[PPT] - Section 33 Finite fields Instructor: Yifan Yang Spring 2007 PowerPoint Presentation

SLIDE 1

Section 33 – Finite fields

Instructor: Yifan Yang Spring 2007

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 2

Review

Corollary (23.6) Let G be a finite subgroup of the multiplicative group of nonzero elements in a field F, then G is cyclic. Theorem (27.19) A field is either of prime characteristic p containing a subfield isomorphic to Zp, or of characteristic 0 containing a subfield isomorphic to Q. In essence, if a field is of characteristic p, then it is an extension field of Zp. If a field is of characteristic 0, then it is an extension field of Q.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 3

Review

Corollary (23.6) Let G be a finite subgroup of the multiplicative group of nonzero elements in a field F, then G is cyclic. Theorem (27.19) A field is either of prime characteristic p containing a subfield isomorphic to Zp, or of characteristic 0 containing a subfield isomorphic to Q. In essence, if a field is of characteristic p, then it is an extension field of Zp. If a field is of characteristic 0, then it is an extension field of Q.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 4

Review

Theorem If E is an extension field of a field F, then E is a vector space

ver F.

Definition If the dimension of E over F is finite, then E is a finite extension

f F, and the dimension of E over F is called the degree of E
ver F and denoted by [E : F].

Theorem Every finite-dimensional vector space over a field contains a basis.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 5

Review

Theorem If E is an extension field of a field F, then E is a vector space

ver F.

Definition If the dimension of E over F is finite, then E is a finite extension

f F, and the dimension of E over F is called the degree of E
ver F and denoted by [E : F].

Theorem Every finite-dimensional vector space over a field contains a basis.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 6

Review

Theorem If E is an extension field of a field F, then E is a vector space

ver F.

Definition If the dimension of E over F is finite, then E is a finite extension

f F, and the dimension of E over F is called the degree of E
ver F and denoted by [E : F].

Theorem Every finite-dimensional vector space over a field contains a basis.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 7

Finite fields

Theorem (33.1) Let E be a finite extension of degree n over a finite field F. If F has q elements, then E has qn elements. Proof. The vector space E contains a basis {α1, . . . , αn} and each element β ∈ E is uniquely expressed as b1α1 + · · · + bnαn. Since each bi has q possible different choices, E has qn elements. Corollary (33.2) If E is a finite field of characteristic p, then the number of elements in E is pn for some positive integer n.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 8

Finite fields

Theorem (33.1) Let E be a finite extension of degree n over a finite field F. If F has q elements, then E has qn elements. Proof. The vector space E contains a basis {α1, . . . , αn} and each element β ∈ E is uniquely expressed as b1α1 + · · · + bnαn. Since each bi has q possible different choices, E has qn elements. Corollary (33.2) If E is a finite field of characteristic p, then the number of elements in E is pn for some positive integer n.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 9

Finite fields

Theorem (33.1) Let E be a finite extension of degree n over a finite field F. If F has q elements, then E has qn elements. Proof. The vector space E contains a basis {α1, . . . , αn} and each element β ∈ E is uniquely expressed as b1α1 + · · · + bnαn. Since each bi has q possible different choices, E has qn elements. Corollary (33.2) If E is a finite field of characteristic p, then the number of elements in E is pn for some positive integer n.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 10

Finite fields

Theorem (33.1) Let E be a finite extension of degree n over a finite field F. If F has q elements, then E has qn elements. Proof. The vector space E contains a basis {α1, . . . , αn} and each element β ∈ E is uniquely expressed as b1α1 + · · · + bnαn. Since each bi has q possible different choices, E has qn elements. Corollary (33.2) If E is a finite field of characteristic p, then the number of elements in E is pn for some positive integer n.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 11

Explicit construction of finite fields

Idea.

Theorem 33.1 asserts that if E is a finite extension of

degree n over Zp, then E has pn elements.

Thus, to construct a field of pn elements, we look for an

irreducible polynomial f(x) over Zp of degree n.

Let α be a zero of f(x). Then Z[α] is a field of pn elements.

(See the slides for Section 29.)

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 12

Explicit construction of finite fields

Idea.

Theorem 33.1 asserts that if E is a finite extension of

degree n over Zp, then E has pn elements.

Thus, to construct a field of pn elements, we look for an

irreducible polynomial f(x) over Zp of degree n.

Let α be a zero of f(x). Then Z[α] is a field of pn elements.

(See the slides for Section 29.)

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 13

Explicit construction of finite fields

Idea.

Theorem 33.1 asserts that if E is a finite extension of

degree n over Zp, then E has pn elements.

Thus, to construct a field of pn elements, we look for an

irreducible polynomial f(x) over Zp of degree n.

Let α be a zero of f(x). Then Z[α] is a field of pn elements.

(See the slides for Section 29.)

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 14

Example

Problem. Construct a field of 16 elements.

Solution.

We look for an irreducible polynomial of degree 4 over Z2.
Such a polynomial takes the form

x4 + a1x3 + a2x2 + a3x + a4, where ai ∈ Z2.

0 cannot be a zero, so a4 = 1.
1 cannot be a zero, so a1 + a2 + a3 + a4 = 0 (in Z2).
Thus, there are only 4 possibilities remained x4 + x3 + 1,

x4 + x2 + 1, x4 + x + 1, and x4 + x3 + x2 + x + 1.

Among them, we have x4 + x2 + 1 = (x2 + x + 1)2. The
thers are irreducible.
Pick any of x4 + x + 1, x4 + x3 + 1, and

x4 + x3 + x2 + x + 1. Let α be a zero of the polynomial. Then Z2[α] is a field of 16 elements.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 15

Example

Problem. Construct a field of 16 elements.

Solution.

We look for an irreducible polynomial of degree 4 over Z2.
Such a polynomial takes the form

x4 + a1x3 + a2x2 + a3x + a4, where ai ∈ Z2.

0 cannot be a zero, so a4 = 1.
1 cannot be a zero, so a1 + a2 + a3 + a4 = 0 (in Z2).
Thus, there are only 4 possibilities remained x4 + x3 + 1,

x4 + x2 + 1, x4 + x + 1, and x4 + x3 + x2 + x + 1.

