔 ,𝜔2

𝑥2 − 𝑥 + 1 𝜆6 𝑥

−𝜔 ,−𝜔2

𝜉 = 112

Galois grups of ℚ 𝒙𝒊 over ℚ, 𝝃 = 𝟏𝒏

Let 𝐺𝑎𝑙 ℚ 𝜉

ℚ = 𝐺

Elements of 𝐺 permute primitive roots of unity and are determine by the image of 𝜉.

So 𝐺 subroup of group of permutations 𝜉𝑘 gcd1≤k<𝑛 𝑘, 𝑛 = 1 i.e. of 𝑆𝜑 𝑛

Let gcd 𝑘, 𝑛 = 1:

𝜉𝜓𝑘 𝜉𝑘 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 𝜉𝑘 = 𝜉𝑘𝑙 = 𝜓𝑘𝑙 𝜉

𝜓𝑙𝜓𝑘 𝜉 = 𝜓𝑙 𝜉𝑘 = 𝜉𝑙𝑘

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.

𝜉 = 112

Note: 𝜉 → 𝜉11 is complex conjugation

Finite Fields If 𝐹 is finite then its characteristics must be some prime 𝑝

And its prime field ≅ ℤ𝑝ℤ .

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, 𝛼2 , 𝛼3 , … 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 𝑥2 + 𝑥 + 1 which is irreducible over ℤ 2ℤ

Extend 𝔽2 to get a field in which 𝑥2 + 𝑥 + 1 has a root.

𝑘 =𝔽2 𝑥

𝑥2 + 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 𝑥2 + 𝑥 + 1 i.e. linear polynomials.

0,1, 𝑥, 𝑥 + 1

+ 0 1 𝑥 𝑥 + 10 0 1 𝑥 𝑥 + 11 1 0 𝑥 + 1 𝑥𝑥 𝑥 𝑥 + 1 0 1

𝑥 + 1 𝑥 + 1 𝑥 1 0

∙ 0 1 𝑥 𝑥 + 10 0 0 0 01 0 1 𝑥 𝑥 + 1𝑥 0 𝑥 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: 𝑥2 + 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.

So 𝐹 ≅ 𝔽𝑝 𝑘

as a vector space and so 𝐹 = 𝑝𝑘

Example:

Look at 𝑥4 + 𝑥3 + 1 over 𝐺𝐹 2

Claim: 𝑥4 + 𝑥3 + 1 is irreducible over 𝐺𝐹 2

Clearly it has no roots.

If it factored as 2 irreducible quadratics then we would have 𝑥4 + 𝑥3 + 1 = 𝑥2 + 𝑥 + 1 2

But 𝑥2 + 𝑥 + 1 2 = 𝑥4 + 𝑥2 + 1

So 𝐺𝐹 2 𝑥 𝑥4 + 𝑥3 + 1

gives an extension of degree 4 and so a field of order 16!

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) 𝑥3 + 𝑥 + 𝑥2 + 𝑥 + 1 = 𝑥3 + 𝑥2 + 1

𝑎𝑥3 + 𝑏𝑥2 + 𝑐𝑥 + 𝑑 ↔

𝑎𝑏𝑐𝑑

Multiplication on the other hand, is harder 𝑥3 + 𝑥 ∙ 𝑥2 + 𝑥 + 1 = 𝑥5 + 𝑥3 + 𝑥4 + 𝑥2 + 𝑥3 + 1 = 𝑥5 + 𝑥4 + 𝑥2 + 𝑥

≡ 𝑥2(𝑚𝑜𝑑 𝑥4 + 𝑥3 + 1

1010

0111

=

0100

Another Notation Let 𝛼 = 𝑥 + 𝑥4 + 𝑥3 + 1 in 𝐹

So 𝛼 root of 𝑥4 + 𝑥3 + 1 in 𝐹. 𝛼4 + 𝛼3 + 1 = 0

1, 𝛼, 𝛼2 , 𝛼3 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

𝛼5 = 𝛼 𝛼3 + 1 = 𝛼4 + 𝛼 = 𝛼3 + 1 + 𝛼 = 𝛼3 + 𝛼 + 1 ≠ 1. Otherwise, 𝛼3 + 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, … , 𝛼14

This notation is convenient for multiplication:

𝛼𝑖 ∙ 𝛼𝑗 = 𝛼𝑖+𝑗 (𝑚𝑜𝑑 15)

(Addition - problematic!)

