So, why are they called ``ideals'' anyway? · Diophantine equations De nition: Diophantine equation...

Origins ofIdeal Theory

R. Garrett

Notation andprerequisitereview

Algebraicnumbers andnumbersystems

SolvingDiophantineequations withZ[α] andfriends

Uniquefactorizationand Kummer’sideal primes

Dedekind’s fixfor Kummer:Ideals

So, why are they called “ideals” anyway?

Reeve Garrett

The Ohio State University

May 17, 2019


R. Garrett






Notation

Our notation will be pretty standard.

* Z = {0, 1,−1, 2,−2, ...} will denote the set of all integers;

* Q will denote the set of all fractions of integers;

* R will denote the set of all real numbers;

* C will denote the set of all complex numbers - that is, allnumbers of the form a + bi , where i =

√−1 and a and b

are real numbers (so R is the subset of C consisting of alla + bi where b = 0);

* Z[x ] will denote the set of all polynomials in the variable xwith integer coefficients - in most future settings, we willuse a specially chosen complex number α instead of avariable. Similarly define Z[x , y ], etc.

* e ≈ 2.71828... is the standard Euler’s constante = 1

1 + 11 + 1

2·1 + 13·2·1 + 1

4·3·2·1 + · · ·


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Diophantine equations

Definition: Diophantine equation

A Diophantine equation is an equation involving polynomialsin 2 or more variables with integer coefficients for which onlyinteger solutions are sought.

Example 1:

x2 + y2 = 1

(the unit circle) may be thought of as a Diophantine equation,and its only integer solutions are (−1, 0), (0, 1), (1, 0), and(0,−1).


R. Garrett






Examples of Diophantine equations

Example 2: Bachet’s equation when k = 2: x2 + 2 = y3

(Generally: x2 + k = y3) We will prove later with clever tricksthat (5, 3) and (−5, 3) are the ONLY solutions to x2 + 2 = y3.

Example 3: The equation from Fermat’s Last Theorem

xn + yn = zn.

When n = 2, you know there are infinitely many solutions,namely in the form of (3, 4, 5), (5, 12, 13), (8, 15, 17),(7, 24, 25), etc. and any triple that is a multiple of a triple youalready know works (e.g. (6, 8, 10) is double (3, 4, 5), so itworks, etc.) When n ≥ 3, there are NO positive integersolutions.


R. Garrett






Primitive roots of unity and DeMoivre’s Theorem

Definition/Notation: Primitive root of unity, ζn

ζn will denote a primitive nth root of unity, i.e. ζn is acomplex number such that ζnn = 1 but for any k < n, ζkn 6= 1.

Examples:

−1 is a primitive 2nd root of unity, and i is a primitive 4th rootof unity (the other being −i), while −1 is NOT a primitive 4th

root of unity despite the fact (−1)4 = 1.

DeMoivre’s Theorem

cos

(2π

n

)+ i sin

(2π

n

)is always primitive nth root of unity, regardless of n.


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Euler’s formula and consequences

Euler’s formula

e iθ = cos(θ) + i sin(θ), for any real number θ.

Consequence of Euler’s formula:

We can use the “neater” form of our primitive nth root of unity

ζn = e i2πn in place of cos

(2πn

)+ i sin

(2πn

), since they’re the

same thing. This gives us the form of ζkn easily: ζkn = e i2kπn .


R. Garrett






Algebraic complex numbers

Definition:

A complex number α is algebraic over the integers Z if it isthe root of a polynomial with integer coefficients; that is, forsuch an α, there is some polynomialf (x) = a0 + a1x + a2x

2 + · · ·+ anxn such that

f (α) = a0 + a1α + a2α2 + · · · anαn = 0.

- Example 1:√

2 is algebraic over Z since it is a root of thepolynomial x2 − 2, which has integer coefficients.

- Example 2: ζn is algebraic over Z since it is a root of thepolynomial xn − 1.

- Example 3: Less trivially, α = 3√

2e iπ3 is algebraic, as it is

a root of the polynomial x3 + 2:(3√

2e iπ/3)3

= (21/3)3e i ·π3·3 = 2e iπ = 2(−1) = −2.

(There are far more complicated examples I’m not discussing)


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Why talk about algebraic numbers anyway?

You can find all solutions to some Diophantine equations bybuilding new number systems containing the integers Z andspecially chosen complex numbers that are algebraic over Z;specifically, for a given equation, we pass to the smallestnumber system that makes each side of an equation factorcompletely. Sometimes, with those new number systems, you’reeven able to prove that the ONLY possible solutions are of theform you found by finding solutions in those number systems!


