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 r be roots of an irreducible polynomial over Q. Then (a) There is an isomorphism : Q(n) Q(2) that fixes Q and with 4(n) = 2. (b) This remains true when Q is replaced with any extension field F, where QCFCC. a b√√2+c√√3 d√6 a+b√2-c√3+0√6 a+b√2-c√3-d√6 a+b+c+d√ba-b√2-c√3+ dvb They form the Galois group of x 5x +6. The multiplication table and Cayley graph cre shown below. Remarks a=√2+√3s a primitive element of F. ie. Q(a) = Q(v2√3). There is a group action of Gal(f(x)) on the set of roots 5 (tvtv√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 QCKCF, 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 r be roots of an irreducible polynomial over Q. Then (a) There is an isomorphism : Q(n) Q(2) that fixes Q and with 4(n) = 2. (b) This remains true when Q is replaced with any extension field F, where QCFCC. a b√√2+c√√3 d√6 a+b√2-c√3+0√6 a+b√2-c√3-d√6 a+b+c+d√ba-b√2-c√3+ dvb They form the Galois group of x 5x +6. The multiplication table and Cayley graph cre shown below. Remarks a=√2+√3s a primitive element of F. ie. Q(a) = Q(v2√3). There is a group action of Gal(f(x)) on the set of roots 5 (tvtv√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 QCKCF, the corresponding subgroup H
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter8: Polynomials
Section8.2: Divisibility And Greatest Common Divisor
Problem 18E
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 r and r be roots of an irreducible polynomial over Q. Then
(a) There is an isomorphism : Q(n) Q(2) that fixes Q and with 4(n) = 2.
(b) This remains true when Q is replaced with any extension field F, where QCFCC.
a b√√2+c√√3 d√6
a+b√2-c√3+0√6 a+b√2-c√3-d√6
a+b+c+d√ba-b√2-c√3+ dvb
They form the Galois group of x 5x +6. The multiplication table and Cayley graph cre
shown below.
Remarks
a=√2+√3s a primitive element of F. ie. Q(a) = Q(v2√3).
There is a group action of Gal(f(x)) on the set of roots 5 (tvtv√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 QCKCF, the corresponding subgroup H<G contains
precisely those automorphisms that fix K.
Problem 2: Solvability by Radicals
Consider the polynomial f(x)=-2-1 € Q[r].
Determine whether this polynomial is solvable by radicals.
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
Q(2) Q(2) Q(35)
Let's see which of its intermediate subfields
Q(C. √2)
■ Q: Trivially normal.
Q(C)
Q(2) Q(32) Q(232)
Q(C. V/2)
(rf)
(124)
are normal extensions of Q.
■Q(C): Splitting field of x²+x+1; roots are C. (2² = 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. 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): Q = 6.
Moreover, you can check that | Gal(Q(2)) =1<[0(2): Q] = 3.
Subfield lattice of Q(C. 2) D
Subgroup lattice of Gal(Q(C. 2)) = D
The automorphisms that fix Q are precisely those in D.
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 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%2F5dab7024-8189-4244-9ed6-4cd878fbe424%2F9e5b2d01-fb21-4557-bde7-1828ef1f1a65%2Fx2w8hyu_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 r be roots of an irreducible polynomial over Q. Then
(a) There is an isomorphism : Q(n) Q(2) that fixes Q and with 4(n) = 2.
(b) This remains true when Q is replaced with any extension field F, where QCFCC.
a b√√2+c√√3 d√6
a+b√2-c√3+0√6 a+b√2-c√3-d√6
a+b+c+d√ba-b√2-c√3+ dvb
They form the Galois group of x 5x +6. The multiplication table and Cayley graph cre
shown below.
Remarks
a=√2+√3s a primitive element of F. ie. Q(a) = Q(v2√3).
There is a group action of Gal(f(x)) on the set of roots 5 (tvtv√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 QCKCF, the corresponding subgroup H<G contains
precisely those automorphisms that fix K.
Problem 2: Solvability by Radicals
Consider the polynomial f(x)=-2-1 € Q[r].
Determine whether this polynomial is solvable by radicals.
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
Q(2) Q(2) Q(35)
Let's see which of its intermediate subfields
Q(C. √2)
■ Q: Trivially normal.
Q(C)
Q(2) Q(32) Q(232)
Q(C. V/2)
(rf)
(124)
are normal extensions of Q.
■Q(C): Splitting field of x²+x+1; roots are C. (2² = 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. 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): Q = 6.
Moreover, you can check that | Gal(Q(2)) =1<[0(2): Q] = 3.
Subfield lattice of Q(C. 2) D
Subgroup lattice of Gal(Q(C. 2)) = D
The automorphisms that fix Q are precisely those in D.
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 Q are: Q. Q(C), and Q(C. 2).
The normal subgroups of D3 are: D3. (r) and (e).
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