Note: Over 𝐹 𝑥4 + 𝑥3 + 1 factors into linear factors and so is a splitting field for this

polynomial over 𝔽2

Notice that: 𝛼4 + 𝛼3 + 1 = 0

(Over ℤ 𝑝ℤ : 𝑥 + 𝑦 𝑝 = 𝑥𝑝 + 𝑦𝑝 )

So 0 = 𝛼4 + 𝛼3 + 1 2 = 𝛼8 + 𝛼6 + 1 ⇒ 𝛼2 is a root of 𝑥4 + 𝑥3 + 1

𝛼8 + 𝛼6 + 1 2 = 𝛼16 + 𝛼12 + 1 ⇒ 𝛼4 is a root of 𝑥4 + 𝑥3 + 1

Same for 𝛼16 + 𝛼12 + 1 2 which leads to 𝛼8 is a root as well

So 𝑥4 + 𝑥3 + 1 = 𝑥 − 𝛼 𝑥 − 𝛼2 𝑥 − 𝛼4 𝑥 − 𝛼8

Theorem: The multiplicative group of a finite field is cyclic.

Proof: next lesson!

Note: If 𝐹 = 𝑞 then all its nonzero elements will satisfy 𝑥𝑞−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 𝑥𝑞−1

If we take 𝑥𝑞 − 𝑥 then every element of 𝐹 (including 0!) is a root and 𝐹 is the splitting field

of 𝑥𝑞 − 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 = 𝐶𝑝

𝑘𝑖1× 𝐶

𝑝𝑘𝑖2

× … × 𝐶𝑝

𝑘𝑖𝑟

Can assume 𝑘1 ≥ ⋯ ≥ 𝑘𝑟

𝐶𝑘 = cyclic of order 𝑘

So every element 𝑎 of 𝐻𝑖 satisfies 𝐴𝑝𝑘1 = 1

So every element of 𝐻𝑖 is a root of the polynomial 𝑥𝑝𝑘1 − 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∗ = 𝐶𝑝1

𝑘1 × … × 𝐶𝑝𝑠

𝑘𝑠

𝑝1 , … , 𝑝𝑠 are distinct primes!

So 𝐹∗ is cyclic generated by the product of the generators of 𝐶𝑝1

𝑘1 , … , 𝐶𝑝𝑠

𝑘𝑠 .

Corollary: If 𝐹 is a finite field of order 𝑞. Then it is the splitting field of 𝑥𝑞 − 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 𝑥𝑞−1 − 1 and so together with 0 all the elements

of 𝐹 are roots of 𝑥𝑞−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 ∈ ℤ𝑝ℤ 𝑥

So 𝑚 𝑥 |𝑥𝑞 − 𝑥

𝐹′ is also a splitting field of 𝑥𝑞 − 𝑥 over ℤ 𝑝ℤ .

So 𝑚 𝑥 has a root 𝛽 in 𝐹′ .

We map 𝛼𝑖 to 𝛽𝑖 ∀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 𝛽𝑟 = 1 for 𝑟 < 𝑞 − 1.

Then 𝛽 is a root of 𝑥𝑟 − 1 in 𝐹′ .

𝑚 𝑥 is the minimal polynomial of 𝛽 so that 𝑚 𝑥 |𝑥𝑟 − 1 over ℤ 𝑝ℤ

So that 𝛼𝑟 = 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 𝛼𝑖 + 𝛼𝑗 = 𝛼𝑘 then need to show 𝛽𝑟 + 𝛽𝑠 = 𝛽𝑡

b) If 𝛼𝑖 + 𝛼𝑗 = 0 then need to show 𝛽𝑟 + 𝛽𝑠 = 0

We shall show (a):

𝛼𝑖 + 𝛼𝑗 = 𝛼𝑘 implies 𝛼 is a root of 𝑥𝑟 + 𝑥𝑠 − 𝑥𝑡 so 𝑚 𝑥 |𝑥𝑟 + 𝑥𝑠 − 𝑥𝑡

So then 𝛽 root of 𝑥𝑟 + 𝑥𝑠 − 𝑥𝑡 and so 𝛽𝑟 + 𝛽𝑠 = 𝛽𝑡 .

