Field Extensions and Splitting Fields Bernd Schr oder logo1 Bernd - - PowerPoint PPT Presentation

logo1 Subfields Splitting Fields Adjoining Elements

Field Extensions and Splitting Fields

Bernd Schr¨

• der

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Introduction

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Introduction

• 1. What do we do when we “use a formula”?

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Introduction

• 1. What do we do when we “use a formula”? We take

coefficients and perform algebraic operations (including root extractions) with them.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Introduction

• 1. What do we do when we “use a formula”? We take

coefficients and perform algebraic operations (including root extractions) with them.

• 2. It turns out that most complex numbers cannot be reached

that way.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Introduction

• 1. What do we do when we “use a formula”? We take

coefficients and perform algebraic operations (including root extractions) with them.

• 2. It turns out that most complex numbers cannot be reached

that way.

• 3. So it makes sense to focus on fields that contain “just

enough” to allow the operations we need.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Introduction

• 1. What do we do when we “use a formula”? We take

coefficients and perform algebraic operations (including root extractions) with them.

• 2. It turns out that most complex numbers cannot be reached

that way.

• 3. So it makes sense to focus on fields that contain “just

enough” to allow the operations we need.

• 4. For a “formula” to solve p(x) = 0 with p ∈ Z[x]

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Introduction

• 1. What do we do when we “use a formula”? We take

coefficients and perform algebraic operations (including root extractions) with them.

• 2. It turns out that most complex numbers cannot be reached

that way.

• 3. So it makes sense to focus on fields that contain “just

enough” to allow the operations we need.

• 4. For a “formula” to solve p(x) = 0 with p ∈ Z[x], we start

with Q.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Introduction

• 1. What do we do when we “use a formula”? We take

coefficients and perform algebraic operations (including root extractions) with them.

• 2. It turns out that most complex numbers cannot be reached

that way.

• 3. So it makes sense to focus on fields that contain “just

enough” to allow the operations we need.

• 4. For a “formula” to solve p(x) = 0 with p ∈ Z[x], we start

with Q.

• 5. Every time we extract a root, we may need to enlarge our

scope.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Introduction

• 1. What do we do when we “use a formula”? We take

coefficients and perform algebraic operations (including root extractions) with them.

• 2. It turns out that most complex numbers cannot be reached

that way.

• 3. So it makes sense to focus on fields that contain “just

enough” to allow the operations we need.

• 4. For a “formula” to solve p(x) = 0 with p ∈ Z[x], we start

with Q.

• 5. Every time we extract a root, we may need to enlarge our

scope.

• 6. This presentation makes the statement in 5 more precise.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Definition.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (E,+,·) be a field

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (E,+,·) be a field and let F ⊆ E be a subset

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (E,+,·) be a field and let F ⊆ E be a subset so

that the restricted operations +|F×F and ·|F×F both map into F

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (E,+,·) be a field and let F ⊆ E be a subset so

that the restricted operations +|F×F and ·|F×F both map into F and so that (F,+,·) is a field whose identity element for addition is 0 ∈ E and whose identity element with respect to multiplication is 1 ∈ E.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (E,+,·) be a field and let F ⊆ E be a subset so

that the restricted operations +|F×F and ·|F×F both map into F and so that (F,+,·) is a field whose identity element for addition is 0 ∈ E and whose identity element with respect to multiplication is 1 ∈ E. Then F is called a subfield of E

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (E,+,·) be a field and let F ⊆ E be a subset so

that the restricted operations +|F×F and ·|F×F both map into F and so that (F,+,·) is a field whose identity element for addition is 0 ∈ E and whose identity element with respect to multiplication is 1 ∈ E. Then F is called a subfield of E and E is called an extension of F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (E,+,·) be a field and let F ⊆ E be a subset so

that the restricted operations +|F×F and ·|F×F both map into F and so that (F,+,·) is a field whose identity element for addition is 0 ∈ E and whose identity element with respect to multiplication is 1 ∈ E. Then F is called a subfield of E and E is called an extension of F. We denote this situation more briefly by F ⊆ E.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Definition.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field of characteristic zero

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field of characteristic zero and let

p ∈ F[x] be a polynomial of positive degree.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field of characteristic zero and let

p ∈ F[x] be a polynomial of positive degree. The equation p(x) = 0 is solvable by radicals iff all its solutions can be calculated from its coefficients in a finite number of steps

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field of characteristic zero and let

p ∈ F[x] be a polynomial of positive degree. The equation p(x) = 0 is solvable by radicals iff all its solutions can be calculated from its coefficients in a finite number of steps using field operations

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field of characteristic zero and let

p ∈ F[x] be a polynomial of positive degree. The equation p(x) = 0 is solvable by radicals iff all its solutions can be calculated from its coefficients in a finite number of steps using field operations (addition, multiplication, additive and multiplicative inversion)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field of characteristic zero and let

p ∈ F[x] be a polynomial of positive degree. The equation p(x) = 0 is solvable by radicals iff all its solutions can be calculated from its coefficients in a finite number of steps using field operations (addition, multiplication, additive and multiplicative inversion) and root extractions.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field of characteristic zero and let

p ∈ F[x] be a polynomial of positive degree. The equation p(x) = 0 is solvable by radicals iff all its solutions can be calculated from its coefficients in a finite number of steps using field operations (addition, multiplication, additive and multiplicative inversion) and root extractions. The root extractions are allowed to yield elements that are not in F, but in an extension field E of F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field of characteristic zero and let