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What do I mean by a “number system”?

We want to formulate a definition for “number system” thatincludes the structures we know are number systems: Z, Q, R,C, so a number system should have most of the properties allof these have:

* We have 0, and for any number a in the number system,an additive inverse −a is in there too. We also have 1,but we don’t always have multiplicative inverses(reciprocals 1

a ) for a 6= 0: 2 is an integer, but 12 isn’t. We

don’t have that problem in Q, R, and C, though.

* Given a and b in the number system, a + b, a− b, anda · b stay in the number system (again, we don’t want tosay the same for division since dividing two integersdoesn’t always get you an integer back.)


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What do I mean by a “number system”? (ctd)

* We also want some other nice properties, but they comefor free because we’re considering real and complexnumbers:

(1) associativity of + and × (i.e. a + (b + c) = (a + b) + c),(2) commutativity of + and × (i.e. a× b = b × a),(3) distributive property of × over + (i.e.

a(b + c) = ab + ac)),(4) if a 6= 0 and b 6= 0, then ab 6= 0.

If the properties listed on the previous slide and (1)-(3) on thisslide hold for ANY collection of objects with addition andmultiplication, we call the collection/structure a commutativering, and if (4) holds too, we call the collection/structure anintegral domain.


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How to build a new number systemwith Z and algebraic α

What we need:

The new number system has to include α, −α, every integer,all possible sums and products you can form with them, andthe corresponding negative numbers. Thus, we must include α,α2, and all other powers αi of α, multiples of any such αi andany chosen integer bi and sums of these integer multiples ofpowers of α.

The result: Z[α]

In building a new number system with Z and α, we getsomething that looks a lot like polynomials in the “variable” α:Our new ring is Z[α], the collection of all complex numbers ofthe form b0 +b1α+b2α

2 +b3α3 + · · ·+bmα

m for some m ≥ 0.


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Why the name “ring”?

Among the first important examples of commutative rings were

the rings Z[ζn], where ζn = e i2πn is a primitive nth root of unity.

Because of Euler’s formula and DeMoivre’s Theorem, we knowthat taking powers of ζn creates a cycle:

ζ0n = 1, ζn, ζ

2n , ζ

3n , ... , ζ

n−1n , ζnn = 1, ζn+1

n = ζnn ζn = ζn, etc.


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Why the name “ring”? (ctd)

If n is large and we connect the points ζkn in the complex plane,the picture looks a lot like a ring in the vernacular sense.


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Divisibility

Definition

A number β in Z[α] divides a number δ in Z[α] if there is anumber γ also in Z[α] such that βγ = δ.

Example

1 +√−5 divides 6 in Z[

√−5] since 1−

√−5 is

also in Z[√−5] and 6 = (1 +

√−5)(1−

√−5)


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Units

Definition

A number γ in Z[α] is a unit if there is a number γ′ also inZ[α] such that γ · γ′ = 1.

Example 1:

i is a unit in Z[i ] sincei · (−i) = −i2 = −(−1) = 1;

Example 2:

ζ5 = e i2π5 is a unit in Z[ζ5] since ζ5 · ζ4

5 = ζ55 = 1.


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Irreducibility and primeness

Definition:

A number β in Z[α] is irreducible if the only numbers thatdivide β are units, β itself, and unit multiples of β.

Definition:

A number β in Z[α] is prime if whenever β divides a givenproduct γδ in Z[α], either β divides γ or β divides δ.

Example: Factorization of 6 in Z[√−5]

6 = 2 · 3 = (1 +√−5)(1−

√−5).

All factors in this equation are irreducible, but none of themare prime since any factor on one side of the equation dividesneither factor on the other side of the equation.


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Okay, but how does building these Z[α] help ussolve Diophantine equations?

The utility of using number systems like Z[α] to solveDiophantine equations is best seen with a few examples.


R. Garrett






An example: Bachet’s equation x2 + 2 = y 3

Consider Bachet’s equation above. We know that the roots ofthe polynomial x2 + 2 are ±

√−2. So, let’s consider this

equation not as a polynomial equation with integer coefficientsbut as a polynomial equation with coefficients in Z[

√−2]. By

doing so, x2 + 2 is no longer irreducible: it factors asx2 + 2 = (x +

√−2)(x −

√−2). Then, our equation becomes

(x +√−2)(x −

√−2) = y3.