Note: It also follows that the roots of 𝑥𝑞 − 𝑥 over ℤ 𝑝ℤ are distinct.

Theorem: For any prime 𝑝 and 1 ≤ 𝑘 ≤ ℕ there exists a field of order 𝑝𝑘 .

Proof: Take ℤ 𝑝ℤ and extend to a splitting field for 𝑥𝑝𝑘− 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))

ℤ𝑝ℤ 𝛼 = 𝐹 and ℤ 𝑝ℤ 𝛼 is a vector space of dimension 𝑙 over ℤ 𝑝ℤ where 𝑙 is the degree

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 𝑥4 + 𝑥3 + 1 over 𝐺𝐹 2 .

Every element in this field is a root of 𝑥16 − 𝑥.

So 𝑥4 + 𝑥3 + 1|𝑥16 − 𝑥 over 𝐺𝐹 2 .

Roots of 𝑥4 + 𝑥3 + 1 in 𝐺𝐹 16 were: 𝛼, 𝛼2 , 𝛼4 , 𝛼16

0 root of 𝑥. (so 𝑥|𝑥16 − 𝑥)

1 root of 𝑥 + 1 (so 𝑥 + 1|𝑥16 − 𝑥)

𝑥16 − 𝑥 = 𝑥 𝑥 + 1 𝑥4 + 𝑥3 + 1 ∙ 𝑕 𝑥 , 𝑕 𝑥 ∈ 𝐺𝐹 2 𝑥 of degree 10.We want to factor

𝑕 𝑥

Definition:

Let 𝑓 𝑥 = polynomial of degree 𝑛.

The reciprocal of 𝑓 𝑥 is 𝑔 𝑥 = 𝑥𝑚𝑓 𝑥−1

Example:

𝑓 𝑥 = 𝑥5 − 2𝑥4 + 3𝑥2 − 7𝑥 + 19

𝑥5𝑓 𝑥−1 = 𝑥5 𝑥−5 − 2𝑥−4 + 3𝑥−2 − 7𝑥−1 + 19 = 1 − 2𝑥 + 3𝑥2 − 7𝑥4 + 19𝑥5

Use question 4 in assignment 4 to get the reciprocal of 𝑥4 + 𝑥3 + 1:

𝑥4 + 𝑥 + 1

So 𝑥4 + 𝑥 + 1 is irreducible and 𝛼−1 =𝛼14

is a root and also 𝛼−2 = 𝛼13 , 𝛼−4 = 𝛼11 , 𝛼−8 = 𝛼7.

We conclude that 𝑥4 + 𝑥 + 1|𝑥16 − 𝑥

So 𝑕 𝑥 has 𝑥4 + 𝑥 + 1 as an irreducible factor over 𝐺𝐹 2

Note also: 𝑥5 − 1|𝑥15 − 1. Since 𝑥5 − 1 𝑥10 + 𝑥5 + 1 = 𝑥15 − 1.

Over 𝐹𝐺 2 we have 𝑥5 − 1 = 𝑥 + 1 𝑥4 + 𝑥3 + 𝑥2 + 𝑥 + 1

So 𝑥4 + 𝑥3 + 𝑥2 + 𝑥 + 1|𝑥16 − 𝑥 and is irreducible (question 1 in assignment 4).

Note also: 1, 𝛼5 , 𝛼10 are roots of 𝑥3 − 1 in 𝐺𝐹 16 : 𝛼3 , 𝛼6 , 𝛼12 , 𝛼24 = 𝛼9

𝑥3 − 1 factors to: 𝑥 − 1 𝑥2 + 𝑥 + 1

So 𝑥2 + 𝑥 + 1 is the minimal polynomial of 𝛼5, 𝛼10 .

