Section 29 Introduction to extension fields Instructor: Yifan Yang - - PowerPoint PPT Presentation

Section 29 – Introduction to extension fields

Instructor: Yifan Yang Spring 2007

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Definition of an extension field

Definition A field E is an extension field of a field if F ≤ E. Example We have C F(x, y) R F(x) F(y) Q F

• ❅

❅ ❅ ❅ ❅ ❅ ❅ ❅

• Instructor: Yifan Yang

Section 29 – Introduction to extension fields

Definition of an extension field

Definition A field E is an extension field of a field if F ≤ E. Example We have C F(x, y) R F(x) F(y) Q F

• ❅

❅ ❅ ❅ ❅ ❅ ❅ ❅

• Instructor: Yifan Yang

Section 29 – Introduction to extension fields

Every polynomial has a zero in some extension field

Remark Theorem 27.19 can be rephrased as follows. If a field is of characteristic p, then it can be regarded as an extension field of

• Zp. If a field is of characteristic 0, then it can be regarded as an

extension field of Q. Theorem (29.3, Kronecker) Let F be a field, and f(x) be a nonconstant polynomial in F[x]. Then there exists an extension field E of F and an α ∈ E such that f(α) = 0.

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Every polynomial has a zero in some extension field

Remark Theorem 27.19 can be rephrased as follows. If a field is of characteristic p, then it can be regarded as an extension field of

• Zp. If a field is of characteristic 0, then it can be regarded as an

extension field of Q. Theorem (29.3, Kronecker) Let F be a field, and f(x) be a nonconstant polynomial in F[x]. Then there exists an extension field E of F and an α ∈ E such that f(α) = 0.

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Proof of Theorem 29.3

Proof.

• Let p(x) be an irreducible factor of f(x) over F, say

f(x) = p(x)g(x) for some g(x) ∈ F[x].

• By Theorems 27.9 and 27.25, F[x]/p(x) is a field.
• The field F is naturally embedded in F[x]/p(x) by

a → a + p(x) for a ∈ F. Thus, we may consider F[x]/p(x) as an extension field of F.

• Now let α = x + p(x) ∈ F[x]/p(x).
• Then we have

f(α) = f(x) + p(x) = p(x)g(x) + p(x) = p(x). That is, α is a zero of f(x) in F[x]/f(x).

• Instructor: Yifan Yang

Section 29 – Introduction to extension fields

Proof of Theorem 29.3

Proof.

• Let p(x) be an irreducible factor of f(x) over F, say

f(x) = p(x)g(x) for some g(x) ∈ F[x].

• By Theorems 27.9 and 27.25, F[x]/p(x) is a field.
• The field F is naturally embedded in F[x]/p(x) by

a → a + p(x) for a ∈ F. Thus, we may consider F[x]/p(x) as an extension field of F.

• Now let α = x + p(x) ∈ F[x]/p(x).
• Then we have

f(α) = f(x) + p(x) = p(x)g(x) + p(x) = p(x). That is, α is a zero of f(x) in F[x]/f(x).

• Instructor: Yifan Yang

Section 29 – Introduction to extension fields

Proof of Theorem 29.3

Proof.

• Let p(x) be an irreducible factor of f(x) over F, say

f(x) = p(x)g(x) for some g(x) ∈ F[x].

• By Theorems 27.9 and 27.25, F[x]/p(x) is a field.
• The field F is naturally embedded in F[x]/p(x) by

a → a + p(x) for a ∈ F. Thus, we may consider F[x]/p(x) as an extension field of F.

• Now let α = x + p(x) ∈ F[x]/p(x).
• Then we have

f(α) = f(x) + p(x) = p(x)g(x) + p(x) = p(x). That is, α is a zero of f(x) in F[x]/f(x).

• Instructor: Yifan Yang

Section 29 – Introduction to extension fields

Proof of Theorem 29.3

Proof.

• Let p(x) be an irreducible factor of f(x) over F, say

f(x) = p(x)g(x) for some g(x) ∈ F[x].

• By Theorems 27.9 and 27.25, F[x]/p(x) is a field.
• The field F is naturally embedded in F[x]/p(x) by

a → a + p(x) for a ∈ F. Thus, we may consider F[x]/p(x) as an extension field of F.

• Now let α = x + p(x) ∈ F[x]/p(x).
• Then we have

f(α) = f(x) + p(x) = p(x)g(x) + p(x) = p(x). That is, α is a zero of f(x) in F[x]/f(x).

• Instructor: Yifan Yang

Section 29 – Introduction to extension fields

Proof of Theorem 29.3

Proof.

• Let p(x) be an irreducible factor of f(x) over F, say

f(x) = p(x)g(x) for some g(x) ∈ F[x].

• By Theorems 27.9 and 27.25, F[x]/p(x) is a field.
• The field F is naturally embedded in F[x]/p(x) by

a → a + p(x) for a ∈ F. Thus, we may consider F[x]/p(x) as an extension field of F.

• Now let α = x + p(x) ∈ F[x]/p(x).
• Then we have

f(α) = f(x) + p(x) = p(x)g(x) + p(x) = p(x). That is, α is a zero of f(x) in F[x]/f(x).

• Instructor: Yifan Yang

Section 29 – Introduction to extension fields

Algebraic and transcendental elements

Definition An element α of an extension field E of a field F is algebraic

• ver F if f(α) = 0 for some nonzero polynomial f(x) ∈ F[x]. If

such a polynomial does not exist, then α is transcendental over F. Example

• √

2, i,

3

√ 3 are algebraic over Q since they are zeros of x2 − 2, x2 + 1, and x3 − 3, respectively.

• α =
• 1 +

√ 2 is algebraic over Q since it satisfies (α2 − 1)2 = 2.

• π and e are transcendental over Q, although the proof is

not easy.

• π is algebraic over R, as it is a zero of x − π ∈ R[x].

