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The Binomial Theorem without middle terms:Putting prime numbers to work in algebra

Tom Marley

University of Nebraska-Lincoln

April 8, 2016

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Rings!

RING

THEORY

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Modular Arithmetic

Clock arithmetic:

8 (o’clock) + 5 (hours) = 1 (o’clock)

8 (hours) + 5 (o’clock) = 1 (o’clock)

So in clock arithmetic, we can simply write:

8 + 5 = 1.

We can also subtract: 4− 6 = 10.

As well as multiply: 4× 7 = 28 = 4.

General rule:

Divide by 12 and take remainder.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Modular Arithmetic

Clock arithmetic:

8 (o’clock) + 5 (hours) = 1 (o’clock)

8 (hours) + 5 (o’clock) = 1 (o’clock)

So in clock arithmetic, we can simply write:

8 + 5 = 1.

We can also subtract: 4− 6 = 10.

As well as multiply: 4× 7 = 28 = 4.

General rule:

Divide by 12 and take remainder.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Modular Arithmetic

Clock arithmetic:

8 (o’clock) + 5 (hours) = 1 (o’clock)

8 (hours) + 5 (o’clock) = 1 (o’clock)

So in clock arithmetic, we can simply write:

8 + 5 = 1.

We can also subtract: 4− 6 = 10.

As well as multiply: 4× 7 = 28 = 4.

General rule:

Divide by 12 and take remainder.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Modular Arithmetic

Clock arithmetic:

8 (o’clock) + 5 (hours) = 1 (o’clock)

8 (hours) + 5 (o’clock) = 1 (o’clock)

So in clock arithmetic, we can simply write:

8 + 5 = 1.

We can also subtract: 4− 6 = 10.

As well as multiply: 4× 7 = 28 = 4.

General rule:

Divide by 12 and take remainder.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Modular Arithmetic

Clock arithmetic:

8 (o’clock) + 5 (hours) = 1 (o’clock)

8 (hours) + 5 (o’clock) = 1 (o’clock)

So in clock arithmetic, we can simply write:

8 + 5 = 1.

We can also subtract: 4− 6 = 10.

As well as multiply: 4× 7 = 28 = 4.

General rule:

Divide by 12 and take remainder.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Modular Arithmetic

Clock arithmetic:

8 (o’clock) + 5 (hours) = 1 (o’clock)

8 (hours) + 5 (o’clock) = 1 (o’clock)

So in clock arithmetic, we can simply write:

8 + 5 = 1.

We can also subtract: 4− 6 =

10.

As well as multiply: 4× 7 = 28 = 4.

General rule:

Divide by 12 and take remainder.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Modular Arithmetic

Clock arithmetic:

8 (o’clock) + 5 (hours) = 1 (o’clock)

8 (hours) + 5 (o’clock) = 1 (o’clock)

So in clock arithmetic, we can simply write:

8 + 5 = 1.

We can also subtract: 4− 6 = 10.

As well as multiply: 4× 7 = 28 = 4.

General rule:

Divide by 12 and take remainder.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Modular Arithmetic

Clock arithmetic:

8 (o’clock) + 5 (hours) = 1 (o’clock)

8 (hours) + 5 (o’clock) = 1 (o’clock)

So in clock arithmetic, we can simply write:

8 + 5 = 1.

We can also subtract: 4− 6 = 10.

As well as multiply: 4× 7

= 28 = 4.

General rule:

Divide by 12 and take remainder.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Modular Arithmetic

Clock arithmetic:

8 (o’clock) + 5 (hours) = 1 (o’clock)

8 (hours) + 5 (o’clock) = 1 (o’clock)

So in clock arithmetic, we can simply write:

8 + 5 = 1.

We can also subtract: 4− 6 = 10.

As well as multiply: 4× 7 = 28

= 4.

General rule:

Divide by 12 and take remainder.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Modular Arithmetic

Clock arithmetic:

8 (o’clock) + 5 (hours) = 1 (o’clock)

8 (hours) + 5 (o’clock) = 1 (o’clock)

So in clock arithmetic, we can simply write:

8 + 5 = 1.

We can also subtract: 4− 6 = 10.

As well as multiply: 4× 7 = 28 = 4.

General rule:

Divide by 12 and take remainder.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Modular Arithmetic

Clock arithmetic:

8 (o’clock) + 5 (hours) = 1 (o’clock)

8 (hours) + 5 (o’clock) = 1 (o’clock)

So in clock arithmetic, we can simply write:

8 + 5 = 1.

We can also subtract: 4− 6 = 10.

As well as multiply: 4× 7 = 28 = 4.

General rule:

Divide by 12 and take remainder.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

The integers modulo n

We denote this number system by Z12, “the integers modulo 12”.

But there is nothing special about a clock with 12 hours.For example, a day on Neptune lasts about 16 hours.

We can do arithmetic in Z16 in the same way:

10 + 10 = 20 = 4

7− 13 = −6 = 10

5× 7 = 35 = 3

Similarly for Zn for any integer n ≥ 1.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

The integers modulo n

We denote this number system by Z12, “the integers modulo 12”.But there is nothing special about a clock with 12 hours.