The factors on the left hand side are relatively prime, i.e. theyhave no irreducible factors in common, since any commonirreducible factor would have to divide their sum 2x anddifference 2

√−2, but x and

√−2 (an irreducible factor of 2)

are irreducible and relatively prime!


R. Garrett






Bachet’s equation example (ctd)

Thus, in the equation

(x +√−2)(x −

√−2) = y3,

we have a cubed factor on the right hand side and two relativelyfactors on the left hand side, so each factor on the left handside must therefore be a cube! In particular, we have that

x +√−2 = (a + b

√−2)3,

where a and b are integers. Recall that by expanding (“FOIL”method or binomial theorem) we have

(a + b√−2)3 = a3 + 3a2b

√−2− 6ab2 − 2b3

√−2


R. Garrett







Thus,

x +√−2 = a3 + 3a2b

√−2− 6ab2 − 2b3

√−2,

and since x is an integer (real), by equating real and imaginaryparts we get

x = a3 − 6ab2 and 3a2b − 2b3 = b(3a2 − 2b2) = 1.

The 2nd equation factors 1 into a product of integers, sob = ±1.

Case 1: b = 1. If b = 1, 3a2−2b2 = 1, so we have 3a2−2 = 1and a = ±1.Case 2: b = −1. If b = −1, 3a2 − 2b2 = −1, so we have3a2 − 2 = −1, yielding 3a2 = 1. No integer a solves this, sob = −1 is impossible.


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Case 1(a): a = b = 1.

In this case,

x = a3 − 6ab2 = 13 − 6(1)(1) = 1− 6 = −5,

and plugging into the original equation x2 + 2 = y3, we gety3 = 27, so then y = 3.

Case 1(b): a = −1 and b = 1.

In this case,

x = a3 − 6ab2 = (−1)3 − 6(−1)(1)2 = −1 + 6 = 5,

and plugging into the original equation x2 + 2 = y3, we getthat y3 = 27 and y = 3.


R. Garrett






What we just did

We considered the equation x2 + 2 = y3 not as an equationwith coefficients in Z but as an equation with coefficients inZ[√−2], for which we very subtly (and correctly) assumed the

definitions of prime and irreducible coincide (more on thatlater). In doing so, we not only found the solutions(x , y) = (±5, 3), but we proved they’re the ONLY integersolutions.


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Fermat’s Last Theorem (abbrev. FLT)

Fermat’s Last Theorem (formulated 1637, proven 1994)

The equation xn + yn = zn

has no integer triple solutions (x , y , z) when n is ANY positiveinteger greater than 2.

Fact

It’s sufficient to prove FLT for (1) n a prime number p ≥ 3 or(2) n = 4 (proven by Fermat); all other cases follow from thesetwo since any n ≥ 3 is either a multiple of 4 or divisible by aprime p ≥ 3. Indeed, if n = km where FLT was already provedfor n = k , we may rewrite xn + yn = zn as(xm)k + (ym)k = (zm)k . So, FLT for k = 4 gets us FLT forany n divisible by 4, and FLT for a prime p ≥ 3 gets us FLT forany n divisible by p.


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Lame’s false proof of FLT (sketch)

In 1847, Lame gave a false proof that FLT holds for any (odd)prime p ≥ 3 by considering the equation xp + yp = zp not asan equation with coefficients in Z but as an equation withcoefficients in Z[ζp], where ζp = e i(2π/p) is a primitive pth rootof unity. In that case, he assumed for sake of contradiction thathe had a positive integer triple (x , y , z) satisfying xp + yp = zp,where z is chosen to be the smallest positive integer for whichsuch a triple (x , y , z) exists. Factoring in Z[ζp], we get

zp = xp + yp = (x + y)(x + yζp)(x + yζ2p) · · · (x + yζp−1

p ).


R. Garrett






Lame’s false proof of FLT (sketch, ctd)

Thus, we have written xp + yp as a product of p distinct andrelatively prime factors, while zp itself is a product of p of thesame factor, z . This means that each such factor (x + yζ ip) for

any choice of i , 0 ≤ i ≤ p − 1, must be a pth power of aproduct of irreducible and prime factors:

x + yζ ip = (βmi,1

i ,1 · · ·βmi,nii ,ni

)p,

where each mi ,j (1 ≤ j ≤ ni ) is a positive integer and each βi ,jis in Z[ζp]; that is, each βi ,j is of the form

βi ,j = bi ,j ,0 + bi ,j ,1ζp + bi ,j ,2ζ2p + · · ·+ bi ,j ,p−1ζ

p−1p ,

where each bi ,j ,k (0 ≤ k ≤ p − 1) is an integer.