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Algebraic and transcendental elements

Definition An element α of an extension field E of a field F is algebraic

• ver F if f(α) = 0 for some nonzero polynomial f(x) ∈ F[x]. If

such a polynomial does not exist, then α is transcendental over F. Example

• √

2, i,

3

√ 3 are algebraic over Q since they are zeros of x2 − 2, x2 + 1, and x3 − 3, respectively.

• α =
• 1 +

√ 2 is algebraic over Q since it satisfies (α2 − 1)2 = 2.

• π and e are transcendental over Q, although the proof is

not easy.

• π is algebraic over R, as it is a zero of x − π ∈ R[x].

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Algebraic and transcendental elements

Definition An element α of an extension field E of a field F is algebraic

• ver F if f(α) = 0 for some nonzero polynomial f(x) ∈ F[x]. If

such a polynomial does not exist, then α is transcendental over F. Example

• √

2, i,

3

√ 3 are algebraic over Q since they are zeros of x2 − 2, x2 + 1, and x3 − 3, respectively.

• α =
• 1 +

√ 2 is algebraic over Q since it satisfies (α2 − 1)2 = 2.

• π and e are transcendental over Q, although the proof is

not easy.

• π is algebraic over R, as it is a zero of x − π ∈ R[x].

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Algebraic and transcendental elements

Definition An element α of an extension field E of a field F is algebraic

• ver F if f(α) = 0 for some nonzero polynomial f(x) ∈ F[x]. If

such a polynomial does not exist, then α is transcendental over F. Example

• √

2, i,

3

√ 3 are algebraic over Q since they are zeros of x2 − 2, x2 + 1, and x3 − 3, respectively.

• α =
• 1 +

√ 2 is algebraic over Q since it satisfies (α2 − 1)2 = 2.

• π and e are transcendental over Q, although the proof is

not easy.

• π is algebraic over R, as it is a zero of x − π ∈ R[x].

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Algebraic and transcendental elements

Definition An element α of an extension field E of a field F is algebraic

• ver F if f(α) = 0 for some nonzero polynomial f(x) ∈ F[x]. If

such a polynomial does not exist, then α is transcendental over F. Example

• √

2, i,

3

√ 3 are algebraic over Q since they are zeros of x2 − 2, x2 + 1, and x3 − 3, respectively.

• α =
• 1 +

√ 2 is algebraic over Q since it satisfies (α2 − 1)2 = 2.

• π and e are transcendental over Q, although the proof is

not easy.

• π is algebraic over R, as it is a zero of x − π ∈ R[x].

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Algebraic and transcendental numbers

Remark The last example shows that algebraicity and transcendence depend on the ground field. So whenever we talk about algebraicity and transcendence, we should specify which field we are talking about. Definition (29.11) An element α ∈ C is an algebraic number if α is algebraic over

• Q. A transcendental number is an element of C that is

transendental over Q. Example 1, √ 2,

• 1 +

√ 2 are algebraic numbers, while π and e are transcendental numbers.

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Algebraic and transcendental numbers

Remark The last example shows that algebraicity and transcendence depend on the ground field. So whenever we talk about algebraicity and transcendence, we should specify which field we are talking about. Definition (29.11) An element α ∈ C is an algebraic number if α is algebraic over

• Q. A transcendental number is an element of C that is

transendental over Q. Example 1, √ 2,

• 1 +

√ 2 are algebraic numbers, while π and e are transcendental numbers.

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Algebraic and transcendental numbers

Remark The last example shows that algebraicity and transcendence depend on the ground field. So whenever we talk about algebraicity and transcendence, we should specify which field we are talking about. Definition (29.11) An element α ∈ C is an algebraic number if α is algebraic over

• Q. A transcendental number is an element of C that is

transendental over Q. Example 1, √ 2,

• 1 +

√ 2 are algebraic numbers, while π and e are transcendental numbers.

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Algebraic and transcendental elements

Theorem (29.12) Let E be an extension field of a field F. Let α ∈ E. Then α is transcendental over F if and only if the evaluation homomorphism φα : F[x] → E is an isomorphism of F[x] with a subdomain of E. Proof. It suffices to show that α ∈ E is transcendental over F if and

• nly if φα is one-to-one, which is obvious.

Remark The theorem says that if α ∈ E is transcendental over F, then F[α] looks just like F[x].

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Algebraic and transcendental elements

Theorem (29.12) Let E be an extension field of a field F. Let α ∈ E. Then α is transcendental over F if and only if the evaluation homomorphism φα : F[x] → E is an isomorphism of F[x] with a subdomain of E. Proof. It suffices to show that α ∈ E is transcendental over F if and

• nly if φα is one-to-one, which is obvious.

Remark The theorem says that if α ∈ E is transcendental over F, then F[α] looks just like F[x].

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Algebraic and transcendental elements

Theorem (29.12) Let E be an extension field of a field F. Let α ∈ E. Then α is transcendental over F if and only if the evaluation homomorphism φα : F[x] → E is an isomorphism of F[x] with a subdomain of E. Proof. It suffices to show that α ∈ E is transcendental over F if and

• nly if φα is one-to-one, which is obvious.

Remark The theorem says that if α ∈ E is transcendental over F, then F[α] looks just like F[x].

Instructor: Yifan Yang Section 29 – Introduction to extension fields

The irreducible polynomial for α over F

• Assume that F ≤ E, and α ∈ E is algebraic over F.
• It is easy to check that the set Iα = {f(x) ∈ F[x] : f(α) = 0}

is an ideal of F[x].

• Since F[x] is a PID, Iα = pα(x) for some pα(x) ∈ F[x].

Lemma The polynomial pα(x) is irreducible.

Proof.

• If pα(x) = r(x)s(x), then we have r(α)s(α) = 0 and thus

r(α) = 0 or s(α) = 0.

