Let f(x) = F[x], let E/F be a splitting the Galois group. (i) If f(x) is irreducible, then G acts transitively on the set of all roots of f(x) (if a and Bare any two roots of f(x) in E, there exists σ € G with a (a)= B). (Hint: Lemma 50.) (ii) If f(x) has no repeated roots and G acts transitively on the roots, then f(x) is irreducible. Conclude, after comparing with Exercise 2, that irreducible polynomials are analogous to regular polygons. (Hint: If = 1; if a is a root of f(x) = g(x)h(x), then the gcd (g(x), h(x)) P(x) such that o (a) is a root of h(x), then a (a) is a common root of
Let f(x) = F[x], let E/F be a splitting the Galois group. (i) If f(x) is irreducible, then G acts transitively on the set of all roots of f(x) (if a and Bare any two roots of f(x) in E, there exists σ € G with a (a)= B). (Hint: Lemma 50.) (ii) If f(x) has no repeated roots and G acts transitively on the roots, then f(x) is irreducible. Conclude, after comparing with Exercise 2, that irreducible polynomials are analogous to regular polygons. (Hint: If = 1; if a is a root of f(x) = g(x)h(x), then the gcd (g(x), h(x)) P(x) such that o (a) is a root of h(x), then a (a) is a common root of
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![(ii) Prove
=
79. Let f(x) E F[x], let E/F be a splitting field, and let G
the Galois group.
Gal(E/F) be
(i) If f(x) is irreducible, then G acts transitively on the set of all roots of
f(x) (if a and B are any two roots of f(x) in E, there exists σ € G
with a (a) = 6). (Hint: Lemma 50.)
(ii) If f(x) has no repeated roots and G acts transitively on the roots, then
f(x) is irreducible. Conclude, after comparing with Exercise 2, that
irreducible polynomials are analogous to regular polygons. (Hint: If
= 1; if a is a root of
f(x) = g(x)h(x), then the gcd (g(x), h(x))
g(x) such that a (a) is a root of h(x), then σ (a) is a common root of
g(x) and h(x).)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F14e5feb4-438d-4d91-8851-92fadf2e52c4%2F9f0ff091-a89a-4be8-8dbf-53a5e16e3189%2Fvrwtl1_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(ii) Prove
=
79. Let f(x) E F[x], let E/F be a splitting field, and let G
the Galois group.
Gal(E/F) be
(i) If f(x) is irreducible, then G acts transitively on the set of all roots of
f(x) (if a and B are any two roots of f(x) in E, there exists σ € G
with a (a) = 6). (Hint: Lemma 50.)
(ii) If f(x) has no repeated roots and G acts transitively on the roots, then
f(x) is irreducible. Conclude, after comparing with Exercise 2, that
irreducible polynomials are analogous to regular polygons. (Hint: If
= 1; if a is a root of
f(x) = g(x)h(x), then the gcd (g(x), h(x))
g(x) such that a (a) is a root of h(x), then σ (a) is a common root of
g(x) and h(x).)
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