Instruction: Do not use AI. : Do not just give outline, Give complete solution with visualizations. : Handwritten is preferred. The "One orbit theorem" Let n 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 (r) = 2. (b) This remains true when Q is replaced with any extension field F, where QCFCC. ea+by+c+d√6 a+b+c+d√ 03:3+b√2-c√3+0√6 a+b√2+c√√3+d√√6 a b√2+ c√√3 d√6 a+b√2-c√3-d√6 a-b√2-c√3+d√6 They form the Galois group of x4 5x²+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 n 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 (r) = 2. (b) This remains true when Q is replaced with any extension field F, where QCFCC. ea+by+c+d√6 a+b+c+d√ 03:3+b√2-c√3+0√6 a+b√2+c√√3+d√√6 a b√2+ c√√3 d√6 a+b√2-c√3-d√6 a-b√2-c√3+d√6 They form the Galois group of x4 5x²+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
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 n 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 (r) = 2.
(b) This remains true when Q is replaced with any extension field F, where QCFCC.
ea+by+c+d√6 a+b+c+d√
03:3+b√2-c√3+0√6
a+b√2+c√√3+d√√6
a b√2+ c√√3 d√6
a+b√2-c√3-d√6
a-b√2-c√3+d√6
They form the Galois group of x4 5x²+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 fix K.
Remarks
a=√2+√3s a primitive element of F, ie, Q(a) = Q(v2, v3)
■There is a group action of Gal(f(x)) on the set of roots 5 = (±√2±√3) of f(x).
Problem 15: Galois Group of a Polynomial with Complex Roots
Let f(x) = x²+z+1€ Q[z].
⚫Find the Galois group of the splitting field of f(z) over Q
Use the fact that f(x) has complex roots and explain the Galois group structure.
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.
QKC)
(3) (c) (c′32)
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. 2): 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.
Q(C)
Q(32) Q(32) Q((2/2)
Q(C. V5)
Subfield lattice of Q(C. 32) = D₂
(125)
Subgroup lattice of Gal(Q(C. 2)) Dr
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. 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%2Fd34b0409-98b9-4a32-bb73-bc20678b9fd8%2Fs1jyboh_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 n 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 (r) = 2.
(b) This remains true when Q is replaced with any extension field F, where QCFCC.
ea+by+c+d√6 a+b+c+d√
03:3+b√2-c√3+0√6
a+b√2+c√√3+d√√6
a b√2+ c√√3 d√6
a+b√2-c√3-d√6
a-b√2-c√3+d√6
They form the Galois group of x4 5x²+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 fix K.
Remarks
a=√2+√3s a primitive element of F, ie, Q(a) = Q(v2, v3)
■There is a group action of Gal(f(x)) on the set of roots 5 = (±√2±√3) of f(x).
Problem 15: Galois Group of a Polynomial with Complex Roots
Let f(x) = x²+z+1€ Q[z].
⚫Find the Galois group of the splitting field of f(z) over Q
Use the fact that f(x) has complex roots and explain the Galois group structure.
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.
QKC)
(3) (c) (c′32)
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. 2): 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.
Q(C)
Q(32) Q(32) Q((2/2)
Q(C. V5)
Subfield lattice of Q(C. 32) = D₂
(125)
Subgroup lattice of Gal(Q(C. 2)) Dr
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. 2).
The normal subgroups of D3 are: D3. (r) and (e).
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