• That is, r(x) ∈ pα(x) or s(x) ∈ pα(x).
• Since pα(x) is a nonzero polynomial of minimal degree in

Iα, we must have deg s(x) = 0 or deg r(x) = 0. That is,

• ne of r(x) and s(x) is a unit.
• Instructor: Yifan Yang

Section 29 – Introduction to extension fields

The irreducible polynomial for α over F

• Assume that F ≤ E, and α ∈ E is algebraic over F.
• It is easy to check that the set Iα = {f(x) ∈ F[x] : f(α) = 0}

is an ideal of F[x].

• Since F[x] is a PID, Iα = pα(x) for some pα(x) ∈ F[x].

Lemma The polynomial pα(x) is irreducible.

Proof.

• If pα(x) = r(x)s(x), then we have r(α)s(α) = 0 and thus

r(α) = 0 or s(α) = 0.

• That is, r(x) ∈ pα(x) or s(x) ∈ pα(x).
• Since pα(x) is a nonzero polynomial of minimal degree in

Iα, we must have deg s(x) = 0 or deg r(x) = 0. That is,

• ne of r(x) and s(x) is a unit.
• Instructor: Yifan Yang

Section 29 – Introduction to extension fields

The irreducible polynomial for α over F

• Assume that F ≤ E, and α ∈ E is algebraic over F.
• It is easy to check that the set Iα = {f(x) ∈ F[x] : f(α) = 0}

is an ideal of F[x].

• Since F[x] is a PID, Iα = pα(x) for some pα(x) ∈ F[x].

Lemma The polynomial pα(x) is irreducible.

Proof.

• If pα(x) = r(x)s(x), then we have r(α)s(α) = 0 and thus

r(α) = 0 or s(α) = 0.

• That is, r(x) ∈ pα(x) or s(x) ∈ pα(x).
• Since pα(x) is a nonzero polynomial of minimal degree in

Iα, we must have deg s(x) = 0 or deg r(x) = 0. That is,

• ne of r(x) and s(x) is a unit.
• Instructor: Yifan Yang

Section 29 – Introduction to extension fields

The irreducible polynomial for α over F

• Assume that F ≤ E, and α ∈ E is algebraic over F.
• It is easy to check that the set Iα = {f(x) ∈ F[x] : f(α) = 0}

is an ideal of F[x].

• Since F[x] is a PID, Iα = pα(x) for some pα(x) ∈ F[x].

Lemma The polynomial pα(x) is irreducible.

Proof.

• If pα(x) = r(x)s(x), then we have r(α)s(α) = 0 and thus

r(α) = 0 or s(α) = 0.

• That is, r(x) ∈ pα(x) or s(x) ∈ pα(x).
• Since pα(x) is a nonzero polynomial of minimal degree in

Iα, we must have deg s(x) = 0 or deg r(x) = 0. That is,

• ne of r(x) and s(x) is a unit.
• Instructor: Yifan Yang

Section 29 – Introduction to extension fields

The irreducible polynomial for α over F

• Assume that F ≤ E, and α ∈ E is algebraic over F.
• It is easy to check that the set Iα = {f(x) ∈ F[x] : f(α) = 0}

is an ideal of F[x].

• Since F[x] is a PID, Iα = pα(x) for some pα(x) ∈ F[x].

Lemma The polynomial pα(x) is irreducible.

Proof.

• If pα(x) = r(x)s(x), then we have r(α)s(α) = 0 and thus

r(α) = 0 or s(α) = 0.

• That is, r(x) ∈ pα(x) or s(x) ∈ pα(x).
• Since pα(x) is a nonzero polynomial of minimal degree in

Iα, we must have deg s(x) = 0 or deg r(x) = 0. That is,

• ne of r(x) and s(x) is a unit.
• Instructor: Yifan Yang

Section 29 – Introduction to extension fields

The irreducible polynomial for α over F

• Assume that F ≤ E, and α ∈ E is algebraic over F.
• It is easy to check that the set Iα = {f(x) ∈ F[x] : f(α) = 0}

is an ideal of F[x].

• Since F[x] is a PID, Iα = pα(x) for some pα(x) ∈ F[x].

Lemma The polynomial pα(x) is irreducible.

Proof.

• If pα(x) = r(x)s(x), then we have r(α)s(α) = 0 and thus

r(α) = 0 or s(α) = 0.

• That is, r(x) ∈ pα(x) or s(x) ∈ pα(x).
• Since pα(x) is a nonzero polynomial of minimal degree in

Iα, we must have deg s(x) = 0 or deg r(x) = 0. That is,

• ne of r(x) and s(x) is a unit.
• Instructor: Yifan Yang

Section 29 – Introduction to extension fields

The irreducible polynomial for α over F

• Assume that F ≤ E, and α ∈ E is algebraic over F.
• It is easy to check that the set Iα = {f(x) ∈ F[x] : f(α) = 0}

is an ideal of F[x].

• Since F[x] is a PID, Iα = pα(x) for some pα(x) ∈ F[x].

Lemma The polynomial pα(x) is irreducible.

Proof.

• If pα(x) = r(x)s(x), then we have r(α)s(α) = 0 and thus

r(α) = 0 or s(α) = 0.

• That is, r(x) ∈ pα(x) or s(x) ∈ pα(x).
• Since pα(x) is a nonzero polynomial of minimal degree in

Iα, we must have deg s(x) = 0 or deg r(x) = 0. That is,

• ne of r(x) and s(x) is a unit.
• Instructor: Yifan Yang

Section 29 – Introduction to extension fields

The irreducible polynomial for α over F

• Assume that F ≤ E, and α ∈ E is algebraic over F.
• It is easy to check that the set Iα = {f(x) ∈ F[x] : f(α) = 0}

is an ideal of F[x].

• Since F[x] is a PID, Iα = pα(x) for some pα(x) ∈ F[x].

