Instruction: Do not use AI. : Do not just give outline, Give complete solution with visualizations. : Handwritten is preferred. The "One orbit theorem" Let 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 4(n) = (b) This remains true when Q is replaced with any extension field F, where QCFCC. e: a+b√2-c√√3+da+b+c+d√ a+b√2-cv3+d√6 a+b√2-cv3-d√6 a+b√2-c+d√ba-bv2-cv3+d√b They form the Galois group of x 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 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 4(n) = (b) This remains true when Q is replaced with any extension field F, where QCFCC. e: a+b√2-c√√3+da+b+c+d√ a+b√2-cv3+d√6 a+b√2-cv3-d√6 a+b√2-c+d√ba-bv2-cv3+d√b They form the Galois group of x 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 35E
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 ry 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) =
(b) This remains true when Q is replaced with any extension field F, where QCFCC.
e: a+b√2-c√√3+da+b+c+d√
a+b√2-cv3+d√6 a+b√2-cv3-d√6
a+b√2-c+d√ba-bv2-cv3+d√b
They form the Galois group of x 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.√3)
There is a group action of Gal(f(x)) on the set of roots 5= (±√2±√3) of f(x).
Problem 11: Galois Group of a Polynomial with Repeated Roots
Let f(x)=³-3x²+32-1€ Q[z]-
Find the Galois group of the splitting field of f() over Q
Explain why the Galois group might be smaller than expected given the structure of the
polynomial
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
Q) Q(2) Q®)
Let's see which of its intermediate subfields
■Q: Trivially normal.
Q(C. V)
are normal extensions of Q.
■Q(C): Splitting field of x²+x+ 1; roots are C. C² = Q(C). Normal.
■Q(2): 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(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.
(2) (૮) Q(cS2)
Q(C. 3/2)
Subfield lattice of Q(C. 32) = D₂
(rf)
(125)
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 (r).
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%2F16813fc3-7e96-4a66-aa97-4b94fe805ccc%2Fc80dcb35-70ce-4944-b980-631f293b4c50%2F5qo20h8_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 ry 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) =
(b) This remains true when Q is replaced with any extension field F, where QCFCC.
e: a+b√2-c√√3+da+b+c+d√
a+b√2-cv3+d√6 a+b√2-cv3-d√6
a+b√2-c+d√ba-bv2-cv3+d√b
They form the Galois group of x 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.√3)
There is a group action of Gal(f(x)) on the set of roots 5= (±√2±√3) of f(x).
Problem 11: Galois Group of a Polynomial with Repeated Roots
Let f(x)=³-3x²+32-1€ Q[z]-
Find the Galois group of the splitting field of f() over Q
Explain why the Galois group might be smaller than expected given the structure of the
polynomial
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
Q) Q(2) Q®)
Let's see which of its intermediate subfields
■Q: Trivially normal.
Q(C. V)
are normal extensions of Q.
■Q(C): Splitting field of x²+x+ 1; roots are C. C² = Q(C). Normal.
■Q(2): 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(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.
(2) (૮) Q(cS2)
Q(C. 3/2)
Subfield lattice of Q(C. 32) = D₂
(rf)
(125)
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 (r).
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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