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
Top Related