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Lame’s false proof of FLT (sketch, ctd)

From there, we plug the given form for each βi ,j into theequation for x + yζ ip in terms of them and equate coefficientsin a similar (but much messier!) way to how we did withBachet’s equation to find all valid choices of x , y and z . Whatends up happening, however, is that we’re able to find anothertriple (x , y , z) solving xp + yp = zp where z is strictly smallerthan the z we chose to be minimal! A contradiction thenoccurs, so no such triple (x , y , z) can exist.


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The problem with Lame’s proof

The notions of “prime” and “irreducible” don’t always coincidedepending on the choice of ζp when Z[ζp] is built!

Kummer’s counterexample:

Consider Z[ζ23]. Ernst Kummer showed that 1− ζ23 + ζ2123 is

irreducible but not prime in Z[ζ23] by means of norms, atechnical technique used in field/Galois theory we’ll discussmomentarily.


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Fundamental Theorem of Arithmetic and UniqueFactorization

The Fundamental Theorem of Arithmetic

Every integer can be written as a unique product of powers ofprime integers; that is, Z exhibits unique factorization.

The notions of prime and irreducible coincide when we haveunique factorization but become distinct when uniquefactorization fails. Lame assumed that unique factorizationholds in Z[ζp] for any choice of p in his proof, but that’s notalways true, as Kummer showed when he came up with theirreducible but non-prime counterexample 1− ζ23 + ζ21

23 inZ[ζ23] given on the previous slide!


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Kummer’s question to save Lame’s proof

Can Lame’s proof be “saved” by finding some other context inwhich a given Z[ζp] exhibits unique factorization if we’reallowed to throw in “ideal prime factors” not actually in Z[ζp]that divide the given factors which we found were irreduciblebut not prime?


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Properties of ideal prime factors

In order to ensure that ideal prime factors behave in a waythat’s analagous to “true” prime factors, Kummer noted thatthe following properties needed to hold:

1. each α and β in Z[ζp] would together have the same idealprime factors as αβ (counting multiplicity, the number oftimes a number or ideal prime factor P divides a numberq);

- Analogue for Z: 72, as the product of 8 and 9, has all thesame prime factors collectively as 8 and 9: 2 and 3.

2. each α in Z[ζp] is divisible n times by a prime q in Z ifand only if it’s divisible nµP times by each of the idealprime factors P of q, where µP is the number of times Pdivides q;

- Analogue for Z: α = 144 is divisible by n = 2 times byq = 12. P1 = 2 divides 12 µP1 = 2 times and P2 = 3divides 12 µP2 = 1 time, so P1 = 2 divides α = 144nµP1 = 2× 2 = 4 times and P2 = 3 divides α = 144nµP2 = 2× 1 = 2 times.


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Properties of ideal prime factors (ctd)

3. each α 6= 0 in Z[ζp] should have a finite number of idealprime factors;

- Analogue for Z: all prime factorizations of integers arefinite.

4. the introduction of ideal prime factors should keepdivisibility in Z[ζp] consistent;

- Example: We don’t want something like an ideal primefactor P dividing 3 µP(3) = 2 times but dividing 9µP(9) = 5 times; 3 divides 9 twice, so P should divide 93× 2 = 6 times.

5. each α and β in Z[ζp] that have the same ideal primefactors (counting multiplicity) must only differ by a unitfactor.

- Analogue for Z: 108 and -108 have the exact same primefactors, which is obvious since one is −1 times the otherand −1 is a unit, as (−1)(−1) = 1.


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Prerequisites for Kummer’s definition

The setup

Let α be an arbitrary number in Z[ζp]. Then, α is obtained byplugging ζp into a polynomial with integer coefficients, sayf (x) = a0 + a1x + a2x

2 + · · ·+ ap−1xp−1; that is,

α = f (ζp) = a0 + a1ζp + a2ζ2p + · · ·+ ap−1ζ

p−1p .

Definitions

The norm of an element α in Z[ζp] is the number

N(α) = f (ζp)f (ζ2p) · · · f (ζp−1

p ),

where f (x) is as above. The numbers f (ζ ip) where i 6= 1 arecalled the conjugates of α = f (ζp).


R. Garrett






“Finding” ideal prime factors for a prime integer q

Kummer began by defining what it means to be an ideal primefactor of a prime integer q 6= p in Z[ζp].

