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 (11) = 12. (b) This remains true when Q is replaced with any extension field F, where QCFCC. ea+b√2+c√3+d√ 02: a+b√√2+c√√3+α√ a+b√2+c√3+0√6 a+b√2+c√√] +√ 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√б 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(a) = Q(v2, v3). There is a group action of Gal(f(x)) on the set of roots 5 = {±√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 18: Splitting Field of a Polynomial and Its Galois Group Let f(x)=-2x²+x-1€ Q[x] Find the splitting field of f(x) over Q and compute the Galois group of this splitting field. 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 QKC) m(x)=x+108, the minimal polynomial of Q(2) Q(2) Q(²) a = √√√√-3. Let's see which of its intermediate subfields are normal extensions of Q. Q: Trivially normal. Q(C. V2) ■Q(C): Splitting field of x²+x+1; roots are C. (2² = Q(C). Normal. ■Q(V2): 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): Q]=2, Moreover, you can check that | Gal(Q(2))=1<[Q(√2): Q] = 3. | Gal(Q(C. 2))| = [Q(C. √2): Q] = 6. (૪) (૮) Q(cTY Q(C. V/2) Subfield lattice of Q(C. 2) = D₂ (124) Subgroup lattice of Gal(Q(C. 2)) Dy 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).

11. Show that defined by is not a homomorphism.

Consider the set S={[0],[2],[4],[6],[8],[10],[12],[14],[16]}18, with addition and multiplication as defined in 18. a. Is S an integral domain? If not, give a reason. b. Is S a field? If not, give a reason. [Type here][Type here]

Find the quartic (fourth-degree) polynomial function f with real coefficients that has 1,2and2i, as zeros, and f(1)=10.

Let be an irreducible polynomial over a field . Prove that is irreducible over for all nonzero in

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…

Find the divergence of the field F. vj-xk (y2+ x31/2 x2- y2 (y2+ x3 3/2 F = ーx2 (y2+ x23/2 x2 (y2+ x23/2

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 (1)=12. (b) This remains true when Q is replaced with any extension field F, where QCFCC. ea+b+c+da+by+c√] + d√6. 02: a+b√√2+c√√3+d√6 a b√2+0√3 d√ a+b√2-c√3+0√6+b√2-c√3-d√6 They form the Galois group of x 5x²+6. The multiplication table and Cayley graph are 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 = (±√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)…

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 4(n) = 2. (b) This remains true when Q is replaced with any extension field F, where QCFCC. eat b√2-c√3+d√ 02: a+b√2+c√√3+0√√6 03:3+b√2-c√3+0√√6 a+b+c+ a+b√2+ c√3+ dvb a b√√2+ c√√3 d√6 a+b√2-cv/3-dv6 -b√2-c√] + dvb 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…

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 (1)=2- (b) This remains true when Q is replaced with any extension field F, where QCFCC. ea+b+c+d+by+ c√] + dvb a+b√2-c√3+0√6 a b√√2+c√√3 d√b a+b√2-c√3-d√6 a+b√√2-c√] +dva-bv2-c+d√б They form the Galois group of x 5x² + 6. The multiplication table and Cayley graph are shown below. Remarks √2+3 is a primitive element of F, ie, Q(o)= Q(√2√3) There is a group action of Gal(f(x)) on the set of roots 5 (±√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.…