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A Question regarding root of Bring quintic not expressible with radicals

  1. Dec 2, 2016 #1
    I became curious about the following problem from a discussion in another thread:
    After a bit of study I concluded that the meaning of the assertion below regarding some specific real number rl P has the meaning which follows it.
    Assertion: "r is not expressible in terms of radicals."
    Meaning: r is not expressible in terms of a finite application of a collection of operators (+, -, ×, and /, together with any of the n-th roots (where n a positive integer) ), where these operators are applied to integer operands, or to expressions of the same kind.​
    If this meaning is incorrect, I hope someone will correct it.

    The article
    contains the following text:
    An example of a quintic whose roots cannot be expressed in terms of radicals is x5x + 1 = 0.​
    This simple example of a Bring-Jerrard quintic equation has one real root, with an approximate value of
    r = -1.16730397783.​
    This can be seen in the attached PNG file.

    My QUESTION IS:
    What would a proof that "r is not expressible in terms of radicals" look like?​
    I have no idea whatever how one would go about proving the non-radical nature of just this one example.

    Any help would be much appreciated.
     

    Attached Files:

  2. jcsd
  3. Dec 2, 2016 #2

    fresh_42

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    It is correct. The starting point isn't any real number ##r## but a solution of ##p(r)=0## with a polynomial ##p(x) \in \mathbb{Q}[x]##, i.e. rational numbers are allowed. Polynomials of degree ##1## are trivial, of degree ##2## is what we learn at school (Vieta), and the formulas for degree ##3## and ##4## are a bit tricky, if not to say rather unpleasant (Cardano, Ferrari). And of course one has to be a bit more careful with a formal definition of "solvable by radicals", since expressions like ##\sqrt[6]{1}## are either useless or ambiguous. The first time it cannot be done is with polynomials of degree ##5##. Of course this isn't true for all polynomials, because some are solvable by radicals: we only needed to multiply ##(x-a_1)(x-a_2)(x-a_3)(x-a_4)(x-a_5)## and have a beautiful expression of roots. The point is, that there is no method, which works in all cases.
    One way to prove it is:
    Compute the splitting field of a given polynomial ##p(x)## with the help of complex numbers and formal roots. Then determine the (finite) group of automorphisms of this field, which leave all rational numbers unchanged. This is a subgroup of ##\mathcal{Sym}(\deg p)##. Finally show, that this group isn't solvable.

    I'm not saying that it is the fastest method, because ##\mathcal{Sym}(5)## has already ##120## elements. Only a principle way of doing it. It also explains, why ##5## is the lowest degree, where we cannot solve ##p(x)=0## anymore: ##A_5 \subseteq \mathcal{Sym}(5)## is the smallest non-solvable group.

    Given a certain polynomial, it might be faster to show that the group isn't solvable, by proving it indirectly and deduce a contradiction to some theorems or the solvability itself.
     
  4. Dec 20, 2016 #3

    Stephen Tashi

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    That's a proof if any of those concepts can be related to definition that Buzz Bloom gave for "solvable by radicals". It would be service to world if someone would offer a simple explanation of the connection!
     
  5. Dec 20, 2016 #4

    fresh_42

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    I'm not sure I understood this correctly.

    The first part is to determine what "solvable by radicals" means. Basically it's a successive adjunction of primitive roots ##\sqrt[p]{r}##. Thus one gets a series of normal field extensions which corresponds to a series of normal subgroups of ##\mathcal{Sym}(\deg f(x))##, which is the definition of a solvable group. The correspondence is proven in Galois' theory. Now ##\mathcal{Sym}(n)\, , \,n \geq 5## isn't solvable anymore, so one only has to find an equation ##f(x)=0##, which actually has the entire group as symmetry group of its roots (Abel), i.e. all permutations define an automorphism of the splitting field.
     
  6. Dec 20, 2016 #5

    Stephen Tashi

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    That is the part where the world needs a good explanation!

    I don't think its hard to explain what it means to adjoin something to a field. But how do we explain the connection between "solvable by radials" given above (i.e. using a restricted repertoire of operations on the coefficients of an equation) and the process of adjoining roots to a field?
     
  7. Dec 20, 2016 #6

    fresh_42

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    If we solve ##f(x)=0## by radicals, it means we look for an expression ##x=r_0 + \sqrt[p_1]{r_1 + \sqrt[p_2]{r_2}+ \ldots}## Since we don't have, e.g. ##\sqrt{2}## in ##\mathbb{Q}##, we have to adjoin it in order to solve ##x^2-2=0##. One also wants to have all choices of ##\sqrt[p]{r}## in the field, ideally also as a solution, and that it always means the same choice, if written this way. In the end, radical is already the same word as root, only Latin. But you might be right. I've looked it up, how van der Waerden explained it and he, too, used a lot of text, examples and terms like "or similar". I start to understand what you might mean. Funnier is probably, how to connect compass and ruler constructions to square roots.
     
  8. Dec 21, 2016 #7

    Stephen Tashi

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    It's easier to see how things go in the reverse direction - how the coefficients of an equation are be created from the roots using a "limited repertoire" of operations.

    For example the equation ##(x - r_1)(x - r_1)(x - r_2) = 0## gives ##(1)x^3 + (-2r_1 - r_2) x^2 + (r_1^2 + 2r_1r_2)x + (-r_1^2r_2) = 0 ##

    So a natural question is whether we can "undo" the things created by limited repertoire of operations by using another (possibly different) repertoire of operations to recover the roots.

    Treatments of Lagrange/ Galois theory that make a concrete connection to polynomial equations always explain that the coefficients of a polynomial equation are symmetric functions of the roots.

    Some connection to group theory is made from the fact that the symmetric functions are invariant under "permutation of the roots". I don't find it easy to explain exactly what that means!

    "Symmetric functions" result from an equation with distinct "symbolic" roots such as ## (x-a)(x-b)(x-c) = 0 ## which says ##x^3 + (-a-b-c)x^2 + (ab + ac + cb)x - abc = 0 ##.

    