Fact

It turns out that p is always a unit multiple of (ζp − 1)p, andζp − 1 is always prime, so there’s no need to introduce “idealprime factors” of p since its lone (repeating) factor is alreadyprime.

In every case he tried with small primes p, when he selectedanother prime integer q 6= p, Kummer found a prime numbergq in Z[ζp]

gq(ζp) = b0,q + b1,qζp + b2,qζ2p + · · ·+ bp−1,qζ

p−1p .

satisfying some desirable properties that allow us to define idealprime factors for the prime q.


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Proceeding with gq

With this “nice” gq found, Kummer specifically chose a subset{g1(ζp), ..., gn(ζp)} of all the conjugates of gq, excluding gqitself and selecting one representative for each orbit in somesense we won’t get into.He then took their product

Φgq = g1 · · · gn.

It turns out that gqΦgq is ALWAYS an integer multiple of q, soevery ideal prime factor of gq must also be an ideal primefactor of q itself, and every ideal prime factor of q must be anideal prime factor of one of the gi or gq itself!


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Defining ideal prime factors by divisibility

Let P be an ideal prime factor of q and gq (choosing anothergi is similar).

Definition

We say that an arbitrary α = f (ζp) is divisible n times by P ifα · Φn

gq is divisible by qn.

Definition

If α is divisible n times by P but not n + 1 times, we sayP divides α with multiplicity n.

We denote by µP(α) the multiplicity by which P divides α.


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Some glaring problems

1 The existence of a gq(ζp) with the desired properties forany pair of primes p 6= q wasn’t proven.

2 We haven’t actually found what the ideal prime factors Pare; we only defined them in terms of what they divide.


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Going for something more general

Considering equations that only have potential use for Z[ζp] forsome prime p is pretty restrictive. Can ideal primes be definedfor more general Z[α] where α is ANY algebraic number?


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Example showing nonexistence of ideal primes

Consider

Z[√−3] = {x + y

√−3 : x , y integers}

and letb = 1 +

√−3.

Notice that (1 +√−3)3 = −8, i.e. b3 = −8, and since any

positive integer k can be decomposed as k = 3j + i where0 ≤ i ≤ 2,

8bk = 8b3j+i = 8(b3)jbi = 8(−8)jbi = (−1)j8j+1bi .

Then, because 8j+1 = (23)j+1 = 23j+3 = 23j+i23−i = 2k23−i ,we see 8bk is always divisible by 2k .


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Example showing nonexistence of ideal primes (ctd)

Since 2 doesn’t divide b = 1 +√−3, Kummer’s theory would

(if it generalized) guarantee an ideal prime factor P of 2 where

1 2 is divisible µP(2) times by P,

2 b is divisible µP(b) times by P, and

3 µP(2) > µP(b).

If item 3 weren’t the case, i.e. µP(2) ≤ µP(b) for all idealprime factors P of b and/or 2, meaning each P divides b atleast as much as P divides 2. That would then mean that 2would divide b, which is false.Thus,

µP(2) > µP(b).


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Example showing nonexistence of ideal primes (ctd)

However, since 2k always divides 8bk , P must divide 8bk atleast as many times as it divides 2k , i.e.

µP(2k) ≤ µP(8bk).

Since µP(ck) = kµP(c) and µP(cd) = µP(c) + µP(d) for allchoices of c and d , this is the same as

kµP(2) ≤ µP(8) + kµP(b) for all k .

Subtracting kµP(b) from both sides and factoring out k fromthe left side, we have

k[µP(2)− µP(b)] ≤ µP(8) for all k .

Then, since 1kµP(8) < 1 for large enough k , we have

µP(2)− µP(b) ≤ 0

andµP(2) ≤ µP(b).


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Our contradiction

We just showed that µP(2) ≤ µP(b), but we assumed thatµP(2) > µP(b) when we chose the ideal prime factor P, andthis P had to exist because 2 doesn’t divide b!This means we have a contradiction, and “divisibility” by Pcannot make sense in Z[

√−3].

We can’t consistently define ideal prime factors of 2 in Z[√−3]!


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What we need to fix

The main problem that broke the example we just did was that2 didn’t divide b = 1 +

√−3. Suppose we expanded Z[

√−3]

to a new number system where 2 actually divides 1 +√−3;

that is, we include β = 1+√−3

2 . Notice that√−3 = 2

(1+√−3

2

)− 1, so

Z[β] = {a + bβ : a, b integers}

is a strictly larger number system containing Z[√−3].