Among them, we have x4 + x2 + 1 = (x2 + x + 1)2. The
thers are irreducible.
Pick any of x4 + x + 1, x4 + x3 + 1, and

x4 + x3 + x2 + x + 1. Let α be a zero of the polynomial. Then Z2[α] is a field of 16 elements.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 16

Example

Problem. Construct a field of 16 elements.

Solution.

We look for an irreducible polynomial of degree 4 over Z2.
Such a polynomial takes the form

x4 + a1x3 + a2x2 + a3x + a4, where ai ∈ Z2.

0 cannot be a zero, so a4 = 1.
1 cannot be a zero, so a1 + a2 + a3 + a4 = 0 (in Z2).
Thus, there are only 4 possibilities remained x4 + x3 + 1,

x4 + x2 + 1, x4 + x + 1, and x4 + x3 + x2 + x + 1.

Among them, we have x4 + x2 + 1 = (x2 + x + 1)2. The
thers are irreducible.
Pick any of x4 + x + 1, x4 + x3 + 1, and

x4 + x3 + x2 + x + 1. Let α be a zero of the polynomial. Then Z2[α] is a field of 16 elements.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 17

Example

Problem. Construct a field of 16 elements.

Solution.

We look for an irreducible polynomial of degree 4 over Z2.
Such a polynomial takes the form

x4 + a1x3 + a2x2 + a3x + a4, where ai ∈ Z2.

0 cannot be a zero, so a4 = 1.
1 cannot be a zero, so a1 + a2 + a3 + a4 = 0 (in Z2).
Thus, there are only 4 possibilities remained x4 + x3 + 1,

x4 + x2 + 1, x4 + x + 1, and x4 + x3 + x2 + x + 1.

Among them, we have x4 + x2 + 1 = (x2 + x + 1)2. The
thers are irreducible.
Pick any of x4 + x + 1, x4 + x3 + 1, and

x4 + x3 + x2 + x + 1. Let α be a zero of the polynomial. Then Z2[α] is a field of 16 elements.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 18

Example

Problem. Construct a field of 16 elements.

Solution.

We look for an irreducible polynomial of degree 4 over Z2.
Such a polynomial takes the form

x4 + a1x3 + a2x2 + a3x + a4, where ai ∈ Z2.

0 cannot be a zero, so a4 = 1.
1 cannot be a zero, so a1 + a2 + a3 + a4 = 0 (in Z2).
Thus, there are only 4 possibilities remained x4 + x3 + 1,

x4 + x2 + 1, x4 + x + 1, and x4 + x3 + x2 + x + 1.

Among them, we have x4 + x2 + 1 = (x2 + x + 1)2. The
thers are irreducible.
Pick any of x4 + x + 1, x4 + x3 + 1, and

x4 + x3 + x2 + x + 1. Let α be a zero of the polynomial. Then Z2[α] is a field of 16 elements.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 19

Example

Problem. Construct a field of 16 elements.

Solution.

We look for an irreducible polynomial of degree 4 over Z2.
Such a polynomial takes the form

x4 + a1x3 + a2x2 + a3x + a4, where ai ∈ Z2.

0 cannot be a zero, so a4 = 1.
1 cannot be a zero, so a1 + a2 + a3 + a4 = 0 (in Z2).
Thus, there are only 4 possibilities remained x4 + x3 + 1,

x4 + x2 + 1, x4 + x + 1, and x4 + x3 + x2 + x + 1.

Among them, we have x4 + x2 + 1 = (x2 + x + 1)2. The
thers are irreducible.
Pick any of x4 + x + 1, x4 + x3 + 1, and

x4 + x3 + x2 + x + 1. Let α be a zero of the polynomial. Then Z2[α] is a field of 16 elements.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 20

Example

Problem. Construct a field of 16 elements.

Solution.

We look for an irreducible polynomial of degree 4 over Z2.
Such a polynomial takes the form

x4 + a1x3 + a2x2 + a3x + a4, where ai ∈ Z2.

0 cannot be a zero, so a4 = 1.
1 cannot be a zero, so a1 + a2 + a3 + a4 = 0 (in Z2).
Thus, there are only 4 possibilities remained x4 + x3 + 1,

x4 + x2 + 1, x4 + x + 1, and x4 + x3 + x2 + x + 1.

Among them, we have x4 + x2 + 1 = (x2 + x + 1)2. The
thers are irreducible.
Pick any of x4 + x + 1, x4 + x3 + 1, and

x4 + x3 + x2 + x + 1. Let α be a zero of the polynomial. Then Z2[α] is a field of 16 elements.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 21

Example

Problem. Construct a field of 16 elements.

Solution.

We look for an irreducible polynomial of degree 4 over Z2.
Such a polynomial takes the form

x4 + a1x3 + a2x2 + a3x + a4, where ai ∈ Z2.

0 cannot be a zero, so a4 = 1.
1 cannot be a zero, so a1 + a2 + a3 + a4 = 0 (in Z2).
Thus, there are only 4 possibilities remained x4 + x3 + 1,

x4 + x2 + 1, x4 + x + 1, and x4 + x3 + x2 + x + 1.

Among them, we have x4 + x2 + 1 = (x2 + x + 1)2. The
thers are irreducible.
Pick any of x4 + x + 1, x4 + x3 + 1, and

x4 + x3 + x2 + x + 1. Let α be a zero of the polynomial. Then Z2[α] is a field of 16 elements.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 22

Example, continued

Problem. Assume that we pick x4 + x + 1 ∈ Z2[x], and let α be

a zero of x4 + x + 1. Find an element of Z2[α] that generates the cyclic group Z2[α]×.

Solution.

Note that the cyclic group of nonzero elements in Z2[α] has
rder 15.
Thus, if an element β satisfies β, β3, β5 = 1, then β has
rder 15.
Let α be a zero of x4 + x + 1 ∈ Z2[x]. We have α, α3 = 1.
Also, α5 = α(α4 + α + 1) + α2 + α = α2 + α = 1.
Thus, α generates the cyclic group of nonzero elements in

Z2[α].

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 23

Example, continued

Problem. Assume that we pick x4 + x + 1 ∈ Z2[x], and let α be

a zero of x4 + x + 1. Find an element of Z2[α] that generates the cyclic group Z2[α]×.

Solution.

