Instruction: Do not use AI. : Do not just give outline, Give complete solution with visualizations. : Handwritten is preferred. The "One orbit theorem" Let and be roots of an irreducible polynomial over Q. Then (a) There is an isomorphism : Q(n) Q(2) that fixes Q and with (1)=2- (b) This remains true when Q is replaced with any extension field F, where QCFCC. ea+b+c+d+by+ c√] + dvb a+b√2-c√3+0√6 a b√√2+c√√3 d√b a+b√2-c√3-d√6 a+b√√2-c√] +dva-bv2-c+d√б They form the Galois group of x 5x² + 6. The multiplication table and Cayley graph are shown below. Remarks √2+3 is a primitive element of F, ie, Q(o)= Q(√2√3) There is a group action of Gal(f(x)) on the set of roots 5 (±√2±√3) of f(x). Fundamental theorem of Galois theory Given f€ Z[x]. let F be the splitting field of f. and G the Galois group. Then the following hold: (a) The subgroup lattice of G is identical to the subfield lattice of F, but upside-down. Moreover, HG if and only if the corresponding subfield is a normal extension of Q. (b) Given an intermediate field QC KCF, the corresponding subgroup H
Instruction: Do not use AI. : Do not just give outline, Give complete solution with visualizations. : Handwritten is preferred. The "One orbit theorem" Let and be roots of an irreducible polynomial over Q. Then (a) There is an isomorphism : Q(n) Q(2) that fixes Q and with (1)=2- (b) This remains true when Q is replaced with any extension field F, where QCFCC. ea+b+c+d+by+ c√] + dvb a+b√2-c√3+0√6 a b√√2+c√√3 d√b a+b√2-c√3-d√6 a+b√√2-c√] +dva-bv2-c+d√б They form the Galois group of x 5x² + 6. The multiplication table and Cayley graph are shown below. Remarks √2+3 is a primitive element of F, ie, Q(o)= Q(√2√3) There is a group action of Gal(f(x)) on the set of roots 5 (±√2±√3) of f(x). Fundamental theorem of Galois theory Given f€ Z[x]. let F be the splitting field of f. and G the Galois group. Then the following hold: (a) The subgroup lattice of G is identical to the subfield lattice of F, but upside-down. Moreover, HG if and only if the corresponding subfield is a normal extension of Q. (b) Given an intermediate field QC KCF, the corresponding subgroup H
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter5: Rings, Integral Domains, And Fields
Section5.3: The Field Of Quotients Of An Integral Domain
Problem 11E
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Question
![Instruction: Do not use AI.
: Do not just give outline, Give complete solution with visualizations.
: Handwritten is preferred.
The "One orbit theorem"
Let and be roots of an irreducible polynomial over Q. Then
(a) There is an isomorphism : Q(n) Q(2) that fixes Q and with (1)=2-
(b) This remains true when Q is replaced with any extension field F, where QCFCC.
ea+b+c+d+by+ c√] + dvb
a+b√2-c√3+0√6
a b√√2+c√√3 d√b
a+b√2-c√3-d√6
a+b√√2-c√] +dva-bv2-c+d√б
They form the Galois group of x 5x² + 6. The multiplication table and Cayley graph are
shown below.
Remarks
√2+3 is a primitive element of F, ie, Q(o)= Q(√2√3)
There is a group action of Gal(f(x)) on the set of roots 5 (±√2±√3) of f(x).
Fundamental theorem of Galois theory
Given f€ Z[x]. let F be the splitting field of f. and G the Galois group. Then the
following hold:
(a) The subgroup lattice of G is identical to the subfield lattice of F, but upside-down.
Moreover, HG if and only if the corresponding subfield is a normal extension of Q.
(b) Given an intermediate field QC KCF, the corresponding subgroup H<G contains
precisely those automorphisms that fix K.
Problem 4: Computation of the Galois Group of a Quartic Polynomial
Let f(z) = -4+3€ Q[2]
Find the Galois group of f(x) over Q.
An example: the Galois correspondence for f(x) = x³-2
Consider Q(C. 2)=Q(a), the splitting field
of f(x)=x³-2.
It is also the splitting field of
Q(2) Q(2) Q(C²/2)
(124)
m(x)=x+108, the minimal polynomial of
Q(2) Q(2) Q²)
Let's see which of its intermediate subfields
Q(C. 2)
Q: Trivially normal.
are normal extensions of Q.
■Q(C): Splitting field of x²+x+1; roots are C.C² = Q(C). Normal.