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Some questions

Does considering Z[

1+√−3

2

]instead of Z[

√−3] fix the

problem?

If not, what other numbers besides β = 1+√−3

2 do we have tothrow in so that we can actually define ideal prime factors inany case if we start with Z[

√−3]?


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More questions

What about other potential choices for Z[α], α a complexnumber that’s algebraic over the integers?Is there a methodical way to identify what additional numbersto include in a number system potentially larger than Z[α] sothat we can use ideal prime factors in a consistent way?


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The quotient field

In the same way we build the rationals Q from Z by just takingfractions where the denominator is nonzero, we can build thecollection

Q(α) = all fractions of numbers in Z[α].

In other words, since all elements of Z[α] are polynomials in asingle variable with integer coefficients but with α plugged infor x , we’re considering all possible fractions you can form fromthese, excluding those where α is a root of the denominatorpolynomial.We call Q(α) the quotient field of Z[α]. As an example, thequotient field of Z[

√−5] is

Q(√−5) =

{a + b

√−5

c + d√−5

: a, b, c , d integers, c2 + d2 6= 0

}.


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Dedekind’s fix

Our desired number system Oα where ideal prime factors mayalways be defined consistently should obviously be contained inQ(α), in addition to containing Z[α]. Dedekind wanted thisOα to be to Q(α) just like the regular integers Z are to Q (inhis words, he wanted “Begriffsbestimmung”). Making thisprecise is beyond the scope of our talk.

What does it:

Oα must be the integral closure of Z[α] in Q(α).


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Integrality over Z

Definition

A monic polynomial is a polynomial with leading coefficient 1.For example, x3 + x + 1 and x5 − 3x + 2 are monic, but2x7 + 5x4 + 1 is not monic.

Definition

A complex number γ is integral over Z if γ satisfies anirreducible monic polynomial with integer coefficients.

Examples:

i is integral over Z since it’s a root of x2 + 1.ζp is integral over Z since it’s a root of xp − 1.1±√−23

2 is integral over Z since it’s a root ofx2 + x + 6. But 1

2 is not integral since it’s the rootof 2x − 1, which isn’t monic.


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Integrality over Z[α], α algebraic

Definition

A complex number γ is integral over Z[α] if γ satisfies anirreducible monic polynomial with coefficients in Z[α].

Examples

11±√−23

2 are integral over Z[√−3] since

they’re the roots of x2 + x + 6.

2−√−3±

√−7

2 are integral over Z[√−3] since

they’re the roots of x2 +√−3x + 1.


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Integral closure

Definition

The integral closure of Z[α] is the collection Oα of allcomplex numbers β in Q(α) that are integral over Z[α].

Example 1:

Even though −√−3±

√−7

2 are roots ofx2 +

√−3x + 1 and are thus integral, they are

NOT in the integral closure O√−3 since they are

not fractions in Q(√−3).

Example 2:

1+√−3

2 IS in O√−3 since it’s in Q(√−3) and

satisfies the monic polynomial x2 − x + 1.

It’s common to refer to Oα as the ring of integers of Q(α).


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Addressing the “vagueness” issue

Dedekind found in 1871 that, by passing to Oα instead of Z[α],Kummer’s ideal prime factors could always make sense again.He also noticed that Kummer’s ideal prime factors could alwaysbe represented by a pair of numbers in Oα, thus addressing thevagueness question posed earlier.


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Representing ideal prime factors with pairs

Definition

A pair (Ψ, p) where Ψ is a number in Oα and p is a primeinteger represents an ideal prime factor of p if

1 Ψ is not divisible by p in Oα, and

2 whenever β and γ are numbers in Oα where βγΨ isdivisible by p in Oα, either βΨ or γΨ must be divisible byp.

Note

The Ψ chosen to represent an ideal prime factor is NOTnecessarily unique; that is, if (Ψ, p) represents the ideal primefactor P of p, P might also be represented by another pair(Φ, p).


R. Garrett






Divisibility by ideal primes and ideal complexnumbers

Definition

We say a given β in Oα is divisible m times by the idealprime factor (Ψ, p) if βΨm is divisible by pm in Oα. Likebefore, multiplicity is defined similarly.

Definition

An ideal complex number in Oα is a formal product (i.e.we’re not trying to make sense of what the multiplicationactually means) of ideal complex numbers (Ψ1, p1), ...,(Ψk , pk) (potentially with some factors repeated).