p ∈ F[x] be a polynomial of positive degree. The equation p(x) = 0 is solvable by radicals iff all its solutions can be calculated from its coefficients in a finite number of steps using field operations (addition, multiplication, additive and multiplicative inversion) and root extractions. The root extractions are allowed to yield elements that are not in F, but in an extension field E of F. Note that solvability by radicals does not mean there is a general formula.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field of characteristic zero and let

p ∈ F[x] be a polynomial of positive degree. The equation p(x) = 0 is solvable by radicals iff all its solutions can be calculated from its coefficients in a finite number of steps using field operations (addition, multiplication, additive and multiplicative inversion) and root extractions. The root extractions are allowed to yield elements that are not in F, but in an extension field E of F. Note that solvability by radicals does not mean there is a general formula. It means that there is some way to express the zeros of the polynomial under investigation.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Proposition.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proposition. Let (E,+,·) be a field and let {Fj}j∈J be a family
• f subfields of E.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proposition. Let (E,+,·) be a field and let {Fj}j∈J be a family
• f subfields of E. Then
• j∈J

Fj is a subfield of E.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proposition. Let (E,+,·) be a field and let {Fj}j∈J be a family
• f subfields of E. Then
• j∈J

Fj is a subfield of E. Proof.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proposition. Let (E,+,·) be a field and let {Fj}j∈J be a family
• f subfields of E. Then
• j∈J

Fj is a subfield of E.

• Proof. By assumption, 0,1 ∈
• j∈J

Fj.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proposition. Let (E,+,·) be a field and let {Fj}j∈J be a family
• f subfields of E. Then
• j∈J

Fj is a subfield of E.

• Proof. By assumption, 0,1 ∈
• j∈J
• Fj. Because the Fj are

subfields, sums and products of elements of Fj are in Fj, too.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proposition. Let (E,+,·) be a field and let {Fj}j∈J be a family
• f subfields of E. Then
• j∈J

Fj is a subfield of E.

• Proof. By assumption, 0,1 ∈
• j∈J
• Fj. Because the Fj are

subfields, sums and products of elements of Fj are in Fj, too. The binary operations on

• j∈J

Fj are the restrictions of the binary

• perations in E

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proposition. Let (E,+,·) be a field and let {Fj}j∈J be a family
• f subfields of E. Then
• j∈J

Fj is a subfield of E.

• Proof. By assumption, 0,1 ∈
• j∈J
• Fj. Because the Fj are

subfields, sums and products of elements of Fj are in Fj, too. The binary operations on

• j∈J

Fj are the restrictions of the binary

• perations in E and they map into
• j∈J

Fj.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proposition. Let (E,+,·) be a field and let {Fj}j∈J be a family
• f subfields of E. Then
• j∈J

Fj is a subfield of E.

• Proof. By assumption, 0,1 ∈
• j∈J
• Fj. Because the Fj are

subfields, sums and products of elements of Fj are in Fj, too. The binary operations on

• j∈J

Fj are the restrictions of the binary

• perations in E and they map into
• j∈J
• Fj. Moreover

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proposition. Let (E,+,·) be a field and let {Fj}j∈J be a family
• f subfields of E. Then
• j∈J

Fj is a subfield of E.

• Proof. By assumption, 0,1 ∈
• j∈J
• Fj. Because the Fj are

subfields, sums and products of elements of Fj are in Fj, too. The binary operations on

• j∈J

Fj are the restrictions of the binary

• perations in E and they map into
• j∈J
• Fj. Moreover,

associativity

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proposition. Let (E,+,·) be a field and let {Fj}j∈J be a family
• f subfields of E. Then
• j∈J

Fj is a subfield of E.

• Proof. By assumption, 0,1 ∈
• j∈J
• Fj. Because the Fj are

subfields, sums and products of elements of Fj are in Fj, too. The binary operations on

• j∈J

Fj are the restrictions of the binary

• perations in E and they map into
• j∈J
• Fj. Moreover,

associativity, commutativity

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proposition. Let (E,+,·) be a field and let {Fj}j∈J be a family
• f subfields of E. Then
• j∈J

Fj is a subfield of E.

• Proof. By assumption, 0,1 ∈
• j∈J
• Fj. Because the Fj are

subfields, sums and products of elements of Fj are in Fj, too. The binary operations on

• j∈J

Fj are the restrictions of the binary

• perations in E and they map into
• j∈J
• Fj. Moreover,

associativity, commutativity and distributivity of multiplication

• ver addition hold in
• j∈J

Fj, too.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proposition. Let (E,+,·) be a field and let {Fj}j∈J be a family
• f subfields of E. Then
• j∈J

Fj is a subfield of E.

• Proof. By assumption, 0,1 ∈
• j∈J
• Fj. Because the Fj are

subfields, sums and products of elements of Fj are in Fj, too. The binary operations on

• j∈J

Fj are the restrictions of the binary

• perations in E and they map into
• j∈J
• Fj. Moreover,

associativity, commutativity and distributivity of multiplication

• ver addition hold in
• j∈J

Fj, too. 0 and 1 are identity elements for addition and multiplication in E.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proposition. Let (E,+,·) be a field and let {Fj}j∈J be a family
• f subfields of E. Then
• j∈J

Fj is a subfield of E.