Lemma The polynomial pα(x) is irreducible.

Proof.

• If pα(x) = r(x)s(x), then we have r(α)s(α) = 0 and thus

r(α) = 0 or s(α) = 0.

• That is, r(x) ∈ pα(x) or s(x) ∈ pα(x).
• Since pα(x) is a nonzero polynomial of minimal degree in

Iα, we must have deg s(x) = 0 or deg r(x) = 0. That is,

• ne of r(x) and s(x) is a unit.
• Instructor: Yifan Yang

Section 29 – Introduction to extension fields

The irreducible polynomial for α over F

Definition A polynomial in F[x] with leading coefficient 1 is called a monic polynomial. Definition The unique monic irreducible polynomial p(x) in Iα is called the irreducible polynomial for α over F, and will be denoted by irr(α, F). The degree of irr(α, F) is the degree of α over F, and is denoted by deg(α, F). Remark

• The irreducible polynomial irr(α, F) can also be defined as

the monic polynomial of smallest degree in F[x] having α as a zero.

• Again, when we speak of the irreducible polynomial of an

element, we need to specify which field we are talking about.

Instructor: Yifan Yang Section 29 – Introduction to extension fields

The irreducible polynomial for α over F

Definition A polynomial in F[x] with leading coefficient 1 is called a monic polynomial. Definition The unique monic irreducible polynomial p(x) in Iα is called the irreducible polynomial for α over F, and will be denoted by irr(α, F). The degree of irr(α, F) is the degree of α over F, and is denoted by deg(α, F). Remark

• The irreducible polynomial irr(α, F) can also be defined as

the monic polynomial of smallest degree in F[x] having α as a zero.

• Again, when we speak of the irreducible polynomial of an

element, we need to specify which field we are talking about.

Instructor: Yifan Yang Section 29 – Introduction to extension fields

The irreducible polynomial for α over F

Definition A polynomial in F[x] with leading coefficient 1 is called a monic polynomial. Definition The unique monic irreducible polynomial p(x) in Iα is called the irreducible polynomial for α over F, and will be denoted by irr(α, F). The degree of irr(α, F) is the degree of α over F, and is denoted by deg(α, F). Remark

• The irreducible polynomial irr(α, F) can also be defined as

the monic polynomial of smallest degree in F[x] having α as a zero.

• Again, when we speak of the irreducible polynomial of an

element, we need to specify which field we are talking about.

Instructor: Yifan Yang Section 29 – Introduction to extension fields

The irreducible polynomial for α over F

Definition A polynomial in F[x] with leading coefficient 1 is called a monic polynomial. Definition The unique monic irreducible polynomial p(x) in Iα is called the irreducible polynomial for α over F, and will be denoted by irr(α, F). The degree of irr(α, F) is the degree of α over F, and is denoted by deg(α, F). Remark

• The irreducible polynomial irr(α, F) can also be defined as

the monic polynomial of smallest degree in F[x] having α as a zero.

• Again, when we speak of the irreducible polynomial of an

element, we need to specify which field we are talking about.

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Examples

1

irr(1/ √ 2, Q) = x2 − 1/2.

2

irr(

4

√ 2, Q[ √ 2]) = x2 − √ 2.

3

The number α =

• 1 +

√ 2 satisfies (α2 − 1)2 = 2. We then check that x4 − 2x2 − 1 is irreducible. Thus, irr(α, Q) = x4 − 2x2 − 1.

4

irr(√π, Q(π)) = x2 − π.

5

Let F = Z3 and E = Z[i]/3. Then the irreducible polynomial for (1 + i) + 3 over Z3 is x2 + x + 2 ∈ Z3[x]. (We have (1 + i)2 + (1 + i) + 2 = 3 + 3i ∈ 3.)

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Examples

1

irr(1/ √ 2, Q) = x2 − 1/2.

2

irr(

4

√ 2, Q[ √ 2]) = x2 − √ 2.

3

The number α =

• 1 +

√ 2 satisfies (α2 − 1)2 = 2. We then check that x4 − 2x2 − 1 is irreducible. Thus, irr(α, Q) = x4 − 2x2 − 1.

4

irr(√π, Q(π)) = x2 − π.

5

Let F = Z3 and E = Z[i]/3. Then the irreducible polynomial for (1 + i) + 3 over Z3 is x2 + x + 2 ∈ Z3[x]. (We have (1 + i)2 + (1 + i) + 2 = 3 + 3i ∈ 3.)

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Examples

1

irr(1/ √ 2, Q) = x2 − 1/2.

2

irr(

4

√ 2, Q[ √ 2]) = x2 − √ 2.

3

The number α =

• 1 +

√ 2 satisfies (α2 − 1)2 = 2. We then check that x4 − 2x2 − 1 is irreducible. Thus, irr(α, Q) = x4 − 2x2 − 1.

4

irr(√π, Q(π)) = x2 − π.

5

Let F = Z3 and E = Z[i]/3. Then the irreducible polynomial for (1 + i) + 3 over Z3 is x2 + x + 2 ∈ Z3[x]. (We have (1 + i)2 + (1 + i) + 2 = 3 + 3i ∈ 3.)

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Examples

1

irr(1/ √ 2, Q) = x2 − 1/2.

2

irr(

4

√ 2, Q[ √ 2]) = x2 − √ 2.

3

The number α =

• 1 +

√ 2 satisfies (α2 − 1)2 = 2. We then check that x4 − 2x2 − 1 is irreducible. Thus, irr(α, Q) = x4 − 2x2 − 1.

4

irr(√π, Q(π)) = x2 − π.

5

Let F = Z3 and E = Z[i]/3. Then the irreducible polynomial for (1 + i) + 3 over Z3 is x2 + x + 2 ∈ Z3[x]. (We have (1 + i)2 + (1 + i) + 2 = 3 + 3i ∈ 3.)

