Instruction: Do not use AI. : Do not just give outline, Give complete solution with visualizations. : Handwritten is preferred. The "One orbit theorem" Let r and ry be roots of an irreducible polynomial over Q. Then (a) There is an isomorphism : Q(n)→Q(2) that fixes Q and with (1)=12. (b) This remains true when Q is replaced with any extension field F, where QCFCC. ea+b√2-c√3+ 02: a+b√2+c√3+d√√ a+b√2-c√√3+0√6 a + b√2 + c√] + d√b a b√√2+ c√3 d√√6 a+b√2-cv3-dv6 They form the Galois group of x 5x +6. The multiplication table and Cayley graph are 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 r and ry be roots of an irreducible polynomial over Q. Then (a) There is an isomorphism : Q(n)→Q(2) that fixes Q and with (1)=12. (b) This remains true when Q is replaced with any extension field F, where QCFCC. ea+b√2-c√3+ 02: a+b√2+c√3+d√√ a+b√2-c√√3+0√6 a + b√2 + c√] + d√b a b√√2+ c√3 d√√6 a+b√2-cv3-dv6 They form the Galois group of x 5x +6. The multiplication table and Cayley graph are 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 r and ry be roots of an irreducible polynomial over Q. Then
(a) There is an isomorphism : Q(n)→Q(2) that fixes Q and with (1)=12.
(b) This remains true when Q is replaced with any extension field F, where QCFCC.
ea+b√2-c√3+
02: a+b√2+c√3+d√√
a+b√2-c√√3+0√6
a + b√2 + c√] + d√b
a b√√2+ c√3 d√√6
a+b√2-cv3-dv6
They form the Galois group of x 5x +6. The multiplication table and Cayley graph are
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 fox K.
Remarks
■=√2+√3 sa primitive element of F. ie.. Q(x) = Q(v2√3).
There is a group action of Gal(f(x)) on the set of roots 5 (tv2±√3) of f(x).
'
1
Problem 3: Fixed Fields and the Galois Correspondence
Let K = Q(√2,√3), the field extension of Q generated by √2 and √3.
Find the Galois group of K/Q and describe the corresponding fixed fields.
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
m(x)=x+108, the minimal polynomial of
a = √√√-3.
Let's see which of its intermediate subfields
are normal extensions of Q.
Q: Trivially normal.
■Q(C): Splitting field of x²+x+1; roots are C.
Q) (2) (3)
Q(C. 32)
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. 32): Q1=6.
Moreover, you can check that | Gal(Q(2)) = 1 <10(2): Q]=3.
(૨) (૮) (cÖ2)
QKC. 32)
Subfield lattice of Q(C. 2) D
(124)
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 (r2).
The automorphisms that fix Q(C. 2) are precisely those in (e).
The normal field extensions of Qare: 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%2F542d64ca-6ae8-4a02-b2c1-9c6231f9bcbe%2F766748ff-be2f-4953-b4d7-8967cce37dce%2Fohiwu6_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 r and ry be roots of an irreducible polynomial over Q. Then
(a) There is an isomorphism : Q(n)→Q(2) that fixes Q and with (1)=12.
(b) This remains true when Q is replaced with any extension field F, where QCFCC.
ea+b√2-c√3+
02: a+b√2+c√3+d√√
a+b√2-c√√3+0√6
a + b√2 + c√] + d√b
a b√√2+ c√3 d√√6
a+b√2-cv3-dv6
They form the Galois group of x 5x +6. The multiplication table and Cayley graph are
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 fox K.
Remarks
■=√2+√3 sa primitive element of F. ie.. Q(x) = Q(v2√3).
There is a group action of Gal(f(x)) on the set of roots 5 (tv2±√3) of f(x).
'
1
Problem 3: Fixed Fields and the Galois Correspondence
Let K = Q(√2,√3), the field extension of Q generated by √2 and √3.
Find the Galois group of K/Q and describe the corresponding fixed fields.
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
m(x)=x+108, the minimal polynomial of
a = √√√-3.
Let's see which of its intermediate subfields
are normal extensions of Q.
Q: Trivially normal.
■Q(C): Splitting field of x²+x+1; roots are C.
Q) (2) (3)
Q(C. 32)
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. 32): Q1=6.
Moreover, you can check that | Gal(Q(2)) = 1 <10(2): Q]=3.
(૨) (૮) (cÖ2)
QKC. 32)
Subfield lattice of Q(C. 2) D
(124)
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 (r2).
The automorphisms that fix Q(C. 2) are precisely those in (e).
The normal field extensions of Qare: Q. Q(C), and Q(C. 2).
The normal subgroups of D3 are: D3. (r) and (e).
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