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amcavoy

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amcavoy

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matt grime

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amcavoy

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Yes, that and also the solutions to cubics and quartics.

Thanks.

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robert Ihnot

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A Dover paperback, 1971, on that is by Allan Clark, "Elements of Abstract Algebra," and he has a section on "Solvability of Equations by Radicals," on page 130-144.

Checking with*Amazon.com*, it has it new for $9.56 and used for $3.49. Also there are six reviews of the text given there. These reviews might prove useful.

Checking with

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HallsofIvy

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mathwonk

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A field of numbers is closed under the operations of +,-, mult and dividing. Hence applying those operations to any number in the field yields another element of the field.

But if one field lies inside another field, it is possible that applying those operations to an element of the larger field can bring it down into the smaller field. for example, squaring sqrt(2) which lies in a larger field than the rationals, brings it down into the rationals.

The problem you ask about concerns determining exactly which elements of larger fields can be brought down, by taking linear combinations of powers, to a smaller field. E.g. which real numbers become rational when an appropriate polynomial operation with rational coefficients, is applied to them?

One answer would be to try to explain how to enlarge the smaller field in such a way that obtains all such elements of larger fields. E.g. taking any root of an element of the smaller field, yields an element of a larger field which does descend back to the smaller field by taking a power.

So the general question is this: how is this recip[rocal relationsship affected by takling linear conbinations?

I.e. by taking successive linear combinations of roots, can we obtain all elements that descend to the smaller field under taking linear combinations of powers?

I.e. can we invert the process of taking polynomial operations by adding in the one process of taking roots?

The answer is found in the theory of symmetries of fields. I.e. the fields obtained from taking roots and also polynomial operations are more symmetric, than are the fields which descend to the smaller field on taking polynomial operations, hence the answer is no.

E.g. adding in a sqrt, like sqrt(2) to the rationals allows elements of form a + bsqrt(2), and exchanging this element with a - bsqrt(2), is a symmetry of the larger field. in this case the group of symmetries is the group of two elements.

Now these symmetries form a group in general, and adding in a root, only gives an abelian group of symmetries, hence field extensions obtained by successively adding roots, have symmetry groups which can be obtained by successive abelian extensions.

But in general the symmetry group of a field containing a root of a polynomial of degree n, may be the full symmetric group S(n) on n letters.

Now as long as this group can be obtained by abelian extensions, such as S(3) or S(4), then the answer to the basic question above of solvability by roots, i.e. radicals, is yes. but for S(5) the first big step in building it as a group, is A(5), which cannot be built up from abelian extensions. hence most polynomials of degree 5 or more cannot be "solved" by radicals.

I copy below the cubic formula from the first page of section 2 of my algebra notes on the topic, with VanderWaerden as one source. If you like I will send you my notes as a pdf file. Section 1 with the proof of abel's theorem, is at the bottom of the webpage

http://www.math.uga.edu/~roy/.

"For the solutions x1, x2 of the quadratic equation x^2+px+q = 0, we have (since at least 825 AD) the formula x = (1/2){-p + D^(1/2)}, in terms of the coefficients p, q, where D = (x1-x2)^2 = p^2-4q. The two solutions are obtained by taking the two square roots of D. For the cubic equation x^3+px+q = 0, years of toil and some intrigue led to the publication, by Cardano in 1545, of the following formula. x =

(1/3)[{(-27q/2)+(3/2)(-3D)^(1/2)}^(1/3) - 3p/{(-27q/2) + (3/2)(-3D)^(1/2)}^(1/3)].

Fixing a value of (-3D)^(1/2), where D = (x1- x2)^2(x1-x3)^2(x2-x3)^2 = -4p^3-27q^2, and varying the cube root gives all three solutions. Eg., in the equation x^3-1 = 0, p = 0, q = -1, D = -27, so we get x = (1/3){27/2 + 27/2}^(1/3) = {1/2 + 1/2}^(1/3) = 1^(1/3), as hoped.

Similarly, for x^3-a = 0, we have p = 0, q = -a, D = -27a^2, hence

x = (1/3){27a/2 + (3/2)(81a^2)^(1/2)}^(1/3) = (1/3){27a}^(1/3) = a^(1/3).

For x^3 - 4x = 0, we get p = -4, q = 0, D = 256, and so

x = (1/3) [{(3/2)(-768)^(1/2)}^(1/3) + 12/{(3/2)(-768)^(1/2)}^(1/3)]

= (1/3) [ {(-27)(64)}^(1/6) + 12/{(-27)(64)}^(1/6) ]

= (1/3) [ 2sqrt(3) i^(1/3) + 12/{2sqrt(3) i^(1/3)} ] = (2/sqrt(3) )( i^(1/3) + i^(-1/3))

= (4/sqrt(3) )Re(i^(1/3)).

Varying the cube roots of i in this formula gives

(4/sqrt(3) )(cos(pi/6)) = (4/sqrt(3) )(sqrt(3) /2) = 2,

(4/sqrt(3) )(cos(5pi/6)) = (4/sqrt(3) )(-sqrt(3) /2) = -2, and

(4/sqrt(3) )(cos(9pi/6)) = (4/sqrt(3) )(0) = 0.