■Q(V2): Contains only one root of x³-2, not the other two. Not normal.
■Q(C2): Contains only one root of x3-2, not the other two. Not normal.
■Q(2): Contains only one root of x³-2, not the other two. Not normal.
■Q(C. 2): Splitting field of x³-2. Normal.
By the normal extension theorem,
Gal(Q(C)) [Q(C): Q]=2.
Gal(Q(C. 2)) [Q(C. √2): Q1=6.
Moreover, you can check that | Gal(Q(2))=1<[Q(2): Q = 3.
Q(C. V/2)
Subfield lattice of Q(C. 2) D
Subgroup lattice of Gal(Q(C. 2)) =D
The automorphisms that fix Q are precisely those in Dz.
■The automorphisms that fix Q(C) are precisely those in (r).
The automorphisms that fix Q(2) are precisely those in (f).
■The automorphisms that fix Q(C2) are precisely those in (rf).
■The automorphisms that fix Q(22) are precisely those in (2).
The automorphisms that fix Q(C. 2) are precisely those in (e).
The normal field extensions of Q are: Q. Q(C). and Q(C. 2).
The normal subgroups of D3 are: D3. (r) and (e).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8779eebc-c13f-4d10-b8ed-21a34bc52206%2Fa9e045af-400e-4fa3-9850-29d9ba3802f8%2F1brd4i4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Instruction: Do not use AI.
: Do not just give outline, Give complete solution with visualizations.
: Handwritten is preferred.
The "One orbit theorem"
Let and be roots of an irreducible polynomial over Q. Then
(a) There is an isomorphism : Q(n) Q(2) that fixes Q and with (1)=2-
(b) This remains true when Q is replaced with any extension field F, where QCFCC.
ea+b+c+d+by+ c√] + dvb
a+b√2-c√3+0√6
a b√√2+c√√3 d√b
a+b√2-c√3-d√6
a+b√√2-c√] +dva-bv2-c+d√б
They form the Galois group of x 5x² + 6. The multiplication table and Cayley graph are
shown below.
Remarks
√2+3 is a primitive element of F, ie, Q(o)= Q(√2√3)
There is a group action of Gal(f(x)) on the set of roots 5 (±√2±√3) of f(x).
Fundamental theorem of Galois theory
Given f€ Z[x]. let F be the splitting field of f. and G the Galois group. Then the
following hold:
(a) The subgroup lattice of G is identical to the subfield lattice of F, but upside-down.
Moreover, HG if and only if the corresponding subfield is a normal extension of Q.
(b) Given an intermediate field QC KCF, the corresponding subgroup H<G contains
precisely those automorphisms that fix K.
Problem 4: Computation of the Galois Group of a Quartic Polynomial
Let f(z) = -4+3€ Q[2]
Find the Galois group of f(x) over Q.
An example: the Galois correspondence for f(x) = x³-2
Consider Q(C. 2)=Q(a), the splitting field
of f(x)=x³-2.
It is also the splitting field of
Q(2) Q(2) Q(C²/2)
(124)
m(x)=x+108, the minimal polynomial of
Q(2) Q(2) Q²)
Let's see which of its intermediate subfields
Q(C. 2)
Q: Trivially normal.
are normal extensions of Q.
■Q(C): Splitting field of x²+x+1; roots are C.C² = Q(C). Normal.
■Q(V2): Contains only one root of x³-2, not the other two. Not normal.
■Q(C2): Contains only one root of x3-2, not the other two. Not normal.
■Q(2): Contains only one root of x³-2, not the other two. Not normal.
■Q(C. 2): Splitting field of x³-2. Normal.
By the normal extension theorem,
Gal(Q(C)) [Q(C): Q]=2.
Gal(Q(C. 2)) [Q(C. √2): Q1=6.
Moreover, you can check that | Gal(Q(2))=1<[Q(2): Q = 3.
Q(C. V/2)
Subfield lattice of Q(C. 2) D
Subgroup lattice of Gal(Q(C. 2)) =D
The automorphisms that fix Q are precisely those in Dz.
■The automorphisms that fix Q(C) are precisely those in (r).
The automorphisms that fix Q(2) are precisely those in (f).
■The automorphisms that fix Q(C2) are precisely those in (rf).
■The automorphisms that fix Q(22) are precisely those in (2).
The automorphisms that fix Q(C. 2) are precisely those in (e).
The normal field extensions of Q are: Q. Q(C). and Q(C. 2).
The normal subgroups of D3 are: D3. (r) and (e).
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