For example, a day on Neptune lasts about 16 hours.

We can do arithmetic in Z16 in the same way:

10 + 10 = 20 = 4

7− 13 = −6 = 10

5× 7 = 35 = 3

Similarly for Zn for any integer n ≥ 1.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

The integers modulo n

We denote this number system by Z12, “the integers modulo 12”.But there is nothing special about a clock with 12 hours.For example, a day on Neptune lasts about 16 hours.

We can do arithmetic in Z16 in the same way:

10 + 10 = 20 = 4

7− 13 = −6 = 10

5× 7 = 35 = 3

Similarly for Zn for any integer n ≥ 1.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

The integers modulo n

We denote this number system by Z12, “the integers modulo 12”.But there is nothing special about a clock with 12 hours.For example, a day on Neptune lasts about 16 hours.

We can do arithmetic in Z16 in the same way:

10 + 10 = 20

= 4

7− 13 = −6 = 10

5× 7 = 35 = 3

Similarly for Zn for any integer n ≥ 1.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

The integers modulo n

We denote this number system by Z12, “the integers modulo 12”.But there is nothing special about a clock with 12 hours.For example, a day on Neptune lasts about 16 hours.

We can do arithmetic in Z16 in the same way:

10 + 10 = 20 = 4

7− 13 = −6 = 10

5× 7 = 35 = 3

Similarly for Zn for any integer n ≥ 1.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

The integers modulo n

We denote this number system by Z12, “the integers modulo 12”.But there is nothing special about a clock with 12 hours.For example, a day on Neptune lasts about 16 hours.

We can do arithmetic in Z16 in the same way:

10 + 10 = 20 = 4

7− 13 = −6

= 10

5× 7 = 35 = 3

Similarly for Zn for any integer n ≥ 1.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

The integers modulo n

We denote this number system by Z12, “the integers modulo 12”.But there is nothing special about a clock with 12 hours.For example, a day on Neptune lasts about 16 hours.

We can do arithmetic in Z16 in the same way:

10 + 10 = 20 = 4

7− 13 = −6 = 10

5× 7 = 35 = 3

Similarly for Zn for any integer n ≥ 1.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

The integers modulo n

We denote this number system by Z12, “the integers modulo 12”.But there is nothing special about a clock with 12 hours.For example, a day on Neptune lasts about 16 hours.

We can do arithmetic in Z16 in the same way:

10 + 10 = 20 = 4

7− 13 = −6 = 10

5× 7 = 35

= 3

Similarly for Zn for any integer n ≥ 1.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

The integers modulo n

We denote this number system by Z12, “the integers modulo 12”.But there is nothing special about a clock with 12 hours.For example, a day on Neptune lasts about 16 hours.

We can do arithmetic in Z16 in the same way:

10 + 10 = 20 = 4

7− 13 = −6 = 10

5× 7 = 35 = 3

Similarly for Zn for any integer n ≥ 1.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Ring axioms

In number systems like Zn, many of the familiar axioms fromarithmetic hold: For example:

a + b = b + a (commutativity of addition)

ab = ba (commutative of multiplication)

a(b + c) = ab + ac (distributive property)

Number systems that satisfy these axioms (and a couple more) arecalled Rings.

One can also subtract in rings, but not necessarily divide.For example, in Z12 we have: 6× 3 = 6× 9.However: 3 6= 9 in Z12.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Ring axioms

In number systems like Zn, many of the familiar axioms fromarithmetic hold: For example:

a + b = b + a (commutativity of addition)

ab = ba (commutative of multiplication)

a(b + c) = ab + ac (distributive property)

Number systems that satisfy these axioms (and a couple more) arecalled Rings.

One can also subtract in rings, but not necessarily divide.For example, in Z12 we have: 6× 3 = 6× 9.However: 3 6= 9 in Z12.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Ring axioms

In number systems like Zn, many of the familiar axioms fromarithmetic hold: For example:

a + b = b + a (commutativity of addition)

ab = ba (commutative of multiplication)

a(b + c) = ab + ac (distributive property)

Number systems that satisfy these axioms (and a couple more) arecalled Rings.

One can also subtract in rings, but not necessarily divide.For example, in Z12 we have: 6× 3 = 6× 9.However: 3 6= 9 in Z12.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Ring axioms

In number systems like Zn, many of the familiar axioms fromarithmetic hold: For example:

a + b = b + a (commutativity of addition)

ab = ba (commutative of multiplication)

a(b + c) = ab + ac (distributive property)

Number systems that satisfy these axioms (and a couple more) arecalled Rings.

One can also subtract in rings, but not necessarily divide.For example, in Z12 we have: 6× 3 = 6× 9.However: 3 6= 9 in Z12.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Ring axioms

In number systems like Zn, many of the familiar axioms fromarithmetic hold: For example:

a + b = b + a (commutativity of addition)

ab = ba (commutative of multiplication)

a(b + c) = ab + ac (distributive property)

Number systems that satisfy these axioms (and a couple more) arecalled Rings.