So over 𝐺𝐹 2 :

𝑥16 − 𝑥 = 𝑥 𝑥 − 1 𝑥2 + 𝑥 + 1 𝑥4 + 𝑥3 + 1 𝑥4 + 𝑥 + 1 𝑥4 + 𝑥3 + 𝑥2 + 𝑥 + 1

Roots (in the appropriate order of the factors):

0, 1, 𝛼5 , 𝛼10 , 𝛼 , 𝛼2 , 𝛼4 , 𝛼8 , 𝛼14 , 𝛼13 , 𝛼11 , 𝛼7 , 𝛼3 , 𝛼6 , 𝛼9, 𝛼12

Note: 𝛼, 𝛼−1 = 𝛼14 are primitives elements (i.e. generators of 𝐺𝐹 16 ∗ but the roots of

𝑥4 + 𝑥3 + 𝑥2 + 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 𝛼3 (but not a power of 𝛼3!)

Every element of 𝐺𝐹 𝑝𝑘 satisfies 𝑥𝑝𝑘−1 = 1.

If 𝑥𝑛 − 1 has a root in 𝐺𝐹 𝑝𝑘 .

Must have 𝑛|𝑝𝑘 − 1

Can see which are the subfields of 𝐺𝐹 16 by looking at the factorization of 𝑥16 − 𝑥.

Possible subfields (are of order 2𝑚 , 𝑚 ≤ 4):

𝐺𝐹 2 - prime field and so a subfield!

𝐺𝐹 4 – 0,1, 𝛼5 , 𝛼10 as 𝐺𝐹 4 splitting field of 𝑥2 + 𝑥 + 1

𝐺𝐹 8 - Don’t have any irreducible polynomials of degree 3 dividing 𝑥16 − 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

𝑥𝑛 − 𝑥 over 𝐺𝐹(2)

- What are the subfields of a given finite field 𝐺𝐹 𝑞 , 𝑞 = 𝑝𝑥 , 𝑝 𝑝𝑟𝑖𝑚𝑒.

Lemma: 𝑥𝑚 − 1|𝑥𝑛 − 1 ⇔ 𝑚|𝑛

Proof: Divide = 𝑥𝑛 − 1 by 𝑥𝑚 − 1 with remainder (over ℤ):

𝑥𝑛 − 1 = 𝑥𝑚 − 1 𝑥𝑛−𝑚 + 𝑥𝑛−2𝑚 + 𝑥𝑛−3𝑚 + ⋯ + 𝑥𝑛−𝑘𝑚 + 𝑥𝑛−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 𝑥𝑚 − 1|𝑥𝑛 − 1

So in particular setting 𝑥 = 𝑝 we get 𝑝𝑚 − 1|𝑝𝑛 − 1

Using the lemma again, we get that 𝑥𝑝𝑚 −1 − 1|𝑥𝑝𝑛−1 − 1

So all the roots of 𝑥𝑝𝑚 −1 − 1 are contained in 𝐺𝐹 𝑝𝑛 ∗ (which is the set of roots of

𝑥𝑝𝑛−1 − 1)

Meaning 𝐺𝐹 𝑝𝑚 ∗ ⊆ 𝐺𝐹 𝑝𝑛 ∗ so 𝐺𝐹 𝑝𝑚 ⊆ 𝐺𝐹 𝑝𝑛

Now assume 𝐺𝐹 𝑝𝑚 𝐿

⊆ 𝐺𝐹 𝑝𝑛 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:

𝑥16 − 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

Then 𝜑𝑚 𝑎 = 𝑎𝑝𝑚

So set if fixed points under

𝜑𝑚 = 𝑎 𝜑𝑚 𝑎 = 𝑎, 𝑎 ∈ 𝐺𝐹 𝑝𝑛 = 𝑎 ∈ 𝐺𝐹 𝑝𝑛 𝑎𝑝𝑚= 𝑎 =

𝑎 ∈ 𝐺𝐹 𝑝𝑛 ∗ 𝑎𝑝𝑛−11 = 0 ∪ 0 = set of roots of 𝑥𝑝𝑛− 𝑥 in 𝐺𝐹 𝑝𝑛

Note: If 𝐹 finite field 𝐹 = 𝑝𝑛 and we look at roots of 𝑥𝑘 − 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 𝑥𝑘 − 1 are only for 𝑘|𝑝𝑛 − 1

(as if gcd 𝑘, 𝑝𝑛 − 1 = 1 only roots will be 1: 𝑥𝑘 − 1 = 𝑥 − 1 𝑥𝑘−1 + ⋯ )

In general, we want to factor 𝑥𝑝𝑛− 𝑥 or 𝑥𝑝𝑛−1 − 1 over 𝐺𝐹 𝑝 .