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Examples

1

irr(1/ √ 2, Q) = x2 − 1/2.

2

irr(

4

√ 2, Q[ √ 2]) = x2 − √ 2.

3

The number α =

• 1 +

√ 2 satisfies (α2 − 1)2 = 2. We then check that x4 − 2x2 − 1 is irreducible. Thus, irr(α, Q) = x4 − 2x2 − 1.

4

irr(√π, Q(π)) = x2 − π.

5

Let F = Z3 and E = Z[i]/3. Then the irreducible polynomial for (1 + i) + 3 over Z3 is x2 + x + 2 ∈ Z3[x]. (We have (1 + i)2 + (1 + i) + 2 = 3 + 3i ∈ 3.)

Instructor: Yifan Yang Section 29 – Introduction to extension fields

α algebraic over F ⇒ F(α) = F[α]

Notation Let F be a field. The notation F[α] denotes the set {f(α) : f(x) ∈ F[x]}, while F(α) is the set {f(α)/g(α) : f(x), g(x) ∈ F[x], g(α) = 0}. Theorem Let F ≤ E, and α ∈ E be a nonzero element algebraic over F. Then F(α) = F[α] and is isomorphic to F[x]/irr(α, F). Theorem (29.18) Let E = F(α) be a simple extension of F. Assume that α is algebraic over F. Then any β ∈ E = F(α) can be uniquely expressed in the form β = b0 + b1α + · · · + bn−1αn−1, where n = deg irr(α, F).

Instructor: Yifan Yang Section 29 – Introduction to extension fields

α algebraic over F ⇒ F(α) = F[α]

Notation Let F be a field. The notation F[α] denotes the set {f(α) : f(x) ∈ F[x]}, while F(α) is the set {f(α)/g(α) : f(x), g(x) ∈ F[x], g(α) = 0}. Theorem Let F ≤ E, and α ∈ E be a nonzero element algebraic over F. Then F(α) = F[α] and is isomorphic to F[x]/irr(α, F). Theorem (29.18) Let E = F(α) be a simple extension of F. Assume that α is algebraic over F. Then any β ∈ E = F(α) can be uniquely expressed in the form β = b0 + b1α + · · · + bn−1αn−1, where n = deg irr(α, F).

Instructor: Yifan Yang Section 29 – Introduction to extension fields

α algebraic over F ⇒ F(α) = F[α]

Notation Let F be a field. The notation F[α] denotes the set {f(α) : f(x) ∈ F[x]}, while F(α) is the set {f(α)/g(α) : f(x), g(x) ∈ F[x], g(α) = 0}. Theorem Let F ≤ E, and α ∈ E be a nonzero element algebraic over F. Then F(α) = F[α] and is isomorphic to F[x]/irr(α, F). Theorem (29.18) Let E = F(α) be a simple extension of F. Assume that α is algebraic over F. Then any β ∈ E = F(α) can be uniquely expressed in the form β = b0 + b1α + · · · + bn−1αn−1, where n = deg irr(α, F).

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Proof of F(α) = F[α]

• It is clear that F[α] ⊂ F(α). We now show that every

element f(α)/g(α) ∈ F(α) with g(α) = 0 can be expressed as h(α) for some h(x) ∈ F[x].

• Since g(α) = 0, g(x) is not divisible by irr(α, F). This in

turn implies that gcd(g(x), irr(α, F)) = 1 because irr(α, F) is irreducible.

• Using the Euclidean algorithm, we can find p(x) and q(x)

such that p(x)g(x) + q(x) irr(α, F) = 1. Then we have p(α)g(α) = 1.

• Then we have

f(α) g(α) = f(α)p(α) g(α)p(α) = f(α)p(α) ∈ F[α].

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Proof of F(α) = F[α]

• It is clear that F[α] ⊂ F(α). We now show that every

element f(α)/g(α) ∈ F(α) with g(α) = 0 can be expressed as h(α) for some h(x) ∈ F[x].

• Since g(α) = 0, g(x) is not divisible by irr(α, F). This in

turn implies that gcd(g(x), irr(α, F)) = 1 because irr(α, F) is irreducible.

• Using the Euclidean algorithm, we can find p(x) and q(x)

such that p(x)g(x) + q(x) irr(α, F) = 1. Then we have p(α)g(α) = 1.

• Then we have

f(α) g(α) = f(α)p(α) g(α)p(α) = f(α)p(α) ∈ F[α].

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Proof of F(α) = F[α]

• It is clear that F[α] ⊂ F(α). We now show that every

element f(α)/g(α) ∈ F(α) with g(α) = 0 can be expressed as h(α) for some h(x) ∈ F[x].

• Since g(α) = 0, g(x) is not divisible by irr(α, F). This in

turn implies that gcd(g(x), irr(α, F)) = 1 because irr(α, F) is irreducible.

• Using the Euclidean algorithm, we can find p(x) and q(x)

such that p(x)g(x) + q(x) irr(α, F) = 1. Then we have p(α)g(α) = 1.

• Then we have

f(α) g(α) = f(α)p(α) g(α)p(α) = f(α)p(α) ∈ F[α].

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Proof of F(α) = F[α]

• It is clear that F[α] ⊂ F(α). We now show that every

element f(α)/g(α) ∈ F(α) with g(α) = 0 can be expressed as h(α) for some h(x) ∈ F[x].

• Since g(α) = 0, g(x) is not divisible by irr(α, F). This in

turn implies that gcd(g(x), irr(α, F)) = 1 because irr(α, F) is irreducible.

• Using the Euclidean algorithm, we can find p(x) and q(x)

such that p(x)g(x) + q(x) irr(α, F) = 1. Then we have p(α)g(α) = 1.

• Then we have

f(α) g(α) = f(α)p(α) g(α)p(α) = f(α)p(α) ∈ F[α].

