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) = 2. (b) This remains true when Q is replaced with any extension field F, where QCFCC. 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. 0:3+b√2-c√3+06 a+b√2-cv3-dv6 They form the Galois group of x 5x²+6. The multiplication table and Cayley graph are shown below. Remarks ■ r = √√2+ √3 is 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 5 (tv2±√3) of f(x). Problem 17: Galois Group and Discriminants Let f(x)=-5x+4€ Q[2]. ⚫ Compute the discriminant of f(x). Use the discriminant to identify the Galois group of the splitting field over Q An example: the Galois correspondence for f(x) = x³-2 Consider QC. 2)=Q(a), the splitting field of f(x)=x³-2. It is also the splitting field of m(x)+108, the minimal polynomial of Q2) Q(2) QC35) Q(C) Q(2) Q(2) Q(<2V/2) 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.C² = 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): Q1=6. Moreover, you can check that | Gal(Q(2))=1<[0(2): Q = 3. Q(C. V/2) Subfield lattice of Q(C. 2) D Subgroup lattice of Gal(Q(C. 3/2)) = 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 Qare: Q. Q(C). and Q(C. 2). The normal subgroups of D3 are: D3. (r) and (e).

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) = 2. (b) This remains true when Q is replaced with any extension field F, where QCFCC. 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. 0:3+b√2-c√3+06 a+b√2-cv3-dv6 They form the Galois group of x 5x²+6. The multiplication table and Cayley graph are shown below. Remarks ■ r = √√2+ √3 is 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 5 (tv2±√3) of f(x). Problem 17: Galois Group and Discriminants Let f(x)=-5x+4€ Q[2]. ⚫ Compute the discriminant of f(x). Use the discriminant to identify the Galois group of the splitting field over Q An example: the Galois correspondence for f(x) = x³-2 Consider QC. 2)=Q(a), the splitting field of f(x)=x³-2. It is also the splitting field of m(x)+108, the minimal polynomial of Q2) Q(2) QC35) Q(C) Q(2) Q(2) Q(<2V/2) 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.C² = 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): Q1=6. Moreover, you can check that | Gal(Q(2))=1<[0(2): Q = 3. Q(C. V/2) Subfield lattice of Q(C. 2) D Subgroup lattice of Gal(Q(C. 3/2)) = 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 Qare: Q. Q(C). and Q(C. 2). The normal subgroups of D3 are: D3. (r) and (e).

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

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