R. Garrett






The ideal represented by an ideal complex number

Definition

Given an ideal complex number A = (Ψ1, p1)m1 · · · (Ψk , pk)mk ,the ideal IA defined by A is the set of all β in Oα that Adivides; that is, if β is in IA, then (Ψ1, p1) divides β m1 times,(Ψ2, p2) divides β m2 times, ..., and (Ψk , pk) divides β mk

times.


R. Garrett






Modern notation/definitions

In an abstract algebra class, one usually encounters these:

Definition

Given a ring Oα, a (necessarily infinite) set of numbers I in Oαis called an ideal if

1 0 is in I; 1 is not in I

2 whenever a and b are in I , a + b and a− b must also be inI ;

3 whenever a is in I and r is ANY number in Oα (notnecessarily in I ), ra must be in I .

Definition

An ideal P in Oα is called a prime ideal if, whenever a and bare numbers in Oα where ab is in P, it must be the case thateither a is in P or b is in P.


R. Garrett






Modern notation/definitions (ctd)

Definition

Given two ideals A, and B, the product ideal AB is thecollection of all possible sums of numbers of the form ab,where a is in A and b is in B.

Definition

Given a collection a1, ..., an of numbers in Oα, the idealgenerated by a1, ..., an in Oα is the set of all numbers of theform r1a1 + r2a2 + · · ·+ rnan, where r1, ..., rn are all arbitrarynumbers in Oα. We denote such an ideal by (a1, ..., an). Anideal (a) generated by a single element a is called a principalideal and would consist of all numbers ra where r is anynumber in Oα.


R. Garrett






Examples of ideals and generators

Example

In O√−5 = Z[√−5], the number 2

√−5 is in the ideal

I = (1 +√−5, 2) since

2√−5 = 2× (1 +

√−5) + (−1)× (2),

so in the notation of the previous slide, with a1 = 1 +√−5 and

a2 = 2, we are setting r1 = 2 and r2 = −1 so then 2√−5 is

expressed in the form r1a1 + r2a2, as desired.However, one can show that

√−5 is not I ; you cannot find r1

and r2 in Z[√−5] such that r1 × (1 +

√−5) + r2 × 2 =

√−5.

Additionally, it can be shown that I is a prime ideal.


R. Garrett






A useful fact about products of finitely generatedideals

Given two finitely generated ideals A = (a1, ..., am) andB = (b1, ..., bn), their product AB is also finitely generated,and AB is generated by the collection of all possible productsaibj , where 1 ≤ i ≤ m and 1 ≤ j ≤ n, though there might besome redundancy.


R. Garrett






Proving the equality of two finitely generated ideals

Given two finitely generated ideals A = (a1, ..., am) andB = (b1, ..., bn), we can prove A = B by showing that

1 each ai may be written in the formr1,ib1 + r2,ib2 + · · ·+ rn,ibn, for specially chosen r1,i , ...,rn,i in Oα; and

2 each bj may be written in the forms1,ja1 + s2,ja2 + · · ·+ sm,jam, for specially chosen s1,j , ...,sm,j in Oα.


R. Garrett






Some handy facts proven by Dedekind

In his work on Oα, Dedekind showed the following handy facts:

1 Every ideal in a ring of integers Oα may be expressed asan ideal generated by either one or two elements.

2 In rings of integers Oα, whenever one ideal B is a subsetof another ideal A, there must be another ideal C suchthat B = AC , so it makes sense to say that A divides Bin that case.

3 Given a ring of integers Oα, there is a collection ofspecially chosen numbers 1, β1, β2, ... , and βn−1 in Oα(called an integral basis for Oα) such that is everynumber γ in Oα may be written in the formγ = a0 · 1 + a1 · β1 + a2 · β2 + · · ·+ an−1 · βn−1, wherea0, a1, ... , an−1 are all integers (in Z).


R. Garrett






The connection with the modern term “ideal”

Dedekind’s crowning achievement, however, is proving thefollowing:

Theorem

Every ideal (in the modern sense) of Oα is the ideal IA definedby an ideal complex number A, and more specifically, everyprime ideal p of Oα is the ideal IP defined by an ideal primefactor P = (Ψ, p). Moreover, this correspondence allows us towrite every ideal of Oα uniquely as a (finite) product of primeideals (potentially with repeated factors).

(A sketch of the proof may be found in the notes on mywebsite for a talk I gave to the OSU Math Graduate StudentSeminar. It is too technical for this talk.)


R. Garrett






But we wanted numbers, not sets!

How can we possibly get these “ideal prime numbers” fromthese sets?Let’s see with an example.