Note that the cyclic group of nonzero elements in Z2[α] has
rder 15.
Thus, if an element β satisfies β, β3, β5 = 1, then β has
rder 15.
Let α be a zero of x4 + x + 1 ∈ Z2[x]. We have α, α3 = 1.
Also, α5 = α(α4 + α + 1) + α2 + α = α2 + α = 1.
Thus, α generates the cyclic group of nonzero elements in

Z2[α].

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 24

Example, continued

Problem. Assume that we pick x4 + x + 1 ∈ Z2[x], and let α be

a zero of x4 + x + 1. Find an element of Z2[α] that generates the cyclic group Z2[α]×.

Solution.

Note that the cyclic group of nonzero elements in Z2[α] has
rder 15.
Thus, if an element β satisfies β, β3, β5 = 1, then β has
rder 15.
Let α be a zero of x4 + x + 1 ∈ Z2[x]. We have α, α3 = 1.
Also, α5 = α(α4 + α + 1) + α2 + α = α2 + α = 1.
Thus, α generates the cyclic group of nonzero elements in

Z2[α].

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 25

Example, continued

Problem. Assume that we pick x4 + x + 1 ∈ Z2[x], and let α be

a zero of x4 + x + 1. Find an element of Z2[α] that generates the cyclic group Z2[α]×.

Solution.

Note that the cyclic group of nonzero elements in Z2[α] has
rder 15.
Thus, if an element β satisfies β, β3, β5 = 1, then β has
rder 15.
Let α be a zero of x4 + x + 1 ∈ Z2[x]. We have α, α3 = 1.
Also, α5 = α(α4 + α + 1) + α2 + α = α2 + α = 1.
Thus, α generates the cyclic group of nonzero elements in

Z2[α].

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 26

Example, continued

Problem. Assume that we pick x4 + x + 1 ∈ Z2[x], and let α be

a zero of x4 + x + 1. Find an element of Z2[α] that generates the cyclic group Z2[α]×.

Solution.

Note that the cyclic group of nonzero elements in Z2[α] has
rder 15.
Thus, if an element β satisfies β, β3, β5 = 1, then β has
rder 15.
Let α be a zero of x4 + x + 1 ∈ Z2[x]. We have α, α3 = 1.
Also, α5 = α(α4 + α + 1) + α2 + α = α2 + α = 1.
Thus, α generates the cyclic group of nonzero elements in

Z2[α].

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 27

Example, continued

Problem. Assume that we pick x4 + x + 1 ∈ Z2[x], and let α be

a zero of x4 + x + 1. Find an element of Z2[α] that generates the cyclic group Z2[α]×.

Solution.

Note that the cyclic group of nonzero elements in Z2[α] has
rder 15.
Thus, if an element β satisfies β, β3, β5 = 1, then β has
rder 15.
Let α be a zero of x4 + x + 1 ∈ Z2[x]. We have α, α3 = 1.
Also, α5 = α(α4 + α + 1) + α2 + α = α2 + α = 1.
Thus, α generates the cyclic group of nonzero elements in

Z2[α].

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 28

In-class exercises

Find an irreducible polynomial of the given degree over the given field.

1

degree 3 over Z3.

2

degree 3 over Z5.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 29

Structure of finite fields

Theorem (33.3) Let E be a field of pn elements contained in an algebraic closure Zp of Zp. Then the elements of E are precisely the zeros of xpn − x in Zp[x]. Proof. By Lagrange’s theorem, every nonzero element of E is a zero

f xpn−1 − 1. Thus, every element of E is a zero of

xpn − x = x(xpn−1 − 1). On the other hand, since Zp is a field, a polynomial of degree pn has at most pn zeros. Therefore, the elements of E are precisely the zeros of xpn − x.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 30

Structure of finite fields

Theorem (33.3) Let E be a field of pn elements contained in an algebraic closure Zp of Zp. Then the elements of E are precisely the zeros of xpn − x in Zp[x]. Proof. By Lagrange’s theorem, every nonzero element of E is a zero

f xpn−1 − 1. Thus, every element of E is a zero of

xpn − x = x(xpn−1 − 1). On the other hand, since Zp is a field, a polynomial of degree pn has at most pn zeros. Therefore, the elements of E are precisely the zeros of xpn − x.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 31

Structure of finite fields

Theorem (33.3) Let E be a field of pn elements contained in an algebraic closure Zp of Zp. Then the elements of E are precisely the zeros of xpn − x in Zp[x]. Proof. By Lagrange’s theorem, every nonzero element of E is a zero

f xpn−1 − 1. Thus, every element of E is a zero of

xpn − x = x(xpn−1 − 1). On the other hand, since Zp is a field, a polynomial of degree pn has at most pn zeros. Therefore, the elements of E are precisely the zeros of xpn − x.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 32

Structure of finite fields

Theorem (33.3) Let E be a field of pn elements contained in an algebraic closure Zp of Zp. Then the elements of E are precisely the zeros of xpn − x in Zp[x]. Proof. By Lagrange’s theorem, every nonzero element of E is a zero

f xpn−1 − 1. Thus, every element of E is a zero of

xpn − x = x(xpn−1 − 1). On the other hand, since Zp is a field, a polynomial of degree pn has at most pn zeros. Therefore, the elements of E are precisely the zeros of xpn − x.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 33

Structure of finite fields

Theorem (33.3) Let E be a field of pn elements contained in an algebraic closure Zp of Zp. Then the elements of E are precisely the zeros of xpn − x in Zp[x]. Proof. By Lagrange’s theorem, every nonzero element of E is a zero

f xpn−1 − 1. Thus, every element of E is a zero of

xpn − x = x(xpn−1 − 1). On the other hand, since Zp is a field, a polynomial of degree pn has at most pn zeros. Therefore, the elements of E are precisely the zeros of xpn − x.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 34

Structure of finite fields

Corollary (23.6) The multiplicative group F × of nonzero elements in a finite field F is cyclic. Definition An element α in a field F is an nth root of unity if αn = 1. It is a primitive nth root of unity if αm = 1 for 0 < m < n. Corollary (33.6) A finite extension E of a finite field F is a simple extension of F. Proof. Let α be a generator of the multiplicative group E×. Then clearly, E = F(α).

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 35

Structure of finite fields

Corollary (23.6) The multiplicative group F × of nonzero elements in a finite field F is cyclic. Definition An element α in a field F is an nth root of unity if αn = 1. It is a primitive nth root of unity if αm = 1 for 0 < m < n. Corollary (33.6) A finite extension E of a finite field F is a simple extension of F. Proof. Let α be a generator of the multiplicative group E×. Then clearly, E = F(α).