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Proof of F(α) = F[α]

• It is clear that F[α] ⊂ F(α). We now show that every

element f(α)/g(α) ∈ F(α) with g(α) = 0 can be expressed as h(α) for some h(x) ∈ F[x].

• Since g(α) = 0, g(x) is not divisible by irr(α, F). This in

turn implies that gcd(g(x), irr(α, F)) = 1 because irr(α, F) is irreducible.

• Using the Euclidean algorithm, we can find p(x) and q(x)

such that p(x)g(x) + q(x) irr(α, F) = 1. Then we have p(α)g(α) = 1.

• Then we have

f(α) g(α) = f(α)p(α) g(α)p(α) = f(α)p(α) ∈ F[α].

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Proof of F(α) = F[α]

• It is clear that F[α] ⊂ F(α). We now show that every

element f(α)/g(α) ∈ F(α) with g(α) = 0 can be expressed as h(α) for some h(x) ∈ F[x].

• Since g(α) = 0, g(x) is not divisible by irr(α, F). This in

turn implies that gcd(g(x), irr(α, F)) = 1 because irr(α, F) is irreducible.

• Using the Euclidean algorithm, we can find p(x) and q(x)

such that p(x)g(x) + q(x) irr(α, F) = 1. Then we have p(α)g(α) = 1.

• Then we have

f(α) g(α) = f(α)p(α) g(α)p(α) = f(α)p(α) ∈ F[α].

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Proof of F[α] ≃ F[x]/irr(α, F)

• Consider the evaluation homomorphism φα : F[x] → F[α]

given by φα(f(x)) = f(α). By the isomorphism theorem (Theorem 34.2 in case of groups), F[x]/Ker(φα) ≃ Im(φα).

• The kernel consists of polynomials f over F satisfying

f(α) = 0. Thus, Ker(φα) = irr(α, F).

• The homomorphism φα is clear onto.
• Then the isomorphism theorem says

F[x]/irr(α, F) ≃ F[α]. .

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Proof of F[α] ≃ F[x]/irr(α, F)

• Consider the evaluation homomorphism φα : F[x] → F[α]

given by φα(f(x)) = f(α). By the isomorphism theorem (Theorem 34.2 in case of groups), F[x]/Ker(φα) ≃ Im(φα).

• The kernel consists of polynomials f over F satisfying

f(α) = 0. Thus, Ker(φα) = irr(α, F).

• The homomorphism φα is clear onto.
• Then the isomorphism theorem says

F[x]/irr(α, F) ≃ F[α]. .

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Proof of F[α] ≃ F[x]/irr(α, F)

• Consider the evaluation homomorphism φα : F[x] → F[α]

given by φα(f(x)) = f(α). By the isomorphism theorem (Theorem 34.2 in case of groups), F[x]/Ker(φα) ≃ Im(φα).

• The kernel consists of polynomials f over F satisfying

f(α) = 0. Thus, Ker(φα) = irr(α, F).

• The homomorphism φα is clear onto.
• Then the isomorphism theorem says

F[x]/irr(α, F) ≃ F[α]. .

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Proof of F[α] ≃ F[x]/irr(α, F)

• Consider the evaluation homomorphism φα : F[x] → F[α]

given by φα(f(x)) = f(α). By the isomorphism theorem (Theorem 34.2 in case of groups), F[x]/Ker(φα) ≃ Im(φα).

• The kernel consists of polynomials f over F satisfying

f(α) = 0. Thus, Ker(φα) = irr(α, F).

• The homomorphism φα is clear onto.
• Then the isomorphism theorem says

F[x]/irr(α, F) ≃ F[α]. .

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Proof of F[α] ≃ F[x]/irr(α, F)

• Consider the evaluation homomorphism φα : F[x] → F[α]

given by φα(f(x)) = f(α). By the isomorphism theorem (Theorem 34.2 in case of groups), F[x]/Ker(φα) ≃ Im(φα).

• The kernel consists of polynomials f over F satisfying

f(α) = 0. Thus, Ker(φα) = irr(α, F).

• The homomorphism φα is clear onto.
• Then the isomorphism theorem says

F[x]/irr(α, F) ≃ F[α]. .

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Proof of F[α] ≃ F[x]/irr(α, F)

• Consider the evaluation homomorphism φα : F[x] → F[α]

given by φα(f(x)) = f(α). By the isomorphism theorem (Theorem 34.2 in case of groups), F[x]/Ker(φα) ≃ Im(φα).

• The kernel consists of polynomials f over F satisfying

f(α) = 0. Thus, Ker(φα) = irr(α, F).

• The homomorphism φα is clear onto.
• Then the isomorphism theorem says

F[x]/irr(α, F) ≃ F[α]. .

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Simple extensions

Proof of Theorem 29.18. By the previous theorem, we have F(α) = F[α] since α is algebraic over F. Thus, β = f(α) for some f(x) ∈ f[x]. By the division algorithm, there exists q(x) and r(x) such that f(x) = q(x) irr(α, F) + r(x) with r(x) = 0 or deg r(x) < n. Then r(α) = f(α) = β. The uniqueness is guaranteed by Theorem 23.1 (division algorithm). Definition Let F ≤ E. If there exists α ∈ E such that E = F(a), then E = F(α) is a simple extension of F.

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Simple extensions

Proof of Theorem 29.18. By the previous theorem, we have F(α) = F[α] since α is algebraic over F. Thus, β = f(α) for some f(x) ∈ f[x]. By the division algorithm, there exists q(x) and r(x) such that f(x) = q(x) irr(α, F) + r(x) with r(x) = 0 or deg r(x) < n. Then r(α) = f(α) = β. The uniqueness is guaranteed by Theorem 23.1 (division algorithm). Definition Let F ≤ E. If there exists α ∈ E such that E = F(a), then E = F(α) is a simple extension of F.