R. Garrett






Example getting ideal numbers from ideals

Consider Z[√−5]. It is in fact THE ring of integers O√−5 in

Q(√−5), and it lacks unique factorization because

6 = 2× 3 = (1 +√−5)(1−

√−5).

Kummer’s setting

We would want an ideal prime factor P that divides both 2 and1 +√−5, an ideal prime factor Q that divides both 3 and

1 +√−5, and an ideal prime factor R that divides both 3 and

1−√−5.

Dedekind’s setting

We define the ideal primes to be P = (1 +√−5, 2),

Q = (1 +√−5, 3), and R = (1−

√−5, 3).


R. Garrett






Example (ctd)

Per Dedekind’s result, P, Q, and R represent what we now callprime ideals, specifically the ideals P = (1 +

√−5, 2),

Q = (1 +√−5, 3), and R = (1−

√−5, 3), each generated by 2

elements. Moreover,

P2 = ((1 +√−5)2, 2(1 +

√−5), 4)

= (−4 + 2√−5, 2 + 2

√−5, 4) = (2),

the principal ideal generated by 2, since all 3 generators in thelast equality are divisible by 2, and we may obtain 2 from themby noticing(2 + 2

√−5) + (−1)(−4 + 2

√−5) + (−1)(4) = 6− 4 = 2.

Therefore, in a sense, we may think of P as√

2.


R. Garrett






Example (ctd)

Second, we similarly see PQ = (1 +√−5) since

PQ = (1 +√−5, 2)(1 +

√−5, 3)

= ((1 +√−5)2, 2(1 +

√−5), 3(1 +

√−5), 6),

all 4 generators are multiples of 1 +√−5 (recall

6 = (1 +√−5)(1−

√−5)), and 1 +

√−5 may be obtained by

taking the difference of the generators 3(1 +√−5) and

2(1 +√−5).

Thirdly, QR = (3) since

QR = (1+√−5, 3)(1−

√−5, 3) = (6, 3(1+

√−5), 3(1−

√−5), 9),

all 4 generators are multiples of 3, and 3 = 9− 6, a differenceof generators.


R. Garrett






Example (ctd)

Lastly,

PR = (1+√−5, 2)(1−

√−5, 3) = (6, 3(1+

√−5), 2(1−

√−5), 6),

and 1−√−5 can be written in terms of the 3 generators given:

1−√−5 = (−1) · [3(1 +

√−5)] + (−1) · [2(1−

√−5)] + 6.

Moreover, 6 and 2(1−√−5) are clearly multiples of

(1−√−5), and 3(1 +

√−5) is too because

3(1 +√−5) = 6− 3(1−

√−5).

Therefore, PR = (1−√−5).


R. Garrett






Example (ctd)

Going from prime ideals to ideal prime factors again, we maythink principal ideals as the numbers they represent, and wemay think of Q as PQ/P, so we’d intuitively get

Q ∼ (1 +√−5)√

2.

Similarly, thinking of R as PR/P, we have

R ∼ (1−√−5)√

2.


R. Garrett






Example (ctd)

These are all the ideal prime factors we need since

6 = 2× 3 = [(√

2)× (√

2)]×[

1 +√−5√

2

]×[

1−√−5√

2

]and

6 = (1+√−5)(1−

√−5) =

[√2× 1 +

√−5√

2

]×[√

2× 1−√−5√

2

].

Both of these factorizations are the same (just in differentorders), meaning we have a unique factorization for 6, asdesired. In fact, unique factorization (with prime ideals) iscompletely restored in this setting, as with all rings of integersOα in algebraic number fields Q(α)!


R. Garrett






References

1 H.M. Edwards, The Genesis of Ideal Theory Archive forHistory of Exact Sciences, Volume 23, Springer Verlag(1980), p. 322-378.

2 I. Kleiner, A History of Abstract Algebra. Birkhauser(2007). Relevant material: p. 48-54.

3 I. Kleiner, From Numbers to Rings: The Early History ofRing Theory. Elem. Math, 53 (1998), p. 18-35.

4 J. J. O’Connor and E.F. Robertson, The Development ofRing Theory. Self-published notes:http://www-groups.dcs.st-and.ac.uk/history/HistTopics/Ring theory.html

5 P. Stevenhagen, Kummer Theory and Reciprocity Laws,Self-published notes:http://websites.math.leidenuniv.nl/algebra/Stevenhagen-Kummer.pdf

So, why are they called ``ideals'' anyway? · Diophantine equations De nition: Diophantine equation...

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