• Proof. By assumption, 0,1 ∈
• j∈J
• Fj. Because the Fj are

subfields, sums and products of elements of Fj are in Fj, too. The binary operations on

• j∈J

Fj are the restrictions of the binary

• perations in E and they map into
• j∈J
• Fj. Moreover,

associativity, commutativity and distributivity of multiplication

• ver addition hold in
• j∈J

Fj, too. 0 and 1 are identity elements for addition and multiplication in E. Hence they are the identity elements for addition and multiplication in

• j∈J

Fj.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Proof (concl.).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Proof (concl.). For inverses, let x ∈

• j∈J

Fj.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Proof (concl.). For inverses, let x ∈

• j∈J
• Fj. Then x has a unique

additive inverse −x in E.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Proof (concl.). For inverses, let x ∈

• j∈J
• Fj. Then x has a unique

additive inverse −x in E. Let j ∈ J and let y be the additive inverse of x in Fj.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Proof (concl.). For inverses, let x ∈

• j∈J
• Fj. Then x has a unique

additive inverse −x in E. Let j ∈ J and let y be the additive inverse of x in Fj. Then y

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Proof (concl.). For inverses, let x ∈

• j∈J
• Fj. Then x has a unique

additive inverse −x in E. Let j ∈ J and let y be the additive inverse of x in Fj. Then y = y+0

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Proof (concl.). For inverses, let x ∈

• j∈J
• Fj. Then x has a unique

additive inverse −x in E. Let j ∈ J and let y be the additive inverse of x in Fj. Then y = y+0 = y+

• x+(−x)
• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Proof (concl.). For inverses, let x ∈

• j∈J
• Fj. Then x has a unique

additive inverse −x in E. Let j ∈ J and let y be the additive inverse of x in Fj. Then y = y+0 = y+

• x+(−x)
• = (y+x)+(−x)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Proof (concl.). For inverses, let x ∈

• j∈J
• Fj. Then x has a unique

additive inverse −x in E. Let j ∈ J and let y be the additive inverse of x in Fj. Then y = y+0 = y+

• x+(−x)
• = (y+x)+(−x) = 0+(−x)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Proof (concl.). For inverses, let x ∈

• j∈J
• Fj. Then x has a unique

additive inverse −x in E. Let j ∈ J and let y be the additive inverse of x in Fj. Then y = y+0 = y+

• x+(−x)
• = (y+x)+(−x) = 0+(−x) = −x.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Proof (concl.). For inverses, let x ∈

• j∈J
• Fj. Then x has a unique

additive inverse −x in E. Let j ∈ J and let y be the additive inverse of x in Fj. Then y = y+0 = y+

• x+(−x)
• = (y+x)+(−x) = 0+(−x) = −x.

Because j ∈ J was arbitrary, we conclude that −x ∈

• j∈J

Fj.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Proof (concl.). For inverses, let x ∈

• j∈J
• Fj. Then x has a unique

additive inverse −x in E. Let j ∈ J and let y be the additive inverse of x in Fj. Then y = y+0 = y+

• x+(−x)
• = (y+x)+(−x) = 0+(−x) = −x.

Because j ∈ J was arbitrary, we conclude that −x ∈

• j∈J
• Fj. Thus
• j∈J

Fj contains additive inverses.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Proof (concl.). For inverses, let x ∈

• j∈J
• Fj. Then x has a unique

additive inverse −x in E. Let j ∈ J and let y be the additive inverse of x in Fj. Then y = y+0 = y+

• x+(−x)
• = (y+x)+(−x) = 0+(−x) = −x.

Because j ∈ J was arbitrary, we conclude that −x ∈

• j∈J
• Fj. Thus
• j∈J

Fj contains additive inverses. Multiplicative inverses are handled similarly.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Proof (concl.). For inverses, let x ∈

• j∈J
• Fj. Then x has a unique

additive inverse −x in E. Let j ∈ J and let y be the additive inverse of x in Fj. Then y = y+0 = y+

• x+(−x)
• = (y+x)+(−x) = 0+(−x) = −x.

Because j ∈ J was arbitrary, we conclude that −x ∈

• j∈J
• Fj. Thus
• j∈J

Fj contains additive inverses. Multiplicative inverses are handled similarly.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Definition.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field, let p ∈ F[x] be a polynomial
• ver F and let E be an extension of F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field, let p ∈ F[x] be a polynomial
• ver F and let E be an extension of F. Then f splits in the

extension field E ⊇ F iff p can be factored into linear factors with coefficients in E[x].

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field, let p ∈ F[x] be a polynomial
• ver F and let E be an extension of F. Then f splits in the

extension field E ⊇ F iff p can be factored into linear factors with coefficients in E[x]. Now let F be a field, let p ∈ F[x] and let E be an extension field in which p splits.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field, let p ∈ F[x] be a polynomial
• ver F and let E be an extension of F. Then f splits in the

extension field E ⊇ F iff p can be factored into linear factors with coefficients in E[x]. Now let F be a field, let p ∈ F[x] and let E be an extension field in which p splits. Then the field S :=

• {D : D is an extension field of F,D ⊆ E,p splits in D}

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field, let p ∈ F[x] be a polynomial
• ver F and let E be an extension of F. Then f splits in the

extension field E ⊇ F iff p can be factored into linear factors with coefficients in E[x]. Now let F be a field, let p ∈ F[x] and let E be an extension field in which p splits. Then the field S :=

• {D : D is an extension field of F,D ⊆ E,p splits in D}

is called the splitting field for p over F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field, let p ∈ F[x] be a polynomial
• ver F and let E be an extension of F. Then f splits in the

extension field E ⊇ F iff p can be factored into linear factors with coefficients in E[x]. Now let F be a field, let p ∈ F[x] and let E be an extension field in which p splits. Then the field S :=

• {D : D is an extension field of F,D ⊆ E,p splits in D}

is called the splitting field for p over F. (We will address the fact that we say “the” in the next presentation.)

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

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Definition.

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field, let E be an extension of F

and let θ1,...,θn ∈ E\F.

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field, let E be an extension of F

and let θ1,...,θn ∈ E\F. We define F(θ1,...,θn) to be the intersection of all subfields of E that contain F and θ1,...,θn.