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Simple extensions

Proof of Theorem 29.18. By the previous theorem, we have F(α) = F[α] since α is algebraic over F. Thus, β = f(α) for some f(x) ∈ f[x]. By the division algorithm, there exists q(x) and r(x) such that f(x) = q(x) irr(α, F) + r(x) with r(x) = 0 or deg r(x) < n. Then r(α) = f(α) = β. The uniqueness is guaranteed by Theorem 23.1 (division algorithm). Definition Let F ≤ E. If there exists α ∈ E such that E = F(a), then E = F(α) is a simple extension of F.

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Simple extensions

Proof of Theorem 29.18. By the previous theorem, we have F(α) = F[α] since α is algebraic over F. Thus, β = f(α) for some f(x) ∈ f[x]. By the division algorithm, there exists q(x) and r(x) such that f(x) = q(x) irr(α, F) + r(x) with r(x) = 0 or deg r(x) < n. Then r(α) = f(α) = β. The uniqueness is guaranteed by Theorem 23.1 (division algorithm). Definition Let F ≤ E. If there exists α ∈ E such that E = F(a), then E = F(α) is a simple extension of F.

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Simple extensions

Proof of Theorem 29.18. By the previous theorem, we have F(α) = F[α] since α is algebraic over F. Thus, β = f(α) for some f(x) ∈ f[x]. By the division algorithm, there exists q(x) and r(x) such that f(x) = q(x) irr(α, F) + r(x) with r(x) = 0 or deg r(x) < n. Then r(α) = f(α) = β. The uniqueness is guaranteed by Theorem 23.1 (division algorithm). Definition Let F ≤ E. If there exists α ∈ E such that E = F(a), then E = F(α) is a simple extension of F.

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Simple extensions

Proof of Theorem 29.18. By the previous theorem, we have F(α) = F[α] since α is algebraic over F. Thus, β = f(α) for some f(x) ∈ f[x]. By the division algorithm, there exists q(x) and r(x) such that f(x) = q(x) irr(α, F) + r(x) with r(x) = 0 or deg r(x) < n. Then r(α) = f(α) = β. The uniqueness is guaranteed by Theorem 23.1 (division algorithm). Definition Let F ≤ E. If there exists α ∈ E such that E = F(a), then E = F(α) is a simple extension of F.

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Example

• Problem. Let α be a zero of x3 + x + 1 in Q[x]. Write

β = (2α2 + 3α − 1)/(α2 − α + 1) in the form a0 + a1α + a2α2.

• Solution. We first use the Euclidean algorithm to find

polynomials g(x) and h(x) such that (x3 + x + 1)g(x) + (x2 − x + 1)h(x) = 1. We have x3 + x + 1 = (x + 1)(x2 − x + 1) + x x2 − x + 1 = (x − 1)x + 1 Then 1 = (x2 −x +1)−(x −1)x = x2(x2 −x +1)−(x −1)(x3 +x +1). It follows that 2α2 + 3α − 1 α2 − α + 1 = (2α2 + 3α − 1)α2 (α2 − α + 1)α2 = 2α4 + 3α3 − α2 = (2α + 3)(α3 + α + 1) − (3α2 + 5α + 3) = −3α2 − 5α − 3.

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Example

• Problem. Let α be a zero of x3 + x + 1 in Q[x]. Write

β = (2α2 + 3α − 1)/(α2 − α + 1) in the form a0 + a1α + a2α2.

• Solution. We first use the Euclidean algorithm to find

polynomials g(x) and h(x) such that (x3 + x + 1)g(x) + (x2 − x + 1)h(x) = 1. We have x3 + x + 1 = (x + 1)(x2 − x + 1) + x x2 − x + 1 = (x − 1)x + 1 Then 1 = (x2 −x +1)−(x −1)x = x2(x2 −x +1)−(x −1)(x3 +x +1). It follows that 2α2 + 3α − 1 α2 − α + 1 = (2α2 + 3α − 1)α2 (α2 − α + 1)α2 = 2α4 + 3α3 − α2 = (2α + 3)(α3 + α + 1) − (3α2 + 5α + 3) = −3α2 − 5α − 3.

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Example

• Problem. Let α be a zero of x3 + x + 1 in Q[x]. Write

β = (2α2 + 3α − 1)/(α2 − α + 1) in the form a0 + a1α + a2α2.

• Solution. We first use the Euclidean algorithm to find

polynomials g(x) and h(x) such that (x3 + x + 1)g(x) + (x2 − x + 1)h(x) = 1. We have x3 + x + 1 = (x + 1)(x2 − x + 1) + x x2 − x + 1 = (x − 1)x + 1 Then 1 = (x2 −x +1)−(x −1)x = x2(x2 −x +1)−(x −1)(x3 +x +1). It follows that 2α2 + 3α − 1 α2 − α + 1 = (2α2 + 3α − 1)α2 (α2 − α + 1)α2 = 2α4 + 3α3 − α2 = (2α + 3)(α3 + α + 1) − (3α2 + 5α + 3) = −3α2 − 5α − 3.

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Example

• Problem. Let α be a zero of x3 + x + 1 in Q[x]. Write

β = (2α2 + 3α − 1)/(α2 − α + 1) in the form a0 + a1α + a2α2.

• Solution. We first use the Euclidean algorithm to find

polynomials g(x) and h(x) such that (x3 + x + 1)g(x) + (x2 − x + 1)h(x) = 1. We have x3 + x + 1 = (x + 1)(x2 − x + 1) + x x2 − x + 1 = (x − 1)x + 1 Then 1 = (x2 −x +1)−(x −1)x = x2(x2 −x +1)−(x −1)(x3 +x +1). It follows that 2α2 + 3α − 1 α2 − α + 1 = (2α2 + 3α − 1)α2 (α2 − α + 1)α2 = 2α4 + 3α3 − α2 = (2α + 3)(α3 + α + 1) − (3α2 + 5α + 3) = −3α2 − 5α − 3.