Theorem: over ℤ 𝑝ℤ = 𝐺𝐹 𝑝 𝑥𝑝𝑛− 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:

𝑥16 − 𝑥 = 𝑥 𝑥 + 1 𝑖𝑟𝑟𝑒𝑑𝑢𝑐𝑖𝑏𝑙𝑒𝑜𝑓 𝑑𝑒𝑔𝑟𝑒𝑒 1

𝑥2 + 𝑥 + 1 𝑖𝑟𝑟𝑒𝑑𝑢𝑐𝑖𝑏𝑙𝑒𝑜𝑟 𝑑𝑒𝑔𝑟𝑒𝑒 2

𝑥4 + 𝑥3 + 1 𝑥4 + 𝑥 + 1 𝑥4 + 𝑥3 + 𝑥2 + 𝑥 + 1 𝑎𝑙𝑙 𝑖𝑟𝑟𝑒𝑑𝑢𝑐𝑖𝑏𝑙𝑒𝑠

𝑜𝑓 𝑑𝑒𝑟𝑒𝑒 4

Proof: Suppose 𝑓 𝑥 ∈ ℤ𝑝ℤ 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 .

And so satisfies 𝛼𝑝𝑛= 𝛼. If 𝛼 = 0, 𝑓 𝑥 = 𝑥 and 𝑥|𝑥𝑝𝑛

− 𝑥!

Otherwise 𝛼 ≠ 0, 𝛼𝑝𝑛−1 − 1 = 0 so 𝛼 root of 𝑥𝑝𝑛−1 − 1

And so its minimal polynomial 𝑓 𝑥 divides 𝑥𝑝𝑛−1 and so 𝑥𝑝𝑛− 𝑥.

Conversely: Suppose now 𝑓 𝑥 |𝑥𝑝𝑛− 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 .

So 𝛼 is also a root of 𝑥𝑝𝑛− 𝑥.

And so 𝐺𝐹 𝑝𝑚 = 𝐺𝐹 𝑝 𝛼

In other words, every element of 𝐺𝐹 𝑝𝑚 is a polynomial in 𝛼.

𝛼 is also a root of 𝑋𝑝𝑛− 𝑥 as 𝑓 𝑥 |𝑥𝑝𝑛

− 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 0 0 0 1 1𝑣2 0 1 0 0 1 0 1𝑣3 0 0 1 0 1 1 0𝑣4 0 0 0 1 1 1 1

Suppose we want to transmit 1101?

Send instead 𝑣1 + 𝑣2 + 𝑣4 = 1101001

Big advantage: Efficient decoding and locates ≤ 1 errors.

Use an analog to inner product/scalar multiplication. Induced by matrix multiplication over

𝐺𝐹 2 .

𝑥1 … 𝑥𝑛

𝑦1

⋮𝑦𝑛

= 𝑥𝑖𝑦𝑖(𝑚𝑜𝑑 2)

7

𝑖=1

It is a bilinear form on 𝐺𝐹 2 4.

Decoding:

Suppose we receive 𝑦∗ = 1 1 0 1 1 1 0

𝑣1 + 𝑣2 = 𝑦 = 1 1 0 0 1 1 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 10 1 0 0 1 0 10 0 1 0 1 1 00 0 0 1 1 1 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 𝑎1𝑣1 + 𝑎2𝑣2 , 𝑤 = 𝑎1𝐵 𝑣1 , 𝑤 + 𝑎2 + 𝐵 𝑣2 , 𝑤

𝐵 𝑣, 𝑎1𝑤2 + 𝑎2𝑤2 = 𝑎1𝐵 𝑣, 𝑤 + 𝑎2 𝑣, 𝑤2

And for any subspace 𝑊 of 𝑉 we can define

𝑊⊥ 𝑂𝑟𝑡𝑕𝑜𝑔𝑜𝑛𝑎𝑙𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑜𝑓 𝑊 𝑤𝑟𝑡 𝐵

= 𝑤 ∈ 𝑉 𝐵 𝑢, 𝑤 = 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑤 ∈ 𝑊

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 = 𝑣𝑇 ∙ 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 1 0 0 0 0 0 is orthogonal

to itself!