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 36

Structure of finite fields

Corollary (23.6) The multiplicative group F × of nonzero elements in a finite field F is cyclic. Definition An element α in a field F is an nth root of unity if αn = 1. It is a primitive nth root of unity if αm = 1 for 0 < m < n. Corollary (33.6) A finite extension E of a finite field F is a simple extension of F. Proof. Let α be a generator of the multiplicative group E×. Then clearly, E = F(α).

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 37

Structure of finite fields

Corollary (23.6) The multiplicative group F × of nonzero elements in a finite field F is cyclic. Definition An element α in a field F is an nth root of unity if αn = 1. It is a primitive nth root of unity if αm = 1 for 0 < m < n. Corollary (33.6) A finite extension E of a finite field F is a simple extension of F. Proof. Let α be a generator of the multiplicative group E×. Then clearly, E = F(α).

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 38

Existence of finite fields of pn elements

Theorem (33.10) A finite field of pn elements exists for every prime power pn.

Main idea of proof.

In Theorem 33.3, we have shown that if E is a field of pk

elements contained in Zp, then the elements of E are precisely the zeros of xpk − x in Zp.

Here we will show the converse is true. That is, the set S
f zeros of xpn − x in Zp forms a field of pn elements.
Note that since Zp is algebraically closed, xpn − x factors

completely into linear factors over Zp.

We will show
The zeros of xpn − x are all distinct so that S has pn

elements.

S is a subdomain in Zp. Since S has finitely many

elements, this implies that S is a field.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 39

Existence of finite fields of pn elements

Theorem (33.10) A finite field of pn elements exists for every prime power pn.

Main idea of proof.

In Theorem 33.3, we have shown that if E is a field of pk

elements contained in Zp, then the elements of E are precisely the zeros of xpk − x in Zp.

Here we will show the converse is true. That is, the set S
f zeros of xpn − x in Zp forms a field of pn elements.
Note that since Zp is algebraically closed, xpn − x factors

completely into linear factors over Zp.

We will show
The zeros of xpn − x are all distinct so that S has pn

elements.

S is a subdomain in Zp. Since S has finitely many

elements, this implies that S is a field.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 40

Existence of finite fields of pn elements

Theorem (33.10) A finite field of pn elements exists for every prime power pn.

Main idea of proof.

In Theorem 33.3, we have shown that if E is a field of pk

elements contained in Zp, then the elements of E are precisely the zeros of xpk − x in Zp.

Here we will show the converse is true. That is, the set S
f zeros of xpn − x in Zp forms a field of pn elements.
Note that since Zp is algebraically closed, xpn − x factors

completely into linear factors over Zp.

We will show
The zeros of xpn − x are all distinct so that S has pn

elements.

S is a subdomain in Zp. Since S has finitely many

elements, this implies that S is a field.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 41

Existence of finite fields of pn elements

Theorem (33.10) A finite field of pn elements exists for every prime power pn.

Main idea of proof.

In Theorem 33.3, we have shown that if E is a field of pk

elements contained in Zp, then the elements of E are precisely the zeros of xpk − x in Zp.

Here we will show the converse is true. That is, the set S
f zeros of xpn − x in Zp forms a field of pn elements.
Note that since Zp is algebraically closed, xpn − x factors

completely into linear factors over Zp.

We will show
The zeros of xpn − x are all distinct so that S has pn

elements.

S is a subdomain in Zp. Since S has finitely many

elements, this implies that S is a field.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 42

Existence of finite fields of pn elements

Theorem (33.10) A finite field of pn elements exists for every prime power pn.

Main idea of proof.

In Theorem 33.3, we have shown that if E is a field of pk

elements contained in Zp, then the elements of E are precisely the zeros of xpk − x in Zp.

Here we will show the converse is true. That is, the set S
f zeros of xpn − x in Zp forms a field of pn elements.
Note that since Zp is algebraically closed, xpn − x factors

completely into linear factors over Zp.

We will show
The zeros of xpn − x are all distinct so that S has pn

elements.

S is a subdomain in Zp. Since S has finitely many

elements, this implies that S is a field.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 43

Existence of finite fields of pn elements

Theorem (33.10) A finite field of pn elements exists for every prime power pn.

Main idea of proof.

In Theorem 33.3, we have shown that if E is a field of pk

elements contained in Zp, then the elements of E are precisely the zeros of xpk − x in Zp.

Here we will show the converse is true. That is, the set S
f zeros of xpn − x in Zp forms a field of pn elements.
Note that since Zp is algebraically closed, xpn − x factors

completely into linear factors over Zp.

We will show
The zeros of xpn − x are all distinct so that S has pn

elements.

S is a subdomain in Zp. Since S has finitely many

elements, this implies that S is a field.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 44

Existence of finite fields of pn elements

Theorem (33.10) A finite field of pn elements exists for every prime power pn.

Main idea of proof.

In Theorem 33.3, we have shown that if E is a field of pk

elements contained in Zp, then the elements of E are precisely the zeros of xpk − x in Zp.

Here we will show the converse is true. That is, the set S
f zeros of xpn − x in Zp forms a field of pn elements.
Note that since Zp is algebraically closed, xpn − x factors

completely into linear factors over Zp.

We will show
The zeros of xpn − x are all distinct so that S has pn

elements.

S is a subdomain in Zp. Since S has finitely many

elements, this implies that S is a field.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 45

Derivation

Definition Let F be a field. The map D : F[x] → F[x] defined by D(a0 + a1x + · · · + anxn) = a1 + 2a2x + · · · + nanxn−1 is called the derivation, and D(f(x)) is called the derivative of f(x). Lemma Let F be a field. The derivation D on F[x] satisfies

D(f + g) = D(f) + D(g).
D(fg) = gD(f) + fD(g).

Proof. By direct verification.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 46

Derivation

Definition Let F be a field. The map D : F[x] → F[x] defined by D(a0 + a1x + · · · + anxn) = a1 + 2a2x + · · · + nanxn−1 is called the derivation, and D(f(x)) is called the derivative of f(x). Lemma Let F be a field. The derivation D on F[x] satisfies

D(f + g) = D(f) + D(g).
D(fg) = gD(f) + fD(g).

