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) Given an intermediate field QC KCF, the corresponding subgroup H < G contains precisely those automorphisms that fix K. Problem 20: Galois Group and Intermediate Fields Let L = Q(√2, √√3) be a field extension of Q. Find the Galois group of L/Q ⚫ Describe the intermediate fields between Q and L, and relate these fields to the structure of the Galois group. 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 m(x)x+108, the minimal polynomial of a = √√√-3. Q(2) Q(2) Q²) Q(C) Q(2) Q(2) Q(C²/2) (128) Let's see which of its intermediate subfields Q: Trivially normal. are normal extensions of Q. 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): Q = 6. Moreover, you can check that | Gal(Q(2)) = 1 <10(2): Q1=3. Q(C. V/2) Subfield lattice of Q(C. 32) = D₂ Subgroup lattice of Gal(Q(C. 2)) D₂ 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(C2) are precisely those in (rf). ■The automorphisms that fix Q(2) are precisely those in (r²). 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 (c).