E.g.

If 𝑊 = 𝑠𝑝𝑎𝑛 1 1 0 0 0 0 0 then 𝑊 ⊊ 𝑊⊥

e.g.

0 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 00 0 0 1 0 0 00 0 0 0 1 0 00 0 0 0 0 1 00 0 0 1 0 0 01 1 0 0 0 0 0

--- end of lesson

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, … , 𝛼14 where 𝛼 is the root of 𝑥4 + 𝑥3 + 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

𝐻 = 𝑏1 𝑏2 … 𝑏15

𝑐1 𝑐2 … 𝑐15

Where 𝑏𝑖 , 𝑐𝑖 ∈ 𝐺𝐹 2 4 row vectors.

We think of also as elements of 𝐺𝐹 16 .

Take 𝑏𝑖 = vector of length 4 corresponding to 𝛼𝑖−1 in the table.

So we have 1, 𝛼, … , 𝛼14 in the top half of the matrix.

𝑐𝑖 ’s will be defined later…

We want: If 𝑥 = 𝑥1 … 𝑥15 codeword, we want:

(1) 𝐻 ∙ 𝑥𝑇 = 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:

𝐻 ∙ 𝑥 = 𝐻𝑥 + 𝐻𝑒𝑖 + 𝐻𝑒𝑗 = 𝐻𝑒𝑖 + 𝐻𝑒𝑗 = 𝑏𝑖 + 𝑏𝑗

𝑐𝑖 + 𝑐𝑗

So we want to choose the 𝑐𝑖 ’s so we can recover from this vector.

Bad choice: 𝑐𝑖 = 𝑏𝑖 . Get 𝐻𝑥 = 𝑏𝑏 - in this case we cannot recover 𝑖 and 𝑗.

If 𝑏 =

1001

. We could have had:

0001

𝑏1

+

1000

𝑏4

But also:

0111

𝑏8

+

1110

𝑏0

And a lot of other

possibilities.

Another bad choice: define 𝑐𝑖 = 𝑏𝑖 2 (thinking of 𝑏𝑖 as an element of 𝐺𝐹 16 so that 𝑐𝑖

corresponding to 𝛼2𝑖−2

So we should then get:

𝐻𝑥 = 𝑏𝑖 + 𝑏𝑗

𝑏𝑖2 + 𝑏𝑗

2 = 𝑏𝑖 + 𝑏𝑗

𝑏𝑖 + 𝑏𝑗 2 =

𝑏𝑏2

If you square you get the same thing….

Definition: Take 𝑐𝑖 = 𝑏𝑖3.

𝑏𝑖 + 𝑏𝑗

𝑏𝑖3 + 𝑏𝑗

3 = 𝑏𝑐 want to show 𝑖 and 𝑗 determined uniquely and how to find them.

𝑐 = 𝑏𝑖3 + 𝑏𝑗

3 = 𝑏𝑖 + 𝑏𝑗 𝑏𝑖2 + 𝑏𝑖𝑏𝑗 + 𝑏𝑗

2 = 𝑏 𝑏𝑖2 + 𝑏𝑖𝑏𝑗 + 𝑏𝑗

2 = 𝑏 𝑏2 + 𝑏𝑖𝑏𝑗

(regarding the elements of 𝐺𝐹 16 )

We first assume we have exactly 2 errors. So 𝑖 ≠ 𝑗 and 𝑏 ≠ 0.