Proof. By direct verification.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 47

Derivation

Definition Let F be a field. The map D : F[x] → F[x] defined by D(a0 + a1x + · · · + anxn) = a1 + 2a2x + · · · + nanxn−1 is called the derivation, and D(f(x)) is called the derivative of f(x). Lemma Let F be a field. The derivation D on F[x] satisfies

D(f + g) = D(f) + D(g).
D(fg) = gD(f) + fD(g).

Proof. By direct verification.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 48

Proof of |S| = pn

Suppose that α is a repeated zero of xpn − x, say,

xpn − x = (x − α)2g(x) for some g(x) ∈ Zp[x].

We have

D((x − α)2g(x)) = 2(x − α)g(x) + (x − α)2D(g(x)). Thus, α is also a zero of D((x − α)2g(x)) = D(xpn − x) = pnxpn−1 − 1.

Since the characteristic is p, the last polynomial is just −1,

which has no zeros at all.

Therefore, xpn − x has no repeated root.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 49

Proof of |S| = pn

Suppose that α is a repeated zero of xpn − x, say,

xpn − x = (x − α)2g(x) for some g(x) ∈ Zp[x].

We have

D((x − α)2g(x)) = 2(x − α)g(x) + (x − α)2D(g(x)). Thus, α is also a zero of D((x − α)2g(x)) = D(xpn − x) = pnxpn−1 − 1.

Since the characteristic is p, the last polynomial is just −1,

which has no zeros at all.

Therefore, xpn − x has no repeated root.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 50

Proof of |S| = pn

Suppose that α is a repeated zero of xpn − x, say,

xpn − x = (x − α)2g(x) for some g(x) ∈ Zp[x].

We have

D((x − α)2g(x)) = 2(x − α)g(x) + (x − α)2D(g(x)). Thus, α is also a zero of D((x − α)2g(x)) = D(xpn − x) = pnxpn−1 − 1.

Since the characteristic is p, the last polynomial is just −1,

which has no zeros at all.

Therefore, xpn − x has no repeated root.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 51

Proof of |S| = pn

Suppose that α is a repeated zero of xpn − x, say,

xpn − x = (x − α)2g(x) for some g(x) ∈ Zp[x].

We have

D((x − α)2g(x)) = 2(x − α)g(x) + (x − α)2D(g(x)). Thus, α is also a zero of D((x − α)2g(x)) = D(xpn − x) = pnxpn−1 − 1.

Since the characteristic is p, the last polynomial is just −1,

which has no zeros at all.

Therefore, xpn − x has no repeated root.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 52

Proof of |S| = pn

Suppose that α is a repeated zero of xpn − x, say,

xpn − x = (x − α)2g(x) for some g(x) ∈ Zp[x].

We have

D((x − α)2g(x)) = 2(x − α)g(x) + (x − α)2D(g(x)). Thus, α is also a zero of D((x − α)2g(x)) = D(xpn − x) = pnxpn−1 − 1.

Since the characteristic is p, the last polynomial is just −1,

which has no zeros at all.

Therefore, xpn − x has no repeated root.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 53

Proof that S is an integral domain

We need to show that
0, 1 ∈ S.
If α ∈ S, then so is −α.
If α, β ∈ S, then so are α + β and αβ.
0 and 1 clearly satisfy xpn − x = 0.
Assume that αpn − α = 0. If p = 2, then −α = α ∈ S. If p is
dd, then (−α)pn − (−α) = −(αpn − α) = 0, and −α ∈ S.
If α and β are zeros of xpn − x, then (αβ)pn = αpnβpn = αβ

and αβ is in S.

Recall that p|

p

k

for 1 ≤ k ≤ p − 1. Thus

(u + v)p = up + vp for all u, v ∈ Zp. Then, (α + β)pn = ((α + β)p)pn−1 = (αp + βp)pn−1 = · · · = αpn + βpn = α + β. Therefore, α + β ∈ S.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 54

Proof that S is an integral domain

We need to show that
0, 1 ∈ S.
If α ∈ S, then so is −α.
If α, β ∈ S, then so are α + β and αβ.
0 and 1 clearly satisfy xpn − x = 0.
Assume that αpn − α = 0. If p = 2, then −α = α ∈ S. If p is
dd, then (−α)pn − (−α) = −(αpn − α) = 0, and −α ∈ S.
If α and β are zeros of xpn − x, then (αβ)pn = αpnβpn = αβ

and αβ is in S.

Recall that p|

p

k

for 1 ≤ k ≤ p − 1. Thus

(u + v)p = up + vp for all u, v ∈ Zp. Then, (α + β)pn = ((α + β)p)pn−1 = (αp + βp)pn−1 = · · · = αpn + βpn = α + β. Therefore, α + β ∈ S.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 55

Proof that S is an integral domain

We need to show that
0, 1 ∈ S.
If α ∈ S, then so is −α.
If α, β ∈ S, then so are α + β and αβ.
0 and 1 clearly satisfy xpn − x = 0.
Assume that αpn − α = 0. If p = 2, then −α = α ∈ S. If p is
dd, then (−α)pn − (−α) = −(αpn − α) = 0, and −α ∈ S.
If α and β are zeros of xpn − x, then (αβ)pn = αpnβpn = αβ

and αβ is in S.

Recall that p|

p

k

for 1 ≤ k ≤ p − 1. Thus

(u + v)p = up + vp for all u, v ∈ Zp. Then, (α + β)pn = ((α + β)p)pn−1 = (αp + βp)pn−1 = · · · = αpn + βpn = α + β. Therefore, α + β ∈ S.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 56

Proof that S is an integral domain

We need to show that
0, 1 ∈ S.
If α ∈ S, then so is −α.
If α, β ∈ S, then so are α + β and αβ.
0 and 1 clearly satisfy xpn − x = 0.
Assume that αpn − α = 0. If p = 2, then −α = α ∈ S. If p is
dd, then (−α)pn − (−α) = −(αpn − α) = 0, and −α ∈ S.
If α and β are zeros of xpn − x, then (αβ)pn = αpnβpn = αβ

and αβ is in S.

