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 (r)=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 d√√6 a+b√2-c√3+0√6 a+b√2-cv3-dv6 a+b√2-c√√√√6a-b√2-c√] + dvb 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 (r)=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 d√√6 a+b√2-c√3+0√6 a+b√2-cv3-dv6 a+b√2-c√√√√6a-b√2-c√] + dvb 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
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 (r)=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 d√√6
a+b√2-c√3+0√6 a+b√2-cv3-dv6
a+b√2-c√√√√6a-b√2-c√] + dvb
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.
Problem 19: Galois Groups of Degree 6 Extensions
Let f(x)=-3-2x²+x+1€ Q[x].
.
Find the Galois group of the splitting field of f(x) over Q.
Remarks
x=√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).
• How does the structure of the Galois group reflect the symmetries of the roots?
An example: the Galois correspondence for f(x) = x³-2
Consider Q(C. 2)=Q(a), the splitting field
of f(x)=x³-2.
QKC)
It is also the splitting field of
m(x)=x+108, the minimal polynomial of
a=√2√-3.
Q2) (2) (3)
Q(C)
Q(2) Q(32) Q(<23/2)
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): Q = 6.
Moreover, you can check that | Gal(Q(2)) =1<[0(2): Q] = 3.
QKC. 2)
Subfield lattice of Q(C. 32) = Dr
Subgroup lattice of Gal(Q(C. 2)) = D
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. V/2).
The normal subgroups of D3 are: D3. (r) and (e).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2802fe82-f96c-4a87-93af-5ea2695b0fe4%2F13f0b230-8844-4e11-bd3a-a723b47b67b7%2F9d7kxod_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 (r)=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 d√√6
a+b√2-c√3+0√6 a+b√2-cv3-dv6
a+b√2-c√√√√6a-b√2-c√] + dvb
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.
Problem 19: Galois Groups of Degree 6 Extensions
Let f(x)=-3-2x²+x+1€ Q[x].
.
Find the Galois group of the splitting field of f(x) over Q.
Remarks
x=√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).
• How does the structure of the Galois group reflect the symmetries of the roots?
An example: the Galois correspondence for f(x) = x³-2
Consider Q(C. 2)=Q(a), the splitting field
of f(x)=x³-2.
QKC)
It is also the splitting field of
m(x)=x+108, the minimal polynomial of
a=√2√-3.
Q2) (2) (3)
Q(C)
Q(2) Q(32) Q(<23/2)
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): Q = 6.
Moreover, you can check that | Gal(Q(2)) =1<[0(2): Q] = 3.
QKC. 2)
Subfield lattice of Q(C. 32) = Dr
Subgroup lattice of Gal(Q(C. 2)) = D
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. V/2).
The normal subgroups of D3 are: D3. (r) and (e).
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