Get 𝑐𝑏−1 + 𝑏2 = 𝑏𝑖𝑏𝑗

So 𝑏𝑖 and 𝑏𝑗 are roots in 𝐺𝐹 16 of the quadratic equation:

𝑥 − 𝑏𝑖 𝑥 − 𝑏𝑗 = 𝑥2 − 𝑏𝑖 + 𝑏𝑗 𝑥 + 𝑏𝑖𝑏𝑗 = 𝑥2 − 𝑏𝑥 + 𝑐𝑏−1 + 𝑏2

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.

𝐻′ = 1 𝛼 𝛼2 … 𝛼14

1 𝛼3 𝛼6 … 𝛼12

Suppose 𝑦 is a received message with errors in positions 𝑖 and 𝑗.

And suppose 𝐻′𝑦 = 𝛼𝑖−1 + 𝛼𝑗−1

𝛼3𝑖−3 + 𝛼3𝑗−3 = 𝛼5

𝛼7

Equivalently: 𝐻 ∙ 𝑦 =

10110111

polynomial will be: 𝑥2 + 𝛼5𝑥 + 𝛼8

Since: 𝑐𝑏−1 + 𝑏2 = 𝛼7 ∙ 𝛼−5 + 𝛼10 = 𝛼2 + 𝛼10 = 𝛼3

Need 𝑖 and 𝑗 such that: 𝛼𝑖−1 + 𝛼𝑗−1 = 𝛼5 and 𝛼𝑖−1 ∙ 𝛼𝑗−1 = 𝛼8

𝑖 + 𝑗 − 2 ≡ 8 𝑚𝑜𝑑 15

𝑖 + 𝑗 ≡ 10 𝑚𝑜𝑑 15

Checking possibilities: Get only 𝑖 = 3, 𝑗 = 7 satisfies 𝛼𝑖−1 + 𝛼𝑗−1 = 𝛼5 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:

It is the only case where we get a vector of the form: 𝑏𝑏3 and then determine 𝑏𝑖 = 𝑏 by

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.

Suppose 𝑎𝑖 𝑏𝑖

𝑏𝑖3

8𝑖=1 =

00 and 𝑎𝑖 ∈ 𝐺𝐹 2

Then we also get 𝑎𝑖 𝛼𝑖−1

𝛼3𝑖−3 8𝑖=1 = 0 ⇒ 𝑎𝑖+1

𝛼𝑖

𝛼3𝑖 7𝑖=0 = 0 ⇔

𝑎𝑖+1𝛼𝑖7

𝑖=0 = 0 and 𝑎𝑖+1𝛼3𝑖7

𝑖=0 = 0

Look at the polynomial 𝑎𝑖+1𝑥𝑖7

𝑖=0 = 0 over 𝐺𝐹 2 And 𝛼 and 𝛼3 are both roots.

So their minimal polynomials both divide 𝑎𝑖+1𝑥𝑖7

𝑖=0

𝑥4 + 𝑥3 + 1, 𝑥4 + 𝑥3 + 𝑥2 + 𝑥 + 1| 𝑎𝑖+1𝑥𝑖

7

𝑖=0

The product 𝑥4 + 𝑥3 + 1 𝑥4 + 𝑥3 + 𝑥2 + 𝑥 + 1 which is a polynomial of degree 8

divides 𝑎𝑖+1𝑥𝑖7

𝑖=0 which is of degree less or equal to 7! So 𝑎𝑖+1𝑥𝑖7

𝑖=0 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.

Take 11 ,

𝛼𝛼3 , … ,

𝛼7

𝛼21 first 8 columns of 𝐻′ .

The 9’th column 𝛼8

𝛼24 is a linear combination of the first 8 columns: 𝑠𝑖 𝛼𝑖

𝛼3𝑖 7𝑖=0

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.

Similarly, column 10: 𝛼9

𝛼27 = linear combination of 8 columns of 𝐻′ .

𝑡0 11 + ⋯ + 𝑡7

𝛼7

𝛼21 = 𝛼9

𝛼27

So

𝑡0 11 + ⋯ + 𝑡7

𝛼7

𝛼21 + 𝛼9

𝛼27 = 00

So take the vector 𝑡0 … 𝑡7 0 1 0 … 0 orthogonal to rows of 𝐻′ take to be row

2 of 𝐶 etc.