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 foxes Q and with (n)=2- (b) This remains true when Q is replaced with any extension field F, where QCFCC. eat b√√2-c√3+d√ a+bv2+ c√] + dvb a+b√2-c√3+0√6 a b√2+ c√√3 d√√6 a+b√2-cv3-dv6 a+b√2-c√] +da-b√2-c√3+d√б They form the Galois group of x 5x +6. The multiplication table and Cayley graph cre shown below. 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 foxes Q and with (n)=2- (b) This remains true when Q is replaced with any extension field F, where QCFCC. eat b√√2-c√3+d√ a+bv2+ c√] + dvb a+b√2-c√3+0√6 a b√2+ c√√3 d√√6 a+b√2-cv3-dv6 a+b√2-c√] +da-b√2-c√3+d√б They form the Galois group of x 5x +6. The multiplication table and Cayley graph cre shown below. 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
Related questions
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 foxes Q and with (n)=2-
(b) This remains true when Q is replaced with any extension field F, where QCFCC.
eat b√√2-c√3+d√ a+bv2+ c√] + dvb
a+b√2-c√3+0√6
a b√2+ c√√3 d√√6
a+b√2-cv3-dv6
a+b√2-c√] +da-b√2-c√3+d√б
They form the Galois group of x 5x +6. The multiplication table and Cayley graph cre
shown below.
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.
Remarks
√2+3 is a primitive element of F. ie, Q(a) = Q(√2√3).
■There is a group action of Gal(f(x)) on the set of roots 5 = (±√2±√3) of f(x).
Problem 16: Solvability of Polynomials and Galois Groups
Let f(x) =
-2x²+2€ Q[a].
• Determine whether f(x) is solvable by radicals.
⚫Find the Galois group of the splitting field of f(x) over Q, and use this to analyze solvability.
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
QKC)
m(x)=x+108, the minimal polynomial of
Q2) Q(2) Q(2)
a = √2√-3.
Let's see which of its intermediate subfields
are normal extensions of Q.
Q: Trivially normal.
Q(C. √2)
■Q(C): Splitting field of x²+x+1; roots are C.(² = 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. V2): Splitting field of x³-2. Normal.
By the normal extension theorem,
Gal(Q(C))
(Q(C): Q1=2,
Gal(Q(C. 2)) = [Q(C. 2): Q1=6.
Moreover, you can check that | Gal(Q(2)) =1<[0(2): Q]=3.
Q(C)
Q(2) Q(32) Q(<23/2)
QKC. 2)
Subfield lattice of Q(C. 32) = Dr
Subgroup lattice of Gal(Q(C. 2)) Dy
The automorphisms that fix Q are precisely those in D3.
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. V2).
The normal subgroups of D3 are: D3. (r) and (e).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6fc8022f-f710-4f32-9597-9251ea77d4c4%2F69deec50-3cf5-469a-9c1b-f68de5061c85%2Fulmrggm_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 foxes Q and with (n)=2-
(b) This remains true when Q is replaced with any extension field F, where QCFCC.
eat b√√2-c√3+d√ a+bv2+ c√] + dvb
a+b√2-c√3+0√6
a b√2+ c√√3 d√√6
a+b√2-cv3-dv6
a+b√2-c√] +da-b√2-c√3+d√б
They form the Galois group of x 5x +6. The multiplication table and Cayley graph cre
shown below.
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.
Remarks
√2+3 is a primitive element of F. ie, Q(a) = Q(√2√3).
■There is a group action of Gal(f(x)) on the set of roots 5 = (±√2±√3) of f(x).
Problem 16: Solvability of Polynomials and Galois Groups
Let f(x) =
-2x²+2€ Q[a].
• Determine whether f(x) is solvable by radicals.
⚫Find the Galois group of the splitting field of f(x) over Q, and use this to analyze solvability.
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
QKC)
m(x)=x+108, the minimal polynomial of
Q2) Q(2) Q(2)
a = √2√-3.
Let's see which of its intermediate subfields
are normal extensions of Q.
Q: Trivially normal.
Q(C. √2)
■Q(C): Splitting field of x²+x+1; roots are C.(² = 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. V2): Splitting field of x³-2. Normal.
By the normal extension theorem,
Gal(Q(C))
(Q(C): Q1=2,
Gal(Q(C. 2)) = [Q(C. 2): Q1=6.
Moreover, you can check that | Gal(Q(2)) =1<[0(2): Q]=3.
Q(C)
Q(2) Q(32) Q(<23/2)
QKC. 2)
Subfield lattice of Q(C. 32) = Dr
Subgroup lattice of Gal(Q(C. 2)) Dy
The automorphisms that fix Q are precisely those in D3.
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. V2).
The normal subgroups of D3 are: D3. (r) and (e).
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