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Example

• Problem. Let α be a zero of x3 + x + 1 in Q[x]. Write

β = (2α2 + 3α − 1)/(α2 − α + 1) in the form a0 + a1α + a2α2.

• Solution. We first use the Euclidean algorithm to find

polynomials g(x) and h(x) such that (x3 + x + 1)g(x) + (x2 − x + 1)h(x) = 1. We have x3 + x + 1 = (x + 1)(x2 − x + 1) + x x2 − x + 1 = (x − 1)x + 1 Then 1 = (x2 −x +1)−(x −1)x = x2(x2 −x +1)−(x −1)(x3 +x +1). It follows that 2α2 + 3α − 1 α2 − α + 1 = (2α2 + 3α − 1)α2 (α2 − α + 1)α2 = 2α4 + 3α3 − α2 = (2α + 3)(α3 + α + 1) − (3α2 + 5α + 3) = −3α2 − 5α − 3.

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Example

• Problem. Let α be a zero of x2 + x + 1 ∈ Z2[x] in some

extension field. Write β = α/(α + 1) in the form a0 + a1α, where a0, a1 ∈ Z2[x].

Solution.

• We have x2 + x + 1 = x(x + 1) + 1. Thus, (α + 1)−1 = α.
• Then, β = α2 = α + 1.

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Example

• Problem. Let α be a zero of x2 + x + 1 ∈ Z2[x] in some

extension field. Write β = α/(α + 1) in the form a0 + a1α, where a0, a1 ∈ Z2[x].

Solution.

• We have x2 + x + 1 = x(x + 1) + 1. Thus, (α + 1)−1 = α.
• Then, β = α2 = α + 1.

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Example

• Problem. Let α be a zero of x2 + x + 1 ∈ Z2[x] in some

extension field. Write β = α/(α + 1) in the form a0 + a1α, where a0, a1 ∈ Z2[x].

Solution.

• We have x2 + x + 1 = x(x + 1) + 1. Thus, (α + 1)−1 = α.
• Then, β = α2 = α + 1.

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Example, continued

• Note that, by Theorem 29.18, every element in Z2[α] can

be uniquely written as a0 + a1α with a0, a1 ∈ Z2.

• Thus, Z2[α] is a field of 4 elements. (Each ai has two

possible values.)

• The characteristic of Z2[α] is clear 2.
• The addition and multiplication tables are

+ 1 α 1 + α 1 α 1 + α 1 1 1 + α α α α 1 + α 1 1 + α 1 + α α 1

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Example, continued

• Note that, by Theorem 29.18, every element in Z2[α] can

be uniquely written as a0 + a1α with a0, a1 ∈ Z2.

• Thus, Z2[α] is a field of 4 elements. (Each ai has two

possible values.)

• The characteristic of Z2[α] is clear 2.
• The addition and multiplication tables are

+ 1 α 1 + α 1 α 1 + α 1 1 1 + α α α α 1 + α 1 1 + α 1 + α α 1

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Example, continued

• Note that, by Theorem 29.18, every element in Z2[α] can

be uniquely written as a0 + a1α with a0, a1 ∈ Z2.

• Thus, Z2[α] is a field of 4 elements. (Each ai has two

possible values.)

• The characteristic of Z2[α] is clear 2.
• The addition and multiplication tables are

+ 1 α 1 + α 1 α 1 + α 1 1 1 + α α α α 1 + α 1 1 + α 1 + α α 1

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Example, continued

• Note that, by Theorem 29.18, every element in Z2[α] can

be uniquely written as a0 + a1α with a0, a1 ∈ Z2.

• Thus, Z2[α] is a field of 4 elements. (Each ai has two

possible values.)

• The characteristic of Z2[α] is clear 2.
• The addition and multiplication tables are

+ 1 α 1 + α 1 α 1 + α 1 1 1 + α α α α 1 + α 1 1 + α 1 + α α 1

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Example, continued

× 1 α 1 + α 1 1 α 1 + α α α 1 + α 1 1 + α 1 + α 1 α Again this field is not isomorphic to Z4, which is not even an integral domain. Warning: In exams, if you are asked to construct a field of pn elements and your answer is Zpn, you will receive double penalty!!

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Example, continued

× 1 α 1 + α 1 1 α 1 + α α α 1 + α 1 1 + α 1 + α 1 α Again this field is not isomorphic to Z4, which is not even an integral domain. Warning: In exams, if you are asked to construct a field of pn elements and your answer is Zpn, you will receive double penalty!!

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Example, continued

× 1 α 1 + α 1 1 α 1 + α α α 1 + α 1 1 + α 1 + α 1 α Again this field is not isomorphic to Z4, which is not even an integral domain. Warning: In exams, if you are asked to construct a field of pn elements and your answer is Zpn, you will receive double penalty!!

Instructor: Yifan Yang Section 29 – Introduction to extension fields

In-class exercise

Let α be a zero of x3 + x + 1 ∈ Q[x]. Express the following elements of Q[α] in the form a0 + a1α + a2α2, ai ∈ Q.

1

(α2 + 2)/(α + 2).

2

(2α2 + α)/(α2 + α + 1). Let α be a zero of x3 + x + 1 ∈ Z2[x] in some extension field of

• Z2. Express the following elements of Z2[α] in the form

a0 + a1α + a2α2, ai ∈ Z2.

1

(α2 + 1)/(α2 + α).

2

(α2 + α + 1)/(α + 1).

3

Give the addition table of Z2[α].

4

Give the multiplication table of Z2[α].

Instructor: Yifan Yang Section 29 – Introduction to extension fields

Homework

Problems 6, 8, 12, 16, 18, 25, 26, 29, 30 of Section 29.

Instructor: Yifan Yang Section 29 – Introduction to extension fields