One can also subtract in rings, but not necessarily divide.For example, in Z12 we have: 6× 3 = 6× 9.However: 3 6= 9 in Z12.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Ring axioms

In number systems like Zn, many of the familiar axioms fromarithmetic hold: For example:

a + b = b + a (commutativity of addition)

ab = ba (commutative of multiplication)

a(b + c) = ab + ac (distributive property)

Number systems that satisfy these axioms (and a couple more) arecalled Rings.

One can also subtract in rings, but not necessarily divide.For example, in Z12 we have: 6× 3 = 6× 9.However: 3 6= 9 in Z12.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Ring axioms

In number systems like Zn, many of the familiar axioms fromarithmetic hold: For example:

a + b = b + a (commutativity of addition)

ab = ba (commutative of multiplication)

a(b + c) = ab + ac (distributive property)

Number systems that satisfy these axioms (and a couple more) arecalled Rings.

One can also subtract in rings, but not necessarily divide.

For example, in Z12 we have: 6× 3 = 6× 9.However: 3 6= 9 in Z12.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Ring axioms

In number systems like Zn, many of the familiar axioms fromarithmetic hold: For example:

a + b = b + a (commutativity of addition)

ab = ba (commutative of multiplication)

a(b + c) = ab + ac (distributive property)

Number systems that satisfy these axioms (and a couple more) arecalled Rings.

One can also subtract in rings, but not necessarily divide.For example, in Z12 we have: 6× 3 = 6× 9.

However: 3 6= 9 in Z12.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Ring axioms

In number systems like Zn, many of the familiar axioms fromarithmetic hold: For example:

a + b = b + a (commutativity of addition)

ab = ba (commutative of multiplication)

a(b + c) = ab + ac (distributive property)

Number systems that satisfy these axioms (and a couple more) arecalled Rings.

One can also subtract in rings, but not necessarily divide.For example, in Z12 we have: 6× 3 = 6× 9.However: 3 6= 9 in Z12.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Polynomial and power series rings

There are many examples of rings out there in addition to Zn:

Z (the integers)

R (the real numbers)

C (the complex numbers)

It’s easy to create new rings from old. One way is to consider allpolynomials in a variable x with coefficients from a given ring R:

R[x ] := {a0 + a1x + · · ·+ anxn | ai ∈ R ∀ i , n ≥ 0}.

We can also consider all power series in x with coefficients from R:

R[[x ]] := {∞∑i=0

aixi | ai ∈ R ∀ i}.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Polynomial and power series rings

There are many examples of rings out there in addition to Zn:

Z (the integers)

R (the real numbers)

C (the complex numbers)

It’s easy to create new rings from old. One way is to consider allpolynomials in a variable x with coefficients from a given ring R:

R[x ] := {a0 + a1x + · · ·+ anxn | ai ∈ R ∀ i , n ≥ 0}.

We can also consider all power series in x with coefficients from R:

R[[x ]] := {∞∑i=0

aixi | ai ∈ R ∀ i}.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Polynomial and power series rings

There are many examples of rings out there in addition to Zn:

Z (the integers)

R (the real numbers)

C (the complex numbers)

It’s easy to create new rings from old. One way is to consider allpolynomials in a variable x with coefficients from a given ring R:

R[x ] := {a0 + a1x + · · ·+ anxn | ai ∈ R ∀ i , n ≥ 0}.

We can also consider all power series in x with coefficients from R:

R[[x ]] := {∞∑i=0

aixi | ai ∈ R ∀ i}.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Polynomial and power series rings

There are many examples of rings out there in addition to Zn:

Z (the integers)

R (the real numbers)

C (the complex numbers)

It’s easy to create new rings from old. One way is to consider allpolynomials in a variable x with coefficients from a given ring R:

R[x ] := {a0 + a1x + · · ·+ anxn | ai ∈ R ∀ i , n ≥ 0}.

We can also consider all power series in x with coefficients from R:

R[[x ]] := {∞∑i=0

aixi | ai ∈ R ∀ i}.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Polynomials and power series (cont.)

We can iterate these processes:

R[x , y ] := (R[x ])[y ]

R[x , y , z ] := (R[x , y ])[z ]

R[[x , y ]] := (R[[x ]])[[y ]]

For example, in Z[[x , y ]], we have:

(1− xy) · (1 + xy + x2y2 + · · ·+ x iy i + · · · ) = 1.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Polynomials and power series (cont.)

We can iterate these processes:

R[x , y ] := (R[x ])[y ]

R[x , y , z ] := (R[x , y ])[z ]

R[[x , y ]] := (R[[x ]])[[y ]]

For example, in Z[[x , y ]], we have:

(1− xy) · (1 + xy + x2y2 + · · ·+ x iy i + · · · ) = 1.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Quotient spaces

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Quotient rings

Given a ring R and elements f , g ∈ R, we can form a quotient ringfrom R by identifying f and g ; or equivalently, by identifyingh = f − g and 0. We write this quotient ring as R/(h).

More generally, given f1, . . . , fn ∈ R, the quotient ring obtained byidentifying f1 = 0, . . . , fn = 0 is written:

R/(f1, . . . , fn).

For example: Z/(n) ∼= Zn and R[x ]/(x2 + 1) ∼= C.