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field, let E be an extension of F

and let θ1,...,θn ∈ E\F. We define F(θ1,...,θn) to be the intersection of all subfields of E that contain F and θ1,...,θn. Then F(θ1,...,θn) is called the field F with the elements θ1,...,θn adjoined.

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field, let E be an extension of F

and let θ1,...,θn ∈ E\F. We define F(θ1,...,θn) to be the intersection of all subfields of E that contain F and θ1,...,θn. Then F(θ1,...,θn) is called the field F with the elements θ1,...,θn adjoined. Example.

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field, let E be an extension of F

and let θ1,...,θn ∈ E\F. We define F(θ1,...,θn) to be the intersection of all subfields of E that contain F and θ1,...,θn. Then F(θ1,...,θn) is called the field F with the elements θ1,...,θn adjoined.

• Example. C = R(i).

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field, let E be an extension of F

and let θ1,...,θn ∈ E\F. We define F(θ1,...,θn) to be the intersection of all subfields of E that contain F and θ1,...,θn. Then F(θ1,...,θn) is called the field F with the elements θ1,...,θn adjoined.

• Example. C = R(i).

Theorem.

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field, let E be an extension of F

and let θ1,...,θn ∈ E\F. We define F(θ1,...,θn) to be the intersection of all subfields of E that contain F and θ1,...,θn. Then F(θ1,...,θn) is called the field F with the elements θ1,...,θn adjoined.

• Example. C = R(i).
• Theorem. Let (F,+,·) be a field, let E be an extension of F and

let θ1,...,θn ∈ E\F.

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field, let E be an extension of F

and let θ1,...,θn ∈ E\F. We define F(θ1,...,θn) to be the intersection of all subfields of E that contain F and θ1,...,θn. Then F(θ1,...,θn) is called the field F with the elements θ1,...,θn adjoined.

• Example. C = R(i).
• Theorem. Let (F,+,·) be a field, let E be an extension of F and

let θ1,...,θn ∈ E\F. Then the elements of F(θ1,...,θn) are rational combinations of the θj

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field, let E be an extension of F

and let θ1,...,θn ∈ E\F. We define F(θ1,...,θn) to be the intersection of all subfields of E that contain F and θ1,...,θn. Then F(θ1,...,θn) is called the field F with the elements θ1,...,θn adjoined.

• Example. C = R(i).
• Theorem. Let (F,+,·) be a field, let E be an extension of F and

let θ1,...,θn ∈ E\F. Then the elements of F(θ1,...,θn) are rational combinations of the θj, where a rational combination is formed from elements of F and the θ1,...,θn using

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field, let E be an extension of F

and let θ1,...,θn ∈ E\F. We define F(θ1,...,θn) to be the intersection of all subfields of E that contain F and θ1,...,θn. Then F(θ1,...,θn) is called the field F with the elements θ1,...,θn adjoined.

• Example. C = R(i).
• Theorem. Let (F,+,·) be a field, let E be an extension of F and

let θ1,...,θn ∈ E\F. Then the elements of F(θ1,...,θn) are rational combinations of the θj, where a rational combination is formed from elements of F and the θ1,...,θn using sums

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field, let E be an extension of F

and let θ1,...,θn ∈ E\F. We define F(θ1,...,θn) to be the intersection of all subfields of E that contain F and θ1,...,θn. Then F(θ1,...,θn) is called the field F with the elements θ1,...,θn adjoined.

• Example. C = R(i).
• Theorem. Let (F,+,·) be a field, let E be an extension of F and

let θ1,...,θn ∈ E\F. Then the elements of F(θ1,...,θn) are rational combinations of the θj, where a rational combination is formed from elements of F and the θ1,...,θn using sums, products

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field, let E be an extension of F

and let θ1,...,θn ∈ E\F. We define F(θ1,...,θn) to be the intersection of all subfields of E that contain F and θ1,...,θn. Then F(θ1,...,θn) is called the field F with the elements θ1,...,θn adjoined.

• Example. C = R(i).
• Theorem. Let (F,+,·) be a field, let E be an extension of F and

let θ1,...,θn ∈ E\F. Then the elements of F(θ1,...,θn) are rational combinations of the θj, where a rational combination is formed from elements of F and the θ1,...,θn using sums, products, additive inversions

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field, let E be an extension of F

and let θ1,...,θn ∈ E\F. We define F(θ1,...,θn) to be the intersection of all subfields of E that contain F and θ1,...,θn. Then F(θ1,...,θn) is called the field F with the elements θ1,...,θn adjoined.

• Example. C = R(i).
• Theorem. Let (F,+,·) be a field, let E be an extension of F and

let θ1,...,θn ∈ E\F. Then the elements of F(θ1,...,θn) are rational combinations of the θj, where a rational combination is formed from elements of F and the θ1,...,θn using sums, products, additive inversions and multiplicative inversions

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Definition. Let (F,+,·) be a field, let E be an extension of F

and let θ1,...,θn ∈ E\F. We define F(θ1,...,θn) to be the intersection of all subfields of E that contain F and θ1,...,θn. Then F(θ1,...,θn) is called the field F with the elements θ1,...,θn adjoined.

• Example. C = R(i).
• Theorem. Let (F,+,·) be a field, let E be an extension of F and

let θ1,...,θn ∈ E\F. Then the elements of F(θ1,...,θn) are rational combinations of the θj, where a rational combination is formed from elements of F and the θ1,...,θn using sums, products, additive inversions and multiplicative inversions (except divisions by zero).

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

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Proof.

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proof. A polynomial combination is formed from the elements
• f F and θ1,...,θn using sums, products and additive

inversions.

