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Saturday, December 7, 2013

Dihedral Galois Closures and Quadratic Extensions (14.6.11-12)

Dummit and Foote Abstract Algebra, section 14.6, exercises 11-12:

MathJax TeX Test Page 11. Let F be a non-Galois extension of degree 4 over . Show that the Galois closure L of F has Galois group either S4, A4, or D8. Further, show that the Galois group is dihedral if and only if F contains a quadratic extension of .

12. Let F be an extension of degree 4 over . Show that F can be generated over by an element with minimal polynomial of the form x4+ax2+b if and only if F contains a quadratic extension.

Proof: (11) Let F=(θ). We observe its minimal polynomial mθ(x) of degree 4. It is irreducible, and since simultaneously we have FL and F is not Galois, by the procedure for quartics only the groups of order greater than 4 remain, which are those above.

() By the fundamental theorem, since every subgroup of order 2 in D8 is contained in a subgroup of order 4, we have F contains a subfield of index 4 in L, i.e. degree 2 over .

() Let (D)F be quadratic. Then we have F=(D)(a+bD)=(a+bD) for any nonsquare a+bD, which exists since quadratic extensions are generally extensions via square roots. Let α=a+bD. We observe x42ax2+a2b2D is the minimal polynomial for α. The Galois closure is the splitting field for this polynomial. We observe the resolvent cubic for this quartic is x3+4ax2+4b2Dx which clearly has a factor of x. By the procedure the Galois group cannot be either S4 or A4, hence must be D8.

(12) () Let α have such a minimal polynomial. Factor x2+ax+b=(xc)(xd) so that x4+ax2+b=(x2c)(x2d). Necessarily c and d are not contained in as the minimal polynomial must be irreducible, so they must generate a quadratic extension. Hence α2 generates a quadratic extension in F.

() As we saw previously, without assuming F is non-Galois, we can arrive at a minimal polynomial x42ax2+a2b2D for a generator of F. 

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