To invert an element a ∈ R, just consider the ring:

R[x ]/(ax − 1).

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Quotient rings

Given a ring R and elements f , g ∈ R, we can form a quotient ringfrom R by identifying f and g ; or equivalently, by identifyingh = f − g and 0. We write this quotient ring as R/(h).

More generally, given f1, . . . , fn ∈ R, the quotient ring obtained byidentifying f1 = 0, . . . , fn = 0 is written:

R/(f1, . . . , fn).

For example: Z/(n) ∼= Zn and R[x ]/(x2 + 1) ∼= C.

To invert an element a ∈ R, just consider the ring:

R[x ]/(ax − 1).

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Quotient rings

Given a ring R and elements f , g ∈ R, we can form a quotient ringfrom R by identifying f and g ; or equivalently, by identifyingh = f − g and 0. We write this quotient ring as R/(h).

More generally, given f1, . . . , fn ∈ R, the quotient ring obtained byidentifying f1 = 0, . . . , fn = 0 is written:

R/(f1, . . . , fn).

For example: Z/(n) ∼=

Zn and R[x ]/(x2 + 1) ∼= C.

To invert an element a ∈ R, just consider the ring:

R[x ]/(ax − 1).

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Quotient rings

Given a ring R and elements f , g ∈ R, we can form a quotient ringfrom R by identifying f and g ; or equivalently, by identifyingh = f − g and 0. We write this quotient ring as R/(h).

More generally, given f1, . . . , fn ∈ R, the quotient ring obtained byidentifying f1 = 0, . . . , fn = 0 is written:

R/(f1, . . . , fn).

For example: Z/(n) ∼= Zn

and R[x ]/(x2 + 1) ∼= C.

To invert an element a ∈ R, just consider the ring:

R[x ]/(ax − 1).

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Quotient rings

Given a ring R and elements f , g ∈ R, we can form a quotient ringfrom R by identifying f and g ; or equivalently, by identifyingh = f − g and 0. We write this quotient ring as R/(h).

More generally, given f1, . . . , fn ∈ R, the quotient ring obtained byidentifying f1 = 0, . . . , fn = 0 is written:

R/(f1, . . . , fn).

For example: Z/(n) ∼= Zn and R[x ]/(x2 + 1) ∼=

C.

To invert an element a ∈ R, just consider the ring:

R[x ]/(ax − 1).

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Quotient rings

Given a ring R and elements f , g ∈ R, we can form a quotient ringfrom R by identifying f and g ; or equivalently, by identifyingh = f − g and 0. We write this quotient ring as R/(h).

More generally, given f1, . . . , fn ∈ R, the quotient ring obtained byidentifying f1 = 0, . . . , fn = 0 is written:

R/(f1, . . . , fn).

For example: Z/(n) ∼= Zn and R[x ]/(x2 + 1) ∼= C.

To invert an element a ∈ R, just consider the ring:

R[x ]/(ax − 1).

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Quotient rings

Given a ring R and elements f , g ∈ R, we can form a quotient ringfrom R by identifying f and g ; or equivalently, by identifyingh = f − g and 0. We write this quotient ring as R/(h).

More generally, given f1, . . . , fn ∈ R, the quotient ring obtained byidentifying f1 = 0, . . . , fn = 0 is written:

R/(f1, . . . , fn).

For example: Z/(n) ∼= Zn and R[x ]/(x2 + 1) ∼= C.

To invert an element a ∈ R, just consider the ring:

R[x ]/(ax − 1).

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

What are rings good for?

They’re precious!

Applications of ring theory:

Cryptography

Error-correcting codes

3D animation

Communication networks

Number theory, algebraicgeometry, invariant theory

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

What are rings good for?

They’re precious!

Applications of ring theory:

Cryptography

Error-correcting codes

3D animation

Communication networks

Number theory, algebraicgeometry, invariant theory

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

What are rings good for?

They’re precious!

Applications of ring theory:

Cryptography

Error-correcting codes

3D animation

Communication networks

Number theory, algebraicgeometry, invariant theory

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

What are rings good for?

They’re precious!

Applications of ring theory:

Cryptography

Error-correcting codes

3D animation

Communication networks

Number theory, algebraicgeometry, invariant theory

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

What are rings good for?

They’re precious!

Applications of ring theory:

Cryptography

Error-correcting codes

3D animation

Communication networks

Number theory, algebraicgeometry, invariant theory

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

What are rings good for?

They’re precious!

Applications of ring theory:

Cryptography

Error-correcting codes

3D animation

Communication networks

Number theory, algebraicgeometry, invariant theory

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

What are rings good for?

They’re precious!

Applications of ring theory:

Cryptography

Error-correcting codes

3D animation

Communication networks

Number theory, algebraicgeometry, invariant theory

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

What are rings good for?

They’re precious!

Applications of ring theory:

Cryptography

Error-correcting codes

3D animation

Communication networks

Number theory, algebraicgeometry, invariant theory

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

The Binomial Theorem

Let p be a (positive) prime integer. For the remainder of this talk,we’ll restrict our attention to rings of the form

R = Zp[[x1, . . . , xn]]/(f1, . . . , fr ).

