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 automorphisms that fix K. Remarks a=√2+√3 sa 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 12: Galois Group of a Degree 6 Polynomial Let f(z) = 2"-3-2x²+z+1 € Q[z]. Find the Galois group of the splitting field of f(z) over Q. Discuss how the symmetry of the roots relates to 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)=x-2. It is also the splitting field of Q(C) Q(2) Q(2) Q((2/2) (rf) 11283 m(x)x+108, the minimal polynomial of a = √√√-3. Q) Q(2) QC32) Let's see which of its intermediate subfields Q(C. √2) Q: Trivially normal. are normal extensions of Q. ■Q(C): Splitting field of x²+x+1; roots are C. C² = 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. 2): Splitting field of x-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 <10(2): Q=3. Q(C. 5) Subfield lattice of Q(C. 2) = D Subgroup lattice of Gal(Q(C. 2)) Dy ■The automorphisms that fix Q are precisely those in Dz. ■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(CV2) 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).