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proof. A polynomial combination is formed from the elements
• f F and θ1,...,θn using sums, products and additive
• inversions. We first prove by induction on the total number k of
• perations (sums, products, additive and multiplicative

inversions) needed to form a rational combination r that r = p q, where p and q are polynomial combinations.

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proof. A polynomial combination is formed from the elements
• f F and θ1,...,θn using sums, products and additive
• inversions. We first prove by induction on the total number k of
• perations (sums, products, additive and multiplicative

inversions) needed to form a rational combination r that r = p q, where p and q are polynomial combinations. k = 0

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proof. A polynomial combination is formed from the elements
• f F and θ1,...,θn using sums, products and additive
• inversions. We first prove by induction on the total number k of
• perations (sums, products, additive and multiplicative

inversions) needed to form a rational combination r that r = p q, where p and q are polynomial combinations. k = 0: Trivial

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proof. A polynomial combination is formed from the elements
• f F and θ1,...,θn using sums, products and additive
• inversions. We first prove by induction on the total number k of
• perations (sums, products, additive and multiplicative

inversions) needed to form a rational combination r that r = p q, where p and q are polynomial combinations. k = 0: Trivial: r ∈ F∪{θ1,...,θn} and r = r 1.

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proof. A polynomial combination is formed from the elements
• f F and θ1,...,θn using sums, products and additive
• inversions. We first prove by induction on the total number k of
• perations (sums, products, additive and multiplicative

inversions) needed to form a rational combination r that r = p q, where p and q are polynomial combinations. k = 0: Trivial: r ∈ F∪{θ1,...,θn} and r = r 1. Induction step, k > 0:

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proof. A polynomial combination is formed from the elements
• f F and θ1,...,θn using sums, products and additive
• inversions. We first prove by induction on the total number k of
• perations (sums, products, additive and multiplicative

inversions) needed to form a rational combination r that r = p q, where p and q are polynomial combinations. k = 0: Trivial: r ∈ F∪{θ1,...,θn} and r = r 1. Induction step, k > 0: Let r be a rational combination.

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proof. A polynomial combination is formed from the elements
• f F and θ1,...,θn using sums, products and additive
• inversions. We first prove by induction on the total number k of
• perations (sums, products, additive and multiplicative

inversions) needed to form a rational combination r that r = p q, where p and q are polynomial combinations. k = 0: Trivial: r ∈ F∪{θ1,...,θn} and r = r 1. Induction step, k > 0: Let r be a rational combination. First case: r = r1 +r2, where r1 and r2 are rational combinations.

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proof. A polynomial combination is formed from the elements
• f F and θ1,...,θn using sums, products and additive
• inversions. We first prove by induction on the total number k of
• perations (sums, products, additive and multiplicative

inversions) needed to form a rational combination r that r = p q, where p and q are polynomial combinations. k = 0: Trivial: r ∈ F∪{θ1,...,θn} and r = r 1. Induction step, k > 0: Let r be a rational combination. First case: r = r1 +r2, where r1 and r2 are rational combinations. Both r1 and r2 were formed using fewer than k operations.

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proof. A polynomial combination is formed from the elements
• f F and θ1,...,θn using sums, products and additive
• inversions. We first prove by induction on the total number k of
• perations (sums, products, additive and multiplicative

inversions) needed to form a rational combination r that r = p q, where p and q are polynomial combinations. k = 0: Trivial: r ∈ F∪{θ1,...,θn} and r = r 1. Induction step, k > 0: Let r be a rational combination. First case: r = r1 +r2, where r1 and r2 are rational combinations. Both r1 and r2 were formed using fewer than k operations. By induction hypothesis, for j = 1,2 we have rj = pj qj , where pj and qj are polynomial combinations.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proof. A polynomial combination is formed from the elements
• f F and θ1,...,θn using sums, products and additive
• inversions. We first prove by induction on the total number k of
• perations (sums, products, additive and multiplicative

inversions) needed to form a rational combination r that r = p q, where p and q are polynomial combinations. k = 0: Trivial: r ∈ F∪{θ1,...,θn} and r = r 1. Induction step, k > 0: Let r be a rational combination. First case: r = r1 +r2, where r1 and r2 are rational combinations. Both r1 and r2 were formed using fewer than k operations. By induction hypothesis, for j = 1,2 we have rj = pj qj , where pj and qj are polynomial combinations. Now r

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proof. A polynomial combination is formed from the elements
• f F and θ1,...,θn using sums, products and additive
• inversions. We first prove by induction on the total number k of
• perations (sums, products, additive and multiplicative

inversions) needed to form a rational combination r that r = p q, where p and q are polynomial combinations. k = 0: Trivial: r ∈ F∪{θ1,...,θn} and r = r 1. Induction step, k > 0: Let r be a rational combination. First case: r = r1 +r2, where r1 and r2 are rational combinations. Both r1 and r2 were formed using fewer than k operations. By induction hypothesis, for j = 1,2 we have rj = pj qj , where pj and qj are polynomial combinations. Now r = r1 +r2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proof. A polynomial combination is formed from the elements
• f F and θ1,...,θn using sums, products and additive
• inversions. We first prove by induction on the total number k of
• perations (sums, products, additive and multiplicative

inversions) needed to form a rational combination r that r = p q, where p and q are polynomial combinations. k = 0: Trivial: r ∈ F∪{θ1,...,θn} and r = r 1. Induction step, k > 0: Let r be a rational combination. First case: r = r1 +r2, where r1 and r2 are rational combinations. Both r1 and r2 were formed using fewer than k operations. By induction hypothesis, for j = 1,2 we have rj = pj qj , where pj and qj are polynomial combinations. Now r = r1 +r2 = p1 q1 + p2 q2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proof. A polynomial combination is formed from the elements
• f F and θ1,...,θn using sums, products and additive
• inversions. We first prove by induction on the total number k of
• perations (sums, products, additive and multiplicative

inversions) needed to form a rational combination r that r = p q, where p and q are polynomial combinations. k = 0: Trivial: r ∈ F∪{θ1,...,θn} and r = r 1. Induction step, k > 0: Let r be a rational combination. First case: r = r1 +r2, where r1 and r2 are rational combinations. Both r1 and r2 were formed using fewer than k operations. By induction hypothesis, for j = 1,2 we have rj = pj qj , where pj and qj are polynomial combinations. Now r = r1 +r2 = p1 q1 + p2 q2 = p1q2 +p2q1 q1q2 .