Of course, factoring x^3-4x = x(x-2)(x+2) = 0 confirms these answers. (Notice a point which fascinated earlier workers, who were not entirely happy with "imaginary" numbers: the solution formula involves imaginaries even though the final answer it gives is real! It can be proved that this cannot be avoided.)"

But if one field lies inside another field, it is possible that applying those operations to an element of the larger field can bring it down into the smaller field. for example, squaring sqrt(2) which lies in a larger field than the rationals, brings it down into the rationals.

The problem you ask about concerns determining exactly which elements of larger fields can be brought down, by taking linear combinations of powers, to a smaller field. E.g. which real numbers become rational when an appropriate polynomial operation with rational coefficients, is applied to them?

One answer would be to try to explain how to enlarge the smaller field in such a way that obtains all such elements of larger fields. E.g. taking any root of an element of the smaller field, yields an element of a larger field which does descend back to the smaller field by taking a power.

So the general question is this: how is this recip[rocal relationsship affected by takling linear conbinations?

I.e. by taking successive linear combinations of roots, can we obtain all elements that descend to the smaller field under taking linear combinations of powers?

I.e. can we invert the process of taking polynomial operations by adding in the one process of taking roots?

The answer is found in the theory of symmetries of fields. I.e. the fields obtained from taking roots and also polynomial operations are more symmetric, than are the fields which descend to the smaller field on taking polynomial operations, hence the answer is no.

E.g. adding in a sqrt, like sqrt(2) to the rationals allows elements of form a + bsqrt(2), and exchanging this element with a - bsqrt(2), is a symmetry of the larger field. in this case the group of symmetries is the group of two elements.

Now these symmetries form a group in general, and adding in a root, only gives an abelian group of symmetries, hence field extensions obtained by successively adding roots, have symmetry groups which can be obtained by successive abelian extensions.

But in general the symmetry group of a field containing a root of a polynomial of degree n, may be the full symmetric group S(n) on n letters.

Now as long as this group can be obtained by abelian extensions, such as S(3) or S(4), then the answer to the basic question above of solvability by roots, i.e. radicals, is yes. but for S(5) the first big step in building it as a group, is A(5), which cannot be built up from abelian extensions. hence most polynomials of degree 5 or more cannot be "solved" by radicals.

I copy below the cubic formula from the first page of section 2 of my algebra notes on the topic, with VanderWaerden as one source. If you like I will send you my notes as a pdf file. Section 1 with the proof of abel's theorem, is at the bottom of the webpage

http://www.math.uga.edu/~roy/.

"For the solutions x1, x2 of the quadratic equation x^2+px+q = 0, we have (since at least 825 AD) the formula x = (1/2){-p + D^(1/2)}, in terms of the coefficients p, q, where D = (x1-x2)^2 = p^2-4q. The two solutions are obtained by taking the two square roots of D. For the cubic equation x^3+px+q = 0, years of toil and some intrigue led to the publication, by Cardano in 1545, of the following formula. x =

(1/3)[{(-27q/2)+(3/2)(-3D)^(1/2)}^(1/3) - 3p/{(-27q/2) + (3/2)(-3D)^(1/2)}^(1/3)].

Fixing a value of (-3D)^(1/2), where D = (x1- x2)^2(x1-x3)^2(x2-x3)^2 = -4p^3-27q^2, and varying the cube root gives all three solutions. Eg., in the equation x^3-1 = 0, p = 0, q = -1, D = -27, so we get x = (1/3){27/2 + 27/2}^(1/3) = {1/2 + 1/2}^(1/3) = 1^(1/3), as hoped.

Similarly, for x^3-a = 0, we have p = 0, q = -a, D = -27a^2, hence

x = (1/3){27a/2 + (3/2)(81a^2)^(1/2)}^(1/3) = (1/3){27a}^(1/3) = a^(1/3).

For x^3 - 4x = 0, we get p = -4, q = 0, D = 256, and so

x = (1/3) [{(3/2)(-768)^(1/2)}^(1/3) + 12/{(3/2)(-768)^(1/2)}^(1/3)]

= (1/3) [ {(-27)(64)}^(1/6) + 12/{(-27)(64)}^(1/6) ]

= (1/3) [ 2sqrt(3) i^(1/3) + 12/{2sqrt(3) i^(1/3)} ] = (2/sqrt(3) )( i^(1/3) + i^(-1/3))

= (4/sqrt(3) )Re(i^(1/3)).

Varying the cube roots of i in this formula gives

(4/sqrt(3) )(cos(pi/6)) = (4/sqrt(3) )(sqrt(3) /2) = 2,

(4/sqrt(3) )(cos(5pi/6)) = (4/sqrt(3) )(-sqrt(3) /2) = -2, and

(4/sqrt(3) )(cos(9pi/6)) = (4/sqrt(3) )(0) = 0.

Of course, factoring x^3-4x = x(x-2)(x+2) = 0 confirms these answers. (Notice a point which fascinated earlier workers, who were not entirely happy with "imaginary" numbers: the solution formula involves imaginaries even though the final answer it gives is real! It can be proved that this cannot be avoided.)"

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