Recall that p|

p

k

for 1 ≤ k ≤ p − 1. Thus

(u + v)p = up + vp for all u, v ∈ Zp. Then, (α + β)pn = ((α + β)p)pn−1 = (αp + βp)pn−1 = · · · = αpn + βpn = α + β. Therefore, α + β ∈ S.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 57

Proof that S is an integral domain

We need to show that
0, 1 ∈ S.
If α ∈ S, then so is −α.
If α, β ∈ S, then so are α + β and αβ.
0 and 1 clearly satisfy xpn − x = 0.
Assume that αpn − α = 0. If p = 2, then −α = α ∈ S. If p is
dd, then (−α)pn − (−α) = −(αpn − α) = 0, and −α ∈ S.
If α and β are zeros of xpn − x, then (αβ)pn = αpnβpn = αβ

and αβ is in S.

Recall that p|

p

k

for 1 ≤ k ≤ p − 1. Thus

(u + v)p = up + vp for all u, v ∈ Zp. Then, (α + β)pn = ((α + β)p)pn−1 = (αp + βp)pn−1 = · · · = αpn + βpn = α + β. Therefore, α + β ∈ S.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 58

Proof that S is an integral domain

We need to show that
0, 1 ∈ S.
If α ∈ S, then so is −α.
If α, β ∈ S, then so are α + β and αβ.
0 and 1 clearly satisfy xpn − x = 0.
Assume that αpn − α = 0. If p = 2, then −α = α ∈ S. If p is
dd, then (−α)pn − (−α) = −(αpn − α) = 0, and −α ∈ S.
If α and β are zeros of xpn − x, then (αβ)pn = αpnβpn = αβ

and αβ is in S.

Recall that p|

p

k

for 1 ≤ k ≤ p − 1. Thus

(u + v)p = up + vp for all u, v ∈ Zp. Then, (α + β)pn = ((α + β)p)pn−1 = (αp + βp)pn−1 = · · · = αpn + βpn = α + β. Therefore, α + β ∈ S.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 59

Proof that S is an integral domain

We need to show that
0, 1 ∈ S.
If α ∈ S, then so is −α.
If α, β ∈ S, then so are α + β and αβ.
0 and 1 clearly satisfy xpn − x = 0.
Assume that αpn − α = 0. If p = 2, then −α = α ∈ S. If p is
dd, then (−α)pn − (−α) = −(αpn − α) = 0, and −α ∈ S.
If α and β are zeros of xpn − x, then (αβ)pn = αpnβpn = αβ

and αβ is in S.

Recall that p|

p

k

for 1 ≤ k ≤ p − 1. Thus

(u + v)p = up + vp for all u, v ∈ Zp. Then, (α + β)pn = ((α + β)p)pn−1 = (αp + βp)pn−1 = · · · = αpn + βpn = α + β. Therefore, α + β ∈ S.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 60

Proof that S is an integral domain

We need to show that
0, 1 ∈ S.
If α ∈ S, then so is −α.
If α, β ∈ S, then so are α + β and αβ.
0 and 1 clearly satisfy xpn − x = 0.
Assume that αpn − α = 0. If p = 2, then −α = α ∈ S. If p is
dd, then (−α)pn − (−α) = −(αpn − α) = 0, and −α ∈ S.
If α and β are zeros of xpn − x, then (αβ)pn = αpnβpn = αβ

and αβ is in S.

Recall that p|

p

k

for 1 ≤ k ≤ p − 1. Thus

(u + v)p = up + vp for all u, v ∈ Zp. Then, (α + β)pn = ((α + β)p)pn−1 = (αp + βp)pn−1 = · · · = αpn + βpn = α + β. Therefore, α + β ∈ S.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 61

Proof that S is an integral domain

We need to show that
0, 1 ∈ S.
If α ∈ S, then so is −α.
If α, β ∈ S, then so are α + β and αβ.
0 and 1 clearly satisfy xpn − x = 0.
Assume that αpn − α = 0. If p = 2, then −α = α ∈ S. If p is
dd, then (−α)pn − (−α) = −(αpn − α) = 0, and −α ∈ S.
If α and β are zeros of xpn − x, then (αβ)pn = αpnβpn = αβ

and αβ is in S.

Recall that p|

p

k

for 1 ≤ k ≤ p − 1. Thus

(u + v)p = up + vp for all u, v ∈ Zp. Then, (α + β)pn = ((α + β)p)pn−1 = (αp + βp)pn−1 = · · · = αpn + βpn = α + β. Therefore, α + β ∈ S.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 62

Proof that S is an integral domain

We need to show that
0, 1 ∈ S.
If α ∈ S, then so is −α.
If α, β ∈ S, then so are α + β and αβ.
0 and 1 clearly satisfy xpn − x = 0.
Assume that αpn − α = 0. If p = 2, then −α = α ∈ S. If p is
dd, then (−α)pn − (−α) = −(αpn − α) = 0, and −α ∈ S.
If α and β are zeros of xpn − x, then (αβ)pn = αpnβpn = αβ

and αβ is in S.

Recall that p|

p

k

for 1 ≤ k ≤ p − 1. Thus

(u + v)p = up + vp for all u, v ∈ Zp. Then, (α + β)pn = ((α + β)p)pn−1 = (αp + βp)pn−1 = · · · = αpn + βpn = α + β. Therefore, α + β ∈ S.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 63

Proof that S is an integral domain

We need to show that
0, 1 ∈ S.
If α ∈ S, then so is −α.
If α, β ∈ S, then so are α + β and αβ.
0 and 1 clearly satisfy xpn − x = 0.
Assume that αpn − α = 0. If p = 2, then −α = α ∈ S. If p is
dd, then (−α)pn − (−α) = −(αpn − α) = 0, and −α ∈ S.
If α and β are zeros of xpn − x, then (αβ)pn = αpnβpn = αβ

and αβ is in S.