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Proof (concl.).

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Proof (concl.). The arguments for r = r1 ·r2, r = −r1 and r = (r1)−1 (for r1 = 0) are similar.

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Proof (concl.). The arguments for r = r1 ·r2, r = −r1 and r = (r1)−1 (for r1 = 0) are similar. It is now easy to verify that the rational combinations are a field

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Proof (concl.). The arguments for r = r1 ·r2, r = −r1 and r = (r1)−1 (for r1 = 0) are similar. It is now easy to verify that the rational combinations are a field and that every subfield of E that contains F∪{θ1,...,θn} contains all rational combinations of the θj.

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Proof (concl.). The arguments for r = r1 ·r2, r = −r1 and r = (r1)−1 (for r1 = 0) are similar. It is now easy to verify that the rational combinations are a field and that every subfield of E that contains F∪{θ1,...,θn} contains all rational combinations of the θj. (Good exercise.)

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Proof (concl.). The arguments for r = r1 ·r2, r = −r1 and r = (r1)−1 (for r1 = 0) are similar. It is now easy to verify that the rational combinations are a field and that every subfield of E that contains F∪{θ1,...,θn} contains all rational combinations of the θj. (Good exercise.)

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Proposition.

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proposition. Let (F,+,·) be a field, let p ∈ F[x] be a

polynomial over F, let E be an extension of F that splits p and let θ1,...,θn ∈ E\F be the zeros of p that are not in F.

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proposition. Let (F,+,·) be a field, let p ∈ F[x] be a

polynomial over F, let E be an extension of F that splits p and let θ1,...,θn ∈ E\F be the zeros of p that are not in F. Then F(θ1,...,θn) is the splitting field for p over F.

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proposition. Let (F,+,·) be a field, let p ∈ F[x] be a

polynomial over F, let E be an extension of F that splits p and let θ1,...,θn ∈ E\F be the zeros of p that are not in F. Then F(θ1,...,θn) is the splitting field for p over F. Proof.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proposition. Let (F,+,·) be a field, let p ∈ F[x] be a

polynomial over F, let E be an extension of F that splits p and let θ1,...,θn ∈ E\F be the zeros of p that are not in F. Then F(θ1,...,θn) is the splitting field for p over F.

• Proof. Let ad ∈ F be the leading coefficient of p

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proposition. Let (F,+,·) be a field, let p ∈ F[x] be a

polynomial over F, let E be an extension of F that splits p and let θ1,...,θn ∈ E\F be the zeros of p that are not in F. Then F(θ1,...,θn) is the splitting field for p over F.

• Proof. Let ad ∈ F be the leading coefficient of p, let θ1,...,θn

be the zeros of p in E\F

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proposition. Let (F,+,·) be a field, let p ∈ F[x] be a

polynomial over F, let E be an extension of F that splits p and let θ1,...,θn ∈ E\F be the zeros of p that are not in F. Then F(θ1,...,θn) is the splitting field for p over F.

• Proof. Let ad ∈ F be the leading coefficient of p, let θ1,...,θn

be the zeros of p in E\F, let mj be the multiplicity of θj

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proposition. Let (F,+,·) be a field, let p ∈ F[x] be a

polynomial over F, let E be an extension of F that splits p and let θ1,...,θn ∈ E\F be the zeros of p that are not in F. Then F(θ1,...,θn) is the splitting field for p over F.

• Proof. Let ad ∈ F be the leading coefficient of p, let θ1,...,θn

be the zeros of p in E\F, let mj be the multiplicity of θj, let ν1,...,νl be the zeros of p in F

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proposition. Let (F,+,·) be a field, let p ∈ F[x] be a

polynomial over F, let E be an extension of F that splits p and let θ1,...,θn ∈ E\F be the zeros of p that are not in F. Then F(θ1,...,θn) is the splitting field for p over F.

• Proof. Let ad ∈ F be the leading coefficient of p, let θ1,...,θn

be the zeros of p in E\F, let mj be the multiplicity of θj, let ν1,...,νl be the zeros of p in F and let Mk be the multiplicity of νk.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proposition. Let (F,+,·) be a field, let p ∈ F[x] be a

polynomial over F, let E be an extension of F that splits p and let θ1,...,θn ∈ E\F be the zeros of p that are not in F. Then F(θ1,...,θn) is the splitting field for p over F.

• Proof. Let ad ∈ F be the leading coefficient of p, let θ1,...,θn

be the zeros of p in E\F, let mj be the multiplicity of θj, let ν1,...,νl be the zeros of p in F and let Mk be the multiplicity of νk. Then p(x) = ad

n

∏

j=1

(x−θj)mj

l

∏

k=1

(x−νk)Mk.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proposition. Let (F,+,·) be a field, let p ∈ F[x] be a

polynomial over F, let E be an extension of F that splits p and let θ1,...,θn ∈ E\F be the zeros of p that are not in F. Then F(θ1,...,θn) is the splitting field for p over F.

