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 foxes Q and with (r)=2- (b) This remains true when Q is replaced with any extension field F, where QCFCC. eat b√√2-c√3+d√ a+bv2+ c√] + dvb a b√2+c√√3 d√√6 a+b√2-c√3+0√6 a+b√2-cv3-dv6 a+b√2-c√√√√6a-b√2-c√] + dvb They form the Galois group of x 5x +6. The multiplication table and Cayley graph cre 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

Prove that if R and S are fields, then the direct sum RS is not a field. [Type here][Type here]

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

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

Suppose that f(x),g(x), and h(x) are polynomials over the field F, each of which has positive degree, and that f(x)=g(x)h(x). Prove that the zeros of f(x) in F consist of the zeros of g(x) in F together with the zeros of h(x) in F.

Prove that [ x ]={ a0+a1x+...+anxna0=2kfork }, the set of all polynomials in [ x ] with even constant term, is an ideal of [ x ]. Show that [ x ] is not a principal ideal; that is, show that there is no f(x)[ x ] such that [ x ]=(f(x))={ f(x)g(x)g(x)[ x ] }. Show that [ x ] is an ideal generated by two elements in [ x ] that is, [ x ]=(x,2)={ xf(x)+2g(x)f(x),g(x)[ x ] }.

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)…