Recall that p|

p

k

for 1 ≤ k ≤ p − 1. Thus

(u + v)p = up + vp for all u, v ∈ Zp. Then, (α + β)pn = ((α + β)p)pn−1 = (αp + βp)pn−1 = · · · = αpn + βpn = α + β. Therefore, α + β ∈ S.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 64

Galois fields GF(pn)

Corollary (33.11) Let F be a field. Then for every positive integer n, there exists an irreducible polynomial of degree n over F. Theorem (33.12) Let F and F ′ be two fields of order pn. Then F and F ′ are isomorphic. Definition The unique field of pn (up to isomorphism) is called the Galois field of order pn.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 65

Galois fields GF(pn)

Corollary (33.11) Let F be a field. Then for every positive integer n, there exists an irreducible polynomial of degree n over F. Theorem (33.12) Let F and F ′ be two fields of order pn. Then F and F ′ are isomorphic. Definition The unique field of pn (up to isomorphism) is called the Galois field of order pn.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 66

Galois fields GF(pn)

Corollary (33.11) Let F be a field. Then for every positive integer n, there exists an irreducible polynomial of degree n over F. Theorem (33.12) Let F and F ′ be two fields of order pn. Then F and F ′ are isomorphic. Definition The unique field of pn (up to isomorphism) is called the Galois field of order pn.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 67

Proof of Theorem 33.12

Assume that F and F ′ are fields of pn elements.
Since F and F ′ are both algebraic extensions of Zp, F and

F ′ are isomorphic to some subfields of order pn in Zp.

However, in Zp, there is only one subfield of order pn,

namely, the set of all zeros of xpn − x in Zp.

In other words, F and F ′ are both isomorphic to this

subfield of pn elements in Zp.

Instructor: Yifan Yang

Section 33 – Finite fields

SLIDE 68

Proof of Theorem 33.12

Assume that F and F ′ are fields of pn elements.
Since F and F ′ are both algebraic extensions of Zp, F and

F ′ are isomorphic to some subfields of order pn in Zp.

However, in Zp, there is only one subfield of order pn,

namely, the set of all zeros of xpn − x in Zp.

In other words, F and F ′ are both isomorphic to this

subfield of pn elements in Zp.

Instructor: Yifan Yang

Section 33 – Finite fields

SLIDE 69

Proof of Theorem 33.12

Assume that F and F ′ are fields of pn elements.
Since F and F ′ are both algebraic extensions of Zp, F and

F ′ are isomorphic to some subfields of order pn in Zp.

However, in Zp, there is only one subfield of order pn,

namely, the set of all zeros of xpn − x in Zp.

In other words, F and F ′ are both isomorphic to this

subfield of pn elements in Zp.

Instructor: Yifan Yang

Section 33 – Finite fields

SLIDE 70

Proof of Theorem 33.12

Assume that F and F ′ are fields of pn elements.
Since F and F ′ are both algebraic extensions of Zp, F and

F ′ are isomorphic to some subfields of order pn in Zp.

However, in Zp, there is only one subfield of order pn,

namely, the set of all zeros of xpn − x in Zp.

In other words, F and F ′ are both isomorphic to this

subfield of pn elements in Zp.

Instructor: Yifan Yang

Section 33 – Finite fields

SLIDE 71

Example

Let α be a zero of x3 + x + 1 in Z2[x].
Since x3 + x + 1 is irreducible over Z2, Z2[α] is a field of 8

elements.

Theorem 33.3 says that α is a zero of x8 − x. This implies

that x3 + x + 1 is a factor of x8 − x.

Indeed, we have x8 −x = x(x −1)(x3 +x2 +1)(x3 +x +1).
Note that x3 + x2 + 1 is also irreducible over Z2, giving rise

to the field Z2[x]/x3 + x2 + 1 of 8 elements.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 72

Example

Let α be a zero of x3 + x + 1 in Z2[x].
Since x3 + x + 1 is irreducible over Z2, Z2[α] is a field of 8

elements.

Theorem 33.3 says that α is a zero of x8 − x. This implies

that x3 + x + 1 is a factor of x8 − x.

Indeed, we have x8 −x = x(x −1)(x3 +x2 +1)(x3 +x +1).
Note that x3 + x2 + 1 is also irreducible over Z2, giving rise

to the field Z2[x]/x3 + x2 + 1 of 8 elements.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 73

Example

Let α be a zero of x3 + x + 1 in Z2[x].
Since x3 + x + 1 is irreducible over Z2, Z2[α] is a field of 8

elements.

Theorem 33.3 says that α is a zero of x8 − x. This implies

that x3 + x + 1 is a factor of x8 − x.

Indeed, we have x8 −x = x(x −1)(x3 +x2 +1)(x3 +x +1).
Note that x3 + x2 + 1 is also irreducible over Z2, giving rise

to the field Z2[x]/x3 + x2 + 1 of 8 elements.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 74

Example

Let α be a zero of x3 + x + 1 in Z2[x].
Since x3 + x + 1 is irreducible over Z2, Z2[α] is a field of 8

elements.

Theorem 33.3 says that α is a zero of x8 − x. This implies

that x3 + x + 1 is a factor of x8 − x.

Indeed, we have x8 −x = x(x −1)(x3 +x2 +1)(x3 +x +1).
Note that x3 + x2 + 1 is also irreducible over Z2, giving rise

to the field Z2[x]/x3 + x2 + 1 of 8 elements.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 75

Example

Let α be a zero of x3 + x + 1 in Z2[x].
Since x3 + x + 1 is irreducible over Z2, Z2[α] is a field of 8

elements.

Theorem 33.3 says that α is a zero of x8 − x. This implies

that x3 + x + 1 is a factor of x8 − x.

Indeed, we have x8 −x = x(x −1)(x3 +x2 +1)(x3 +x +1).
Note that x3 + x2 + 1 is also irreducible over Z2, giving rise

to the field Z2[x]/x3 + x2 + 1 of 8 elements.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 76

Example

Let α be a zero of x3 + x + 1 in Z2[x].
Since x3 + x + 1 is irreducible over Z2, Z2[α] is a field of 8

elements.

Theorem 33.3 says that α is a zero of x8 − x. This implies

that x3 + x + 1 is a factor of x8 − x.

Indeed, we have x8 −x = x(x −1)(x3 +x2 +1)(x3 +x +1).
Note that x3 + x2 + 1 is also irreducible over Z2, giving rise

to the field Z2[x]/x3 + x2 + 1 of 8 elements.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 77

Example

Question. Observe that over Z2 we have x16 − x =

x(x −1)(x2+x +1)(x4+x +1)(x4+x3+1)(x4+x3+x2+x +1), where the factors are all irreducible. Why isn’t there any polynomial of degree 3 in the factorization?

Answer.

Suppose that an irreducible factor f(x) of x16 − x has

degree 3.