• Proof. Let ad ∈ F be the leading coefficient of p, let θ1,...,θn

be the zeros of p in E\F, let mj be the multiplicity of θj, let ν1,...,νl be the zeros of p in F and let Mk be the multiplicity of νk. Then p(x) = ad

n

∏

j=1

(x−θj)mj

l

∏

k=1

(x−νk)Mk. Because ad,θ1,...,θn,ν1,...,νl ∈ F(θ1,...,θn), the polynomial p splits in F(θ1,...,θn).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proposition. Let (F,+,·) be a field, let p ∈ F[x] be a

polynomial over F, let E be an extension of F that splits p and let θ1,...,θn ∈ E\F be the zeros of p that are not in F. Then F(θ1,...,θn) is the splitting field for p over F.

• Proof. Let ad ∈ F be the leading coefficient of p, let θ1,...,θn

be the zeros of p in E\F, let mj be the multiplicity of θj, let ν1,...,νl be the zeros of p in F and let Mk be the multiplicity of νk. Then p(x) = ad

n

∏

j=1

(x−θj)mj

l

∏

k=1

(x−νk)Mk. Because ad,θ1,...,θn,ν1,...,νl ∈ F(θ1,...,θn), the polynomial p splits in F(θ1,...,θn). Moreover, every field G with F ⊆ G ⊆ E in which p splits must contain θ1,...,θn.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proposition. Let (F,+,·) be a field, let p ∈ F[x] be a

polynomial over F, let E be an extension of F that splits p and let θ1,...,θn ∈ E\F be the zeros of p that are not in F. Then F(θ1,...,θn) is the splitting field for p over F.

• Proof. Let ad ∈ F be the leading coefficient of p, let θ1,...,θn

be the zeros of p in E\F, let mj be the multiplicity of θj, let ν1,...,νl be the zeros of p in F and let Mk be the multiplicity of νk. Then p(x) = ad

n

∏

j=1

(x−θj)mj

l

∏

k=1

(x−νk)Mk. Because ad,θ1,...,θn,ν1,...,νl ∈ F(θ1,...,θn), the polynomial p splits in F(θ1,...,θn). Moreover, every field G with F ⊆ G ⊆ E in which p splits must contain θ1,...,θn. Hence it must contain F(θ1,...,θn).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proposition. Let (F,+,·) be a field, let p ∈ F[x] be a

polynomial over F, let E be an extension of F that splits p and let θ1,...,θn ∈ E\F be the zeros of p that are not in F. Then F(θ1,...,θn) is the splitting field for p over F.

• Proof. Let ad ∈ F be the leading coefficient of p, let θ1,...,θn

be the zeros of p in E\F, let mj be the multiplicity of θj, let ν1,...,νl be the zeros of p in F and let Mk be the multiplicity of νk. Then p(x) = ad

n

∏

j=1

(x−θj)mj

l

∏

k=1

(x−νk)Mk. Because ad,θ1,...,θn,ν1,...,νl ∈ F(θ1,...,θn), the polynomial p splits in F(θ1,...,θn). Moreover, every field G with F ⊆ G ⊆ E in which p splits must contain θ1,...,θn. Hence it must contain F(θ1,...,θn). Thus F(θ1,...,θn) is the splitting field for p over F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Proposition. Let (F,+,·) be a field, let p ∈ F[x] be a

polynomial over F, let E be an extension of F that splits p and let θ1,...,θn ∈ E\F be the zeros of p that are not in F. Then F(θ1,...,θn) is the splitting field for p over F.

• Proof. Let ad ∈ F be the leading coefficient of p, let θ1,...,θn

be the zeros of p in E\F, let mj be the multiplicity of θj, let ν1,...,νl be the zeros of p in F and let Mk be the multiplicity of νk. Then p(x) = ad

n

∏

j=1

(x−θj)mj

l

∏

k=1

(x−νk)Mk. Because ad,θ1,...,θn,ν1,...,νl ∈ F(θ1,...,θn), the polynomial p splits in F(θ1,...,θn). Moreover, every field G with F ⊆ G ⊆ E in which p splits must contain θ1,...,θn. Hence it must contain F(θ1,...,θn). Thus F(θ1,...,θn) is the splitting field for p over F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

Example.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Example. Q

√ 2

• is the splitting field for p(x) = x2 −2 over

Q.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Example. Q

√ 2

• is the splitting field for p(x) = x2 −2 over
• Q. Because

√ 2 2 = 2, the elements of Q √ 2

• are of the

form a+b √ 2 c+d √ 2 , with a,b,c,d ∈ Q.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Example. Q

√ 2

• is the splitting field for p(x) = x2 −2 over
• Q. Because

√ 2 2 = 2, the elements of Q √ 2

• are of the

form a+b √ 2 c+d √ 2 , with a,b,c,d ∈ Q. For each of these elements we have c ∈ Q and d √ 2 ∈ Q.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Example. Q

√ 2

• is the splitting field for p(x) = x2 −2 over
• Q. Because

√ 2 2 = 2, the elements of Q √ 2

• are of the

form a+b √ 2 c+d √ 2 , with a,b,c,d ∈ Q. For each of these elements we have c ∈ Q and d √ 2 ∈ Q. Therefore c ∈

• ±d

√ 2

• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Example. Q

√ 2

• is the splitting field for p(x) = x2 −2 over
• Q. Because

√ 2 2 = 2, the elements of Q √ 2

• are of the

form a+b √ 2 c+d √ 2 , with a,b,c,d ∈ Q. For each of these elements we have c ∈ Q and d √ 2 ∈ Q. Therefore c ∈

• ±d

√ 2

• , and

hence c2 −2d2 = 0.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Example. Q

√ 2

• is the splitting field for p(x) = x2 −2 over
• Q. Because

√ 2 2 = 2, the elements of Q √ 2

• are of the

form a+b √ 2 c+d √ 2 , with a,b,c,d ∈ Q. For each of these elements we have c ∈ Q and d √ 2 ∈ Q. Therefore c ∈

• ±d

√ 2

• , and

hence c2 −2d2 = 0. Therefore a+b √ 2 c+d √ 2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Example. Q

√ 2

• is the splitting field for p(x) = x2 −2 over
• Q. Because

√ 2 2 = 2, the elements of Q √ 2

• are of the

form a+b √ 2 c+d √ 2 , with a,b,c,d ∈ Q. For each of these elements we have c ∈ Q and d √ 2 ∈ Q. Therefore c ∈

• ±d

√ 2

• , and

hence c2 −2d2 = 0. Therefore a+b √ 2 c+d √ 2 = a+b √ 2 c+d √ 2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Example. Q

√ 2

• is the splitting field for p(x) = x2 −2 over
• Q. Because

√ 2 2 = 2, the elements of Q √ 2

• are of the

form a+b √ 2 c+d √ 2 , with a,b,c,d ∈ Q. For each of these elements we have c ∈ Q and d √ 2 ∈ Q. Therefore c ∈

• ±d

√ 2

• , and

hence c2 −2d2 = 0. Therefore a+b √ 2 c+d √ 2 = a+b √ 2 c+d √ 2 · c−d √ 2 c−d √ 2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Example. Q

√ 2

• is the splitting field for p(x) = x2 −2 over
• Q. Because

√ 2 2 = 2, the elements of Q √ 2

• are of the

form a+b √ 2 c+d √ 2 , with a,b,c,d ∈ Q. For each of these elements we have c ∈ Q and d √ 2 ∈ Q. Therefore c ∈

• ±d

√ 2

• , and

hence c2 −2d2 = 0. Therefore a+b √ 2 c+d √ 2 = a+b √ 2 c+d √ 2 · c−d √ 2 c−d √ 2 = ac−2bd +(bc−ad) √ 2 c2 −2d2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Example. Q

√ 2

• is the splitting field for p(x) = x2 −2 over
• Q. Because

√ 2 2 = 2, the elements of Q √ 2

• are of the

form a+b √ 2 c+d √ 2 , with a,b,c,d ∈ Q. For each of these elements we have c ∈ Q and d √ 2 ∈ Q. Therefore c ∈

• ±d

√ 2

• , and

hence c2 −2d2 = 0. Therefore a+b √ 2 c+d √ 2 = a+b √ 2 c+d √ 2 · c−d √ 2 c−d √ 2 = ac−2bd +(bc−ad) √ 2 c2 −2d2 = ac−2bd c2 −2d2 + bc−ad c2 −2d2 √ 2.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Example. Q

√ 2

• is the splitting field for p(x) = x2 −2 over
• Q. Because

√ 2 2 = 2, the elements of Q √ 2

• are of the

form a+b √ 2 c+d √ 2 , with a,b,c,d ∈ Q. For each of these elements we have c ∈ Q and d √ 2 ∈ Q. Therefore c ∈

• ±d

√ 2

• , and

hence c2 −2d2 = 0. Therefore a+b √ 2 c+d √ 2 = a+b √ 2 c+d √ 2 · c−d √ 2 c−d √ 2 = ac−2bd +(bc−ad) √ 2 c2 −2d2 = ac−2bd c2 −2d2 + bc−ad c2 −2d2 √ 2. Let x := ac−2bd c2 −2d2 ∈ Q and y := bc−ad c2 −2d2 ∈ Q.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Example. Q

√ 2

• is the splitting field for p(x) = x2 −2 over
• Q. Because

√ 2 2 = 2, the elements of Q √ 2

• are of the

form a+b √ 2 c+d √ 2 , with a,b,c,d ∈ Q. For each of these elements we have c ∈ Q and d √ 2 ∈ Q. Therefore c ∈

• ±d

√ 2

• , and

hence c2 −2d2 = 0. Therefore a+b √ 2 c+d √ 2 = a+b √ 2 c+d √ 2 · c−d √ 2 c−d √ 2 = ac−2bd +(bc−ad) √ 2 c2 −2d2 = ac−2bd c2 −2d2 + bc−ad c2 −2d2 √ 2. Let x := ac−2bd c2 −2d2 ∈ Q and y := bc−ad c2 −2d2 ∈ Q. The elements of Q √ 2

• are of the form x+y

√ 2 with x,y ∈ Q.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields

logo1 Subfields Splitting Fields Adjoining Elements

• Example. Q

√ 2

• is the splitting field for p(x) = x2 −2 over
• Q. Because

√ 2 2 = 2, the elements of Q √ 2

• are of the

form a+b √ 2 c+d √ 2 , with a,b,c,d ∈ Q. For each of these elements we have c ∈ Q and d √ 2 ∈ Q. Therefore c ∈

• ±d

√ 2

• , and

hence c2 −2d2 = 0. Therefore a+b √ 2 c+d √ 2 = a+b √ 2 c+d √ 2 · c−d √ 2 c−d √ 2 = ac−2bd +(bc−ad) √ 2 c2 −2d2 = ac−2bd c2 −2d2 + bc−ad c2 −2d2 √ 2. Let x := ac−2bd c2 −2d2 ∈ Q and y := bc−ad c2 −2d2 ∈ Q. The elements of Q √ 2

• are of the form x+y

√ 2 with x,y ∈ Q.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Field Extensions and Splitting Fields