Consturct a splitting field over of . Find its dimensionality over .
Let is the primitive 5th root of unity in , whose minimal polynomial is .
has roots in . Then a splitting field of is . And .
To see this, , because is irreducible by Eisenstein criterion. Furthermore is irreducible in . Suppose not, since , must split into two quadratic polynomial, namely . So . While satisfies , which has roots and . So . This is impossible because does not divide 5.
We can consider this problem in another way. , is irreducible in . Otherwise can be imbedded in . This is impossible because .
Find a primitive element of a splitting field of over .
The Galois group of over are permutations of roots. Every permutation is uniquely determined by and . Let , the only fixes is . So is the splitting field of and is a primitive element.

Search

Recent

Categories
 Algebra (22)
 Galois Theory (20)
 Group Theory (1)
 Rings, Modules, Homology (2)
 Analysis (41)
 Complex Analysis (11)
 Harmonic Analysis (9)
 Inequalities (8)
 Real Analysis (16)
 Geometry (30)
 Hyperbolic Geo (3)
 Riemannian Geo (26)
 PDE (69)
 Elliptic PDE (45)
 Mean curvature flow (6)
 NavierStokes Eqns (9)
 Parabolic PDE (2)
 Schrodinger Eqn (1)
 Yamabe flow (4)
 Uncategorized (7)
 Algebra (22)

Tags
Algebraically closed alternating group barrier function Besicovitch caccioppoli ineq conformally invariant convergence radius 1 covering lemma cubic equation cyclic extension cyclotomic field dihedral group Dirichlet problem discriminant distance function Eisenstein criterion field of characteristic p finite extension fourier transform fully nonlinear galois group Gamma function geodesic green formula green function harmonic functions harnack inequality Hibert's Satz 90 Hilbert transform holder hypersurface Hölder Continuity injective intermediate subfield intersection invariant field irreducible Jacobi field line segment maximum principle Mean Oscillation Minkowski inequality Neumman problem newtonian potential nonnegative coefficients nonextentable poincare ineq pole polynomial primitive element primitive root of unity projective module regularity Riemann mapping Riemann surface root tower rouche separable simply connected singular point sobolev space solenoidal vector field solvable group splitting field subgroup symmetric group torus trace and norm transitive uniform continuous uniqueness continuation univalent weak derivative yamabe zero 
Blog Stats
 29,196 hits

Meta
Comments
I don’t understand why you can conclude that $Q(\sqrt[5]{2})$ can be embedded in $Q(\omega)$ when $x^5 – 2$ is not irreducible over $Q(\omega)$. I know that if one root is in $Q(\omega)$ then so are the rest, but how do you show that $x^5 – 2$ cannot be factored into an irreducible cubic and an irreducible quadratic over $Q(\omega)$? Is there a standard trick that one uses in such circumstances?
I think all the roots of $x^52$ are $\sqrt[5]{2}w^i$, $i=0,1,2,3,4$. The multiplication of any two of roots are not in $Q(\omega)$. So you can not factorize to irreducible quadratic and cubic
Okay, that makes sense. Thank you.