- #1
bham10246
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Something I've been reflecting about:
Question 1: Given any degree 5 polynomial f over the rationals, what are the conditions on the roots of f if I want its Galois group to be isomorphic to the alternating group [itex]A_5[/itex]?
I know that if f is an irreducible degree 5 polynomial with 3 real roots and 2 complex roots, then its Galois group contains a 2-cycle and a 5-cycle, which implies Gal(f) is isomorphic to [itex]S_5[/itex]. But what about the conditions on the roots in order for the Galois group to be [itex]A_5[/itex]?
Question 2: If f is an irreducible degree 5 polynomial whose Galois group is solvable, then are the following true?
Case 1. If f has only two complex roots and three real roots, then Gal(f) is isomorphic to [itex]\mathbb{Z}_5 (semidirect\: product)\: \mathbb{Z}_2[/itex].
Case 2. If f has four complex roots, then Gal(f) has a 5-cycle and two transpositions. So Gal(f) is isomorphic to [itex]\mathbb{Z}_5 (semidirect\: product)\: \mathbb{Z}_2\times \mathbb{Z}_2 [/itex]?
Question 3: But if the Galois group is abelian and f is an irreducible polynomial of degree 5 with at least 2 complex roots, isn't the field extension corresponding to this Galois group isomorphic to a subgroup of the cyclotomic extension? Thus if the polynomial is irreducible of degree 5 with at least 2 complex roots, then could Gal(f) be isomorphic to [itex]\mathbb{Z}_5[/itex]? I think that it cannot, but I cannot come up with a valid argument.
Thank you.
Question 1: Given any degree 5 polynomial f over the rationals, what are the conditions on the roots of f if I want its Galois group to be isomorphic to the alternating group [itex]A_5[/itex]?
I know that if f is an irreducible degree 5 polynomial with 3 real roots and 2 complex roots, then its Galois group contains a 2-cycle and a 5-cycle, which implies Gal(f) is isomorphic to [itex]S_5[/itex]. But what about the conditions on the roots in order for the Galois group to be [itex]A_5[/itex]?
Question 2: If f is an irreducible degree 5 polynomial whose Galois group is solvable, then are the following true?
Case 1. If f has only two complex roots and three real roots, then Gal(f) is isomorphic to [itex]\mathbb{Z}_5 (semidirect\: product)\: \mathbb{Z}_2[/itex].
Case 2. If f has four complex roots, then Gal(f) has a 5-cycle and two transpositions. So Gal(f) is isomorphic to [itex]\mathbb{Z}_5 (semidirect\: product)\: \mathbb{Z}_2\times \mathbb{Z}_2 [/itex]?
Question 3: But if the Galois group is abelian and f is an irreducible polynomial of degree 5 with at least 2 complex roots, isn't the field extension corresponding to this Galois group isomorphic to a subgroup of the cyclotomic extension? Thus if the polynomial is irreducible of degree 5 with at least 2 complex roots, then could Gal(f) be isomorphic to [itex]\mathbb{Z}_5[/itex]? I think that it cannot, but I cannot come up with a valid argument.
Thank you.