HW12_selected_solns

HW12_selected_solns - College of the Holy Cross Fall 2009...

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Unformatted text preview: College of the Holy Cross, Fall 2009 Math 351 – Homework 12 selected solutions Chapter 32, #4. Let E = Q ( √ 2 , √ 5 , √ 7). The problem tells us to assume that Gal( E/ Q ) ∼ = Z 2 ⊕ Z 2 ⊕ Z 2 . By the Fundamental Theorem of Galois Theory, a subgroup H ≤ Gal / ( E/ Q ) has fixed field E H with [ E H : Q ] = | Gal( E/ Q ) | / | H | , and every subfield of E containing Q is of the form E H for some H . Thus to find subfield K of E with [ K : Q ] = 4, we need to find all subgroups of Gal / ( E/ Q ) with order 8/4 = 2. There is exactly one subgroup of order 2 for each element of order 2, and there are 7 elements of order 2 in Z 2 ⊕ Z 2 ⊕ Z 2 . Hence there are 7 subfields of E with [ K : Q ] = 4. Chapter 32, #5. To show E H is a subfield of E , we show it is a subring and that all non- zero elements of E H have inverses in E H . So let a,b ∈ E H , which means that for all φ ∈ H , we have φ ( a ) = a and φ ( b ) = b . Thus for all φ ∈ H we have φ ( a- b ) = φ ( a )- φ ( b ) (since φ is a ring homomorphism), and this latter expression is a- b . So a- b ∈ E H . Similarly, we have φ ( ab ) = φ ( a ) φ ( b ) = ab . So E H is closed under subtraction and multiplication and thus is a subring of E . Now let a ∈ E H with a 6 = 0. Then a- 1 is an element in E since E is a field. Thus for all φ ∈ H , φ ( a- 1 ) = φ ( a )- 1 (since φ is a ring homomorphism), and this equals a- 1 . So a- 1 is fixed by all elements of H and hence is in E H ....
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HW12_selected_solns - College of the Holy Cross Fall 2009...

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