Consider the Binomial Theorem in such rings:

(a + b)p =

p∑i=0

(p

i

)ap−ibi

= ap + pap−1b +p(p − 1)

2ap−2b2 +

p(p − 1)((p − 2)

6ap−3b3 + · · ·

= ap + bp

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

The Binomial Theorem

Let p be a (positive) prime integer. For the remainder of this talk,we’ll restrict our attention to rings of the form

R = Zp[[x1, . . . , xn]]/(f1, . . . , fr ).

Consider the Binomial Theorem in such rings:

(a + b)p =

p∑i=0

(p

i

)ap−ibi

= ap + pap−1b +p(p − 1)

2ap−2b2 +

p(p − 1)((p − 2)

6ap−3b3 + · · ·

= ap + bp

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

The Binomial Theorem

Let p be a (positive) prime integer. For the remainder of this talk,we’ll restrict our attention to rings of the form

R = Zp[[x1, . . . , xn]]/(f1, . . . , fr ).

Consider the Binomial Theorem in such rings:

(a + b)p =

p∑i=0

(p

i

)ap−ibi

= ap + pap−1b +p(p − 1)

2ap−2b2 +

p(p − 1)((p − 2)

6ap−3b3 + · · ·

= ap + bp

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

The Binomial Theorem

Let p be a (positive) prime integer. For the remainder of this talk,we’ll restrict our attention to rings of the form

R = Zp[[x1, . . . , xn]]/(f1, . . . , fr ).

Consider the Binomial Theorem in such rings:

(a + b)p =

p∑i=0

(p

i

)ap−ibi

= ap + pap−1b +p(p − 1)

2ap−2b2 +

p(p − 1)((p − 2)

6ap−3b3 + · · ·

= ap + bp

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Fermat’s (Little) Theorem

Fermat’s Theorem

Let p be a prime. Then ap = a for any a ∈ Zp.

Proof

The theorem clearly holds when a = 0 and a = 1. Suppose ap = afor some a ∈ Zp. Then

(a + 1)p = ap + 1p

= a + 1.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Fermat’s (Little) Theorem

Fermat’s Theorem

Let p be a prime. Then ap = a for any a ∈ Zp.

Proof

The theorem clearly holds when a = 0 and a = 1. Suppose ap = afor some a ∈ Zp.

Then

(a + 1)p = ap + 1p

= a + 1.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Fermat’s (Little) Theorem

Fermat’s Theorem

Let p be a prime. Then ap = a for any a ∈ Zp.

Proof

The theorem clearly holds when a = 0 and a = 1. Suppose ap = afor some a ∈ Zp. Then

(a + 1)p = ap + 1p

= a + 1.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Fermat’s (Little) Theorem

Fermat’s Theorem

Let p be a prime. Then ap = a for any a ∈ Zp.

Proof

The theorem clearly holds when a = 0 and a = 1. Suppose ap = afor some a ∈ Zp. Then

(a + 1)p = ap + 1p

= a + 1.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

A special subring

Key Observation

The setRp := {ap | a ∈ R}

is a subring of R.

Proof

First observe 0 = 0p and 1 = 1p, so 0, 1 ∈ Rp. Now letap, bp ∈ Rp. Then ap + bp = (a + b)p ∈ Rp andapbp = (ab)p ∈ Rp. Finally, −(ap) = (−a)p ∈ Rp.

The relationship between the ring R and the subring Rp providesimportant clues about the structure of R.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

A special subring

Key Observation

The setRp := {ap | a ∈ R}

is a subring of R.

Proof

First observe 0 = 0p and 1 = 1p, so 0, 1 ∈ Rp.

Now letap, bp ∈ Rp. Then ap + bp = (a + b)p ∈ Rp andapbp = (ab)p ∈ Rp. Finally, −(ap) = (−a)p ∈ Rp.

The relationship between the ring R and the subring Rp providesimportant clues about the structure of R.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

A special subring

Key Observation

The setRp := {ap | a ∈ R}

is a subring of R.

Proof

First observe 0 = 0p and 1 = 1p, so 0, 1 ∈ Rp. Now letap, bp ∈ Rp. Then ap + bp = (a + b)p ∈ Rp andapbp = (ab)p ∈ Rp.

Finally, −(ap) = (−a)p ∈ Rp.

The relationship between the ring R and the subring Rp providesimportant clues about the structure of R.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

A special subring

Key Observation

The setRp := {ap | a ∈ R}

is a subring of R.

Proof

First observe 0 = 0p and 1 = 1p, so 0, 1 ∈ Rp. Now letap, bp ∈ Rp. Then ap + bp = (a + b)p ∈ Rp andapbp = (ab)p ∈ Rp. Finally, −(ap) = (−a)p ∈ Rp.

The relationship between the ring R and the subring Rp providesimportant clues about the structure of R.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

A special subring

Key Observation

The setRp := {ap | a ∈ R}

is a subring of R.

