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 (n) = 12. (b) This remains true when Q is replaced with any extension field F, where QCFCC. ea+b+c+d√√ a+b√√2+c√√3+dve a+b√2-c√3+0√6 a+bv2+ c√3+ dvb a b√√2+c√√3 d√6 a+bv2-cv3-dv6 They form the Galois group of x4 5x²+6. The multiplication table and Cayley graph are shown below. Remarks ■a=√2+√3s a primitive element of F. ie. Q(x) = Q(v2√3) There is a group action of Gal(f(x)) on the set of roots S = (±√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 (n) = 12. (b) This remains true when Q is replaced with any extension field F, where QCFCC. ea+b+c+d√√ a+b√√2+c√√3+dve a+b√2-c√3+0√6 a+bv2+ c√3+ dvb a b√√2+c√√3 d√6 a+bv2-cv3-dv6 They form the Galois group of x4 5x²+6. The multiplication table and Cayley graph are shown below. Remarks ■a=√2+√3s a primitive element of F. ie. Q(x) = Q(v2√3) There is a group action of Gal(f(x)) on the set of roots S = (±√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
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 fixes Q and with (n) = 12.
(b) This remains true when Q is replaced with any extension field F, where QCFCC.
ea+b+c+d√√
a+b√√2+c√√3+dve
a+b√2-c√3+0√6
a+bv2+ c√3+ dvb
a b√√2+c√√3 d√6
a+bv2-cv3-dv6
They form the Galois group of x4 5x²+6. The multiplication table and Cayley graph are
shown below.
Remarks
■a=√2+√3s a primitive element of F. ie. Q(x) = Q(v2√3)
There is a group action of Gal(f(x)) on the set of roots S = (±√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 13: Galois Group of a Quartic with Nonreal Roots
Consider the polynomial f(x) = x² + 2x² + 1 € Q[x].
Find the Galois group of the splitting field of f(z) over Q.
An example: the Galois correspondence for f(x)=x³-2
Consider Q(C. 2) Q(a), the splitting field
of f(x)=x-2.
It is also the splitting field of
QKC)
m(x)=x+108, the minimal polynomial of
Q2) Q(2) Q²)
a = √√√√-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(2): 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): Q]=2,
Gal(Q(C. 2)) [Q(C. √2): Q] = 6.
Moreover, you can check that | Gal(Q(2)) = 1 <10(2): Q = 3.
Q(C)
Q(2) Q(2) Q((2/2)
Q(C. 2)
Subfield lattice of Q(C. 2) = D₂
(125)
Subgroup lattice of Gal(Q(C. 2)) = Dr
■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%2F5d91b8f2-a246-483d-979b-f617f36ccbd2%2Fe515784c-7271-4be2-bf2f-c3d8a33ef193%2Frfwfb6c_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 (n) = 12.
(b) This remains true when Q is replaced with any extension field F, where QCFCC.
ea+b+c+d√√
a+b√√2+c√√3+dve
a+b√2-c√3+0√6
a+bv2+ c√3+ dvb
a b√√2+c√√3 d√6
a+bv2-cv3-dv6
They form the Galois group of x4 5x²+6. The multiplication table and Cayley graph are
shown below.
Remarks
■a=√2+√3s a primitive element of F. ie. Q(x) = Q(v2√3)
There is a group action of Gal(f(x)) on the set of roots S = (±√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 13: Galois Group of a Quartic with Nonreal Roots
Consider the polynomial f(x) = x² + 2x² + 1 € Q[x].
Find the Galois group of the splitting field of f(z) over Q.
An example: the Galois correspondence for f(x)=x³-2
Consider Q(C. 2) Q(a), the splitting field
of f(x)=x-2.
It is also the splitting field of
QKC)
m(x)=x+108, the minimal polynomial of
Q2) Q(2) Q²)
a = √√√√-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(2): 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): Q]=2,
Gal(Q(C. 2)) [Q(C. √2): Q] = 6.
Moreover, you can check that | Gal(Q(2)) = 1 <10(2): Q = 3.
Q(C)
Q(2) Q(2) Q((2/2)
Q(C. 2)
Subfield lattice of Q(C. 2) = D₂
(125)
Subgroup lattice of Gal(Q(C. 2)) = Dr
■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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