Exercise: The aim of this exercise is to see how you might derive the cubic formula from Galois theory. Note that the formula was found several centuries before Galois theory was developed, so this is not how it was done originally. But this should give you an idea how Galois theory can be used in reverse to find solutions in radicals for those polynomials that have solvable Galois groups. Throughout, let E be a field of characteristic 0 containing a primitive cube root of unity we'll call C. Note that satisfies ² + + 1 = 0. [For example, we might have E = Q(e²πi/³), but do not assume that this is what E is.] Let K = E(t₁, t2, t3) and let S₁ = t₁ +t₂ +t3 S2 = t₁t₂ + t₁t3 + t2t3 S3 = t₁t₂t3 = KS3. be the elementary symmetric polynomials. Let F E(81, 82, 83), So F Then t₁, t2 and t3 are the roots of f(x) = x³ – $₁x² + $2x − 83 € F[x] and K is the splitting field for f(x) over F. We will find a formula for t₁ in terms of 8₁, 82, and 83. (Formulas for t2 and t3 can be found similarly.) = Let S3 be the symmetric group, generated by the permutations o = (1 2 3) and T = (1 2). Let A3 = {e, 0, 0²}. Let S3 act on K by permuting t₁, t2, and t3.
Exercise: The aim of this exercise is to see how you might derive the cubic formula from Galois theory. Note that the formula was found several centuries before Galois theory was developed, so this is not how it was done originally. But this should give you an idea how Galois theory can be used in reverse to find solutions in radicals for those polynomials that have solvable Galois groups. Throughout, let E be a field of characteristic 0 containing a primitive cube root of unity we'll call C. Note that satisfies ² + + 1 = 0. [For example, we might have E = Q(e²πi/³), but do not assume that this is what E is.] Let K = E(t₁, t2, t3) and let S₁ = t₁ +t₂ +t3 S2 = t₁t₂ + t₁t3 + t2t3 S3 = t₁t₂t3 = KS3. be the elementary symmetric polynomials. Let F E(81, 82, 83), So F Then t₁, t2 and t3 are the roots of f(x) = x³ – $₁x² + $2x − 83 € F[x] and K is the splitting field for f(x) over F. We will find a formula for t₁ in terms of 8₁, 82, and 83. (Formulas for t2 and t3 can be found similarly.) = Let S3 be the symmetric group, generated by the permutations o = (1 2 3) and T = (1 2). Let A3 = {e, 0, 0²}. Let S3 act on K by permuting t₁, t2, and t3.
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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![Exercise: The aim of this exercise is to see how you might derive the cubic formula
from Galois theory. Note that the formula was found several centuries before Galois
theory was developed, so this is not how it was done originally. But this should give
you an idea how Galois theory can be used in reverse to find solutions in radicals
for those polynomials that have solvable Galois groups.
Throughout, let E be a field of characteristic 0 containing a primitive cube root
of unity we'll call C. Note that satisfies 2 + + 1 = 0. [For example, we might
have E = Q(e²πi/3), but do not assume that this is what E is.] Let K = E(t₁, t2, t3)
and let
$₁ = t₁+t₂ + t3
82 t₁t2 + t₁t3 + t2t3
S3 = t₁t₂t3
=
=
be the elementary symmetric polynomials. Let F = E(81, 82, 83), So F = KS³.
Then t₁, t2 and t3 are the roots of f(x)
x³ - s₁x² + s₂x
83 € F[x] and K is the
splitting field for f(x) over F. We will find a formula for t₁ in terms of 8₁, 82, and
$3. (Formulas for t2 and t3 can be found similarly.)
Let S3 be the symmetric group, generated by the permutations o= (1 2 3) and
(1 2). Let A3 = {e,0,0²}. Let S3 act on K by permuting t₁, t2, and t3.
T =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F081e66de-98dd-47c4-ad89-12bf6ce96be8%2Fe301ed9d-50e4-4dbd-99f3-73ebaeb6200b%2F35v6tmc_processed.png&w=3840&q=75)
Transcribed Image Text:Exercise: The aim of this exercise is to see how you might derive the cubic formula
from Galois theory. Note that the formula was found several centuries before Galois
theory was developed, so this is not how it was done originally. But this should give
you an idea how Galois theory can be used in reverse to find solutions in radicals
for those polynomials that have solvable Galois groups.
Throughout, let E be a field of characteristic 0 containing a primitive cube root
of unity we'll call C. Note that satisfies 2 + + 1 = 0. [For example, we might
have E = Q(e²πi/3), but do not assume that this is what E is.] Let K = E(t₁, t2, t3)
and let
$₁ = t₁+t₂ + t3
82 t₁t2 + t₁t3 + t2t3
S3 = t₁t₂t3
=
=
be the elementary symmetric polynomials. Let F = E(81, 82, 83), So F = KS³.
Then t₁, t2 and t3 are the roots of f(x)
x³ - s₁x² + s₂x
83 € F[x] and K is the
splitting field for f(x) over F. We will find a formula for t₁ in terms of 8₁, 82, and
$3. (Formulas for t2 and t3 can be found similarly.)
Let S3 be the symmetric group, generated by the permutations o= (1 2 3) and
(1 2). Let A3 = {e,0,0²}. Let S3 act on K by permuting t₁, t2, and t3.
T =
Transcribed Image Text:(i) Use the preceding parts to write t₁ explicitly in terms of $1, 82, and 83 (and
and radicals). (It's traditional here to bring all constants multiplying a
radical inside the radical.)
=
(j) Suppose you have a cubic over Q of the form f(x) = x³ + px +q (called a
depressed cubic), so s₁ = 0, $2 = P, and S3 = -q. Write out a formula for
a root of f(x) in terms of p and q. Compare your answer to the formula
given on the first page of this semester's lecture notes.
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