Proof

First observe 0 = 0p and 1 = 1p, so 0, 1 ∈ Rp. Now letap, bp ∈ Rp. Then ap + bp = (a + b)p ∈ Rp andapbp = (ab)p ∈ Rp. Finally, −(ap) = (−a)p ∈ Rp.

The relationship between the ring R and the subring Rp providesimportant clues about the structure of R.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Curve singularities

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Coordinate rings

For the parabola:

R[[x , y ]]/(y − x2) ∼= R[[x , x2]]

∼= R[[x ]].

For the cusp: R[[x , y ]]/(y2 − x3) ∼= R[[t2, t3]].

For the node:

R[[x , y ]]/(y2 − x3 − x2) ∼= R[[x , y ]]/(y2 − x2(x + 1))∼= R[[x , y ]]/((y − cx)(y + cx)).

where c =√x + 1 = 1 + 1

2x −18x

2 + · · · .

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Coordinate rings

For the parabola:

R[[x , y ]]/(y − x2) ∼= R[[x , x2]]∼= R[[x ]].

For the cusp: R[[x , y ]]/(y2 − x3) ∼= R[[t2, t3]].

For the node:

R[[x , y ]]/(y2 − x3 − x2) ∼= R[[x , y ]]/(y2 − x2(x + 1))∼= R[[x , y ]]/((y − cx)(y + cx)).

where c =√x + 1 = 1 + 1

2x −18x

2 + · · · .

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Coordinate rings

For the parabola:

R[[x , y ]]/(y − x2) ∼= R[[x , x2]]∼= R[[x ]].

For the cusp: R[[x , y ]]/(y2 − x3) ∼= R[[t2, t3]].

For the node:

R[[x , y ]]/(y2 − x3 − x2) ∼= R[[x , y ]]/(y2 − x2(x + 1))∼= R[[x , y ]]/((y − cx)(y + cx)).

where c =√x + 1 = 1 + 1

2x −18x

2 + · · · .

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Coordinate rings

For the parabola:

R[[x , y ]]/(y − x2) ∼= R[[x , x2]]∼= R[[x ]].

For the cusp: R[[x , y ]]/(y2 − x3) ∼= R[[t2, t3]].

For the node:

R[[x , y ]]/(y2 − x3 − x2) ∼= R[[x , y ]]/(y2 − x2(x + 1))

∼= R[[x , y ]]/((y − cx)(y + cx)).

where c =√x + 1 = 1 + 1

2x −18x

2 + · · · .

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Coordinate rings

For the parabola:

R[[x , y ]]/(y − x2) ∼= R[[x , x2]]∼= R[[x ]].

For the cusp: R[[x , y ]]/(y2 − x3) ∼= R[[t2, t3]].

For the node:

R[[x , y ]]/(y2 − x3 − x2) ∼= R[[x , y ]]/(y2 − x2(x + 1))∼= R[[x , y ]]/((y − cx)(y + cx)).

where c =√x + 1 = 1 + 1

2x −18x

2 + · · · .

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Smoothness

Definition

Let f (x , y) = 0 define a curve C . We say C is smooth ornonsingular at the origin if its coordinate ring k[[x , y ]]/(f ) isisomorphic to a polynomial ring in one variable. Otherwise, C issingular at the origin.

Remark

If k = R then C is smooth at (0, 0) if and only if ∂f∂x and ∂f

∂y don’tboth vanish at the origin. That is, C is smooth at (0, 0) if and onlyif there is a unique well-defined tangent line to C at the origin.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Smoothness

Definition

Let f (x , y) = 0 define a curve C . We say C is smooth ornonsingular at the origin if its coordinate ring k[[x , y ]]/(f ) isisomorphic to a polynomial ring in one variable. Otherwise, C issingular at the origin.

Remark

If k = R then C is smooth at (0, 0) if and only if ∂f∂x and ∂f

∂y don’tboth vanish at the origin. That is, C is smooth at (0, 0) if and onlyif there is a unique well-defined tangent line to C at the origin.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

An illustrative example

Question

Given a ring R = Zp[[x1, . . . , xd ]]/(f1, . . . , fr ), how can we decide ifR is smooth at the origin? That is, how can we tell ifR ∼= Zp[[y1, . . . , ys ]]?

Consider R = Zp[[x ]]. Then Rp = Zp[[xp]].

Note that every element in f ∈ R can be written uniquely in theform:

f = c0 + c1x + c2x2 + · · ·+ cp−1x

p−1

for some c0, . . . , cp−1 ∈ Rp.For example, let p = 3 and f = 2 + x + 2x3 + x4 + 2x5 + x7. Then

f = (2 + 2x3) · 1 + (1 + x3 + x6)x + (2x3)x2.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

An illustrative example

Question

Given a ring R = Zp[[x1, . . . , xd ]]/(f1, . . . , fr ), how can we decide ifR is smooth at the origin? That is, how can we tell ifR ∼= Zp[[y1, . . . , ys ]]?

Consider R = Zp[[x ]].

Then Rp = Zp[[xp]].