Let α be a zero of f(x). Then [Z2[α] : Z2] = deg f = 3.
However, by Theorem 31.4, we must have 4 = [GF(16) :

Z2] = [GF(16) : Z2[α]][Z2[α] : Z2] = 3[GF(16) : Z2[α]], which is absurd.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 78

Example

Question. Observe that over Z2 we have x16 − x =

x(x −1)(x2+x +1)(x4+x +1)(x4+x3+1)(x4+x3+x2+x +1), where the factors are all irreducible. Why isn’t there any polynomial of degree 3 in the factorization?

Answer.

Suppose that an irreducible factor f(x) of x16 − x has

degree 3.

Let α be a zero of f(x). Then [Z2[α] : Z2] = deg f = 3.
However, by Theorem 31.4, we must have 4 = [GF(16) :

Z2] = [GF(16) : Z2[α]][Z2[α] : Z2] = 3[GF(16) : Z2[α]], which is absurd.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 79

Example

Question. Observe that over Z2 we have x16 − x =

x(x −1)(x2+x +1)(x4+x +1)(x4+x3+1)(x4+x3+x2+x +1), where the factors are all irreducible. Why isn’t there any polynomial of degree 3 in the factorization?

Answer.

Suppose that an irreducible factor f(x) of x16 − x has

degree 3.

Let α be a zero of f(x). Then [Z2[α] : Z2] = deg f = 3.
However, by Theorem 31.4, we must have 4 = [GF(16) :

Z2] = [GF(16) : Z2[α]][Z2[α] : Z2] = 3[GF(16) : Z2[α]], which is absurd.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 80

Example

Question. Observe that over Z2 we have x16 − x =

x(x −1)(x2+x +1)(x4+x +1)(x4+x3+1)(x4+x3+x2+x +1), where the factors are all irreducible. Why isn’t there any polynomial of degree 3 in the factorization?

Answer.

Suppose that an irreducible factor f(x) of x16 − x has

degree 3.

Let α be a zero of f(x). Then [Z2[α] : Z2] = deg f = 3.
However, by Theorem 31.4, we must have 4 = [GF(16) :

Z2] = [GF(16) : Z2[α]][Z2[α] : Z2] = 3[GF(16) : Z2[α]], which is absurd.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 81

Galois fields

Theorem In Zp, GF(pm) is contained in GF(pn) if and only if m|n. Proof. The argument in the above example shows that GF(pm) ≤ GF(pn) implies m|n. Conversely, assume that m|n. It suffices to prove that if α ∈ Zp is a zero of xpm − x, then it is also a zero of xpn − x. Now αpn = (αpm)pn−m = αpn−m = αpn−2m = · · · . Since m|n, eventually we arrive at αpn = α. Corollary The polynomial xpn − x is equal to the product of all monic irreducible polynomial over Zp with degree d dividing n.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 82

Galois fields

Theorem In Zp, GF(pm) is contained in GF(pn) if and only if m|n. Proof. The argument in the above example shows that GF(pm) ≤ GF(pn) implies m|n. Conversely, assume that m|n. It suffices to prove that if α ∈ Zp is a zero of xpm − x, then it is also a zero of xpn − x. Now αpn = (αpm)pn−m = αpn−m = αpn−2m = · · · . Since m|n, eventually we arrive at αpn = α. Corollary The polynomial xpn − x is equal to the product of all monic irreducible polynomial over Zp with degree d dividing n.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 83

Galois fields

Theorem In Zp, GF(pm) is contained in GF(pn) if and only if m|n. Proof. The argument in the above example shows that GF(pm) ≤ GF(pn) implies m|n. Conversely, assume that m|n. It suffices to prove that if α ∈ Zp is a zero of xpm − x, then it is also a zero of xpn − x. Now αpn = (αpm)pn−m = αpn−m = αpn−2m = · · · . Since m|n, eventually we arrive at αpn = α. Corollary The polynomial xpn − x is equal to the product of all monic irreducible polynomial over Zp with degree d dividing n.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 84

Galois fields

Theorem In Zp, GF(pm) is contained in GF(pn) if and only if m|n. Proof. The argument in the above example shows that GF(pm) ≤ GF(pn) implies m|n. Conversely, assume that m|n. It suffices to prove that if α ∈ Zp is a zero of xpm − x, then it is also a zero of xpn − x. Now αpn = (αpm)pn−m = αpn−m = αpn−2m = · · · . Since m|n, eventually we arrive at αpn = α. Corollary The polynomial xpn − x is equal to the product of all monic irreducible polynomial over Zp with degree d dividing n.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 85

Galois fields

Theorem In Zp, GF(pm) is contained in GF(pn) if and only if m|n. Proof. The argument in the above example shows that GF(pm) ≤ GF(pn) implies m|n. Conversely, assume that m|n. It suffices to prove that if α ∈ Zp is a zero of xpm − x, then it is also a zero of xpn − x. Now αpn = (αpm)pn−m = αpn−m = αpn−2m = · · · . Since m|n, eventually we arrive at αpn = α. Corollary The polynomial xpn − x is equal to the product of all monic irreducible polynomial over Zp with degree d dividing n.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 86

Galois fields

Theorem In Zp, GF(pm) is contained in GF(pn) if and only if m|n. Proof. The argument in the above example shows that GF(pm) ≤ GF(pn) implies m|n. Conversely, assume that m|n. It suffices to prove that if α ∈ Zp is a zero of xpm − x, then it is also a zero of xpn − x. Now αpn = (αpm)pn−m = αpn−m = αpn−2m = · · · . Since m|n, eventually we arrive at αpn = α. Corollary The polynomial xpn − x is equal to the product of all monic irreducible polynomial over Zp with degree d dividing n.

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 87

Example

Over Z2, we have x64 − x = x(x + 1)(x2 + x + 1)(x3 + x + 1)(x3 + x2 + 1) × (x6 + x + 1)(x6 + x3 + 1)(x6 + x5 + 1) × (x6 + x4 + x2 + x + 1)(x6 + x4 + x3 + x + 1) × (x6 + x5 + x2 + x + 1)(x6 + x5 + x3 + x2 + 1) × (x6 + x5 + x4 + x + 1)(x6 + x5 + x4 + x2 + 1).

Instructor: Yifan Yang Section 33 – Finite fields

SLIDE 88

Homework

Problems 4, 6, 9, 10, 13, 14 of Section 33.

Instructor: Yifan Yang Section 33 – Finite fields