Note that every element in f ∈ R can be written uniquely in theform:

f = c0 + c1x + c2x2 + · · ·+ cp−1x

p−1

for some c0, . . . , cp−1 ∈ Rp.For example, let p = 3 and f = 2 + x + 2x3 + x4 + 2x5 + x7. Then

f = (2 + 2x3) · 1 + (1 + x3 + x6)x + (2x3)x2.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

An illustrative example

Question

Given a ring R = Zp[[x1, . . . , xd ]]/(f1, . . . , fr ), how can we decide ifR is smooth at the origin? That is, how can we tell ifR ∼= Zp[[y1, . . . , ys ]]?

Consider R = Zp[[x ]]. Then Rp = Zp[[xp]].

Note that every element in f ∈ R can be written uniquely in theform:

f = c0 + c1x + c2x2 + · · ·+ cp−1x

p−1

for some c0, . . . , cp−1 ∈ Rp.For example, let p = 3 and f = 2 + x + 2x3 + x4 + 2x5 + x7. Then

f = (2 + 2x3) · 1 + (1 + x3 + x6)x + (2x3)x2.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

An illustrative example

Question

Given a ring R = Zp[[x1, . . . , xd ]]/(f1, . . . , fr ), how can we decide ifR is smooth at the origin? That is, how can we tell ifR ∼= Zp[[y1, . . . , ys ]]?

Consider R = Zp[[x ]]. Then Rp = Zp[[xp]].

Note that every element in f ∈ R can be written uniquely in theform:

f = c0 + c1x + c2x2 + · · ·+ cp−1x

p−1

for some c0, . . . , cp−1 ∈ Rp.

For example, let p = 3 and f = 2 + x + 2x3 + x4 + 2x5 + x7. Then

f = (2 + 2x3) · 1 + (1 + x3 + x6)x + (2x3)x2.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

An illustrative example

Question

Given a ring R = Zp[[x1, . . . , xd ]]/(f1, . . . , fr ), how can we decide ifR is smooth at the origin? That is, how can we tell ifR ∼= Zp[[y1, . . . , ys ]]?

Consider R = Zp[[x ]]. Then Rp = Zp[[xp]].

Note that every element in f ∈ R can be written uniquely in theform:

f = c0 + c1x + c2x2 + · · ·+ cp−1x

p−1

for some c0, . . . , cp−1 ∈ Rp.For example, let p = 3 and f = 2 + x + 2x3 + x4 + 2x5 + x7.

Then

f = (2 + 2x3) · 1 + (1 + x3 + x6)x + (2x3)x2.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

An illustrative example

Question

Given a ring R = Zp[[x1, . . . , xd ]]/(f1, . . . , fr ), how can we decide ifR is smooth at the origin? That is, how can we tell ifR ∼= Zp[[y1, . . . , ys ]]?

Consider R = Zp[[x ]]. Then Rp = Zp[[xp]].

Note that every element in f ∈ R can be written uniquely in theform:

f = c0 + c1x + c2x2 + · · ·+ cp−1x

p−1

for some c0, . . . , cp−1 ∈ Rp.For example, let p = 3 and f = 2 + x + 2x3 + x4 + 2x5 + x7. Then

f = (2 + 2x3) · 1 + (1 + x3 + x6)x + (2x3)x2.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

An illustrative example, cont.

Thus R = Zp[[x ]] has a basis over Rp = Zp[[xp]]. Namely,

{1, x , x2, . . . , xp−1}.

The same holds for R = Zp[[x1, . . . , xd ]].

Now consider the coordinate ring of the cusp over Z2:R = Z2[[x , y ]]/(y2 − x3) ∼= Z2[[t2, t3]].Then

R2 ∼= Z2[[t4, t6]].

Does R have a basis over R2? No!

Suppose {1, t2} is part of the basis. Then

t6 · 1 + t4 · t2 = 0.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

An illustrative example, cont.

Thus R = Zp[[x ]] has a basis over Rp = Zp[[xp]]. Namely,

{1, x , x2, . . . , xp−1}.

The same holds for R = Zp[[x1, . . . , xd ]].

Now consider the coordinate ring of the cusp over Z2:R = Z2[[x , y ]]/(y2 − x3) ∼= Z2[[t2, t3]].Then

R2 ∼= Z2[[t4, t6]].

Does R have a basis over R2? No!

Suppose {1, t2} is part of the basis. Then

t6 · 1 + t4 · t2 = 0.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

An illustrative example, cont.

Thus R = Zp[[x ]] has a basis over Rp = Zp[[xp]]. Namely,

{1, x , x2, . . . , xp−1}.

The same holds for R = Zp[[x1, . . . , xd ]].

Now consider the coordinate ring of the cusp over Z2:R = Z2[[x , y ]]/(y2 − x3) ∼= Z2[[t2, t3]].

ThenR2 ∼= Z2[[t4, t6]].

Does R have a basis over R2? No!

Suppose {1, t2} is part of the basis. Then

t6 · 1 + t4 · t2 = 0.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

An illustrative example, cont.

Thus R = Zp[[x ]] has a basis over Rp = Zp[[xp]]. Namely,

{1, x , x2, . . . , xp−1}.

The same holds for R = Zp[[x1, . . . , xd ]].

Now consider the coordinate ring of the cusp over Z2:R = Z2[[x , y ]]/(y2 − x3) ∼= Z2[[t2, t3]].Then

R2 ∼= Z2[[t4, t6]].

Does R have a basis over R2? No!

Suppose {1, t2} is part of the basis. Then

t6 · 1 + t4 · t2 = 0.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

An illustrative example, cont.

Thus R = Zp[[x ]] has a basis over Rp = Zp[[xp]]. Namely,

{1, x , x2, . . . , xp−1}.

The same holds for R = Zp[[x1, . . . , xd ]].

Now consider the coordinate ring of the cusp over Z2:R = Z2[[x , y ]]/(y2 − x3) ∼= Z2[[t2, t3]].Then

R2 ∼= Z2[[t4, t6]].

Does R have a basis over R2?

No!

Suppose {1, t2} is part of the basis. Then

t6 · 1 + t4 · t2 = 0.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

An illustrative example, cont.

Thus R = Zp[[x ]] has a basis over Rp = Zp[[xp]]. Namely,

{1, x , x2, . . . , xp−1}.

The same holds for R = Zp[[x1, . . . , xd ]].

Now consider the coordinate ring of the cusp over Z2:R = Z2[[x , y ]]/(y2 − x3) ∼= Z2[[t2, t3]].Then

R2 ∼= Z2[[t4, t6]].

Does R have a basis over R2? No!

Suppose {1, t2} is part of the basis. Then

t6 · 1 + t4 · t2 = 0.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

An illustrative example, cont.

Thus R = Zp[[x ]] has a basis over Rp = Zp[[xp]]. Namely,

{1, x , x2, . . . , xp−1}.

The same holds for R = Zp[[x1, . . . , xd ]].

Now consider the coordinate ring of the cusp over Z2:R = Z2[[x , y ]]/(y2 − x3) ∼= Z2[[t2, t3]].Then

R2 ∼= Z2[[t4, t6]].

Does R have a basis over R2? No!

Suppose {1, t2} is part of the basis.

Then

t6 · 1 + t4 · t2 = 0.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

An illustrative example, cont.

Thus R = Zp[[x ]] has a basis over Rp = Zp[[xp]]. Namely,

{1, x , x2, . . . , xp−1}.

The same holds for R = Zp[[x1, . . . , xd ]].

Now consider the coordinate ring of the cusp over Z2:R = Z2[[x , y ]]/(y2 − x3) ∼= Z2[[t2, t3]].Then

R2 ∼= Z2[[t4, t6]].

Does R have a basis over R2? No!

Suppose {1, t2} is part of the basis. Then

t6 · 1 + t4 · t2 = 0.

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Kunz’s Theorem

Theorem (Ernst Kunz, 1969)

R has a basis over Rp if and only if R ∼= Zp[[x1, . . . , xr ]].

That is, a curve, surface, solid, etc. over Zp is smooth at theorigin if and only if its coordinate ring R has a basis over Rp.

Kunz’s Theorem is in some sense the very beginning of the story ofthe study of singularities over Zp. This continues to be a veryactive area of research today.

Theorem (Avramov-Hochster-Iyengar-Yao, 2012)

If there exists any nonzero R-module which has a finite basis overRp then R ∼= Zp[[x1, . . . , xr ]].

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Kunz’s Theorem

Theorem (Ernst Kunz, 1969)

R has a basis over Rp if and only if R ∼= Zp[[x1, . . . , xr ]].

That is, a curve, surface, solid, etc. over Zp is smooth at theorigin if and only if its coordinate ring R has a basis over Rp.

Kunz’s Theorem is in some sense the very beginning of the story ofthe study of singularities over Zp. This continues to be a veryactive area of research today.

Theorem (Avramov-Hochster-Iyengar-Yao, 2012)

If there exists any nonzero R-module which has a finite basis overRp then R ∼= Zp[[x1, . . . , xr ]].

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Kunz’s Theorem

Theorem (Ernst Kunz, 1969)

R has a basis over Rp if and only if R ∼= Zp[[x1, . . . , xr ]].

That is, a curve, surface, solid, etc. over Zp is smooth at theorigin if and only if its coordinate ring R has a basis over Rp.

Kunz’s Theorem is in some sense the very beginning of the story ofthe study of singularities over Zp. This continues to be a veryactive area of research today.

Theorem (Avramov-Hochster-Iyengar-Yao, 2012)

If there exists any nonzero R-module which has a finite basis overRp then R ∼= Zp[[x1, . . . , xr ]].

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Kunz’s Theorem

Theorem (Ernst Kunz, 1969)

R has a basis over Rp if and only if R ∼= Zp[[x1, . . . , xr ]].

That is, a curve, surface, solid, etc. over Zp is smooth at theorigin if and only if its coordinate ring R has a basis over Rp.

Kunz’s Theorem is in some sense the very beginning of the story ofthe study of singularities over Zp. This continues to be a veryactive area of research today.

Theorem (Avramov-Hochster-Iyengar-Yao, 2012)

If there exists any nonzero R-module which has a finite basis overRp then R ∼= Zp[[x1, . . . , xr ]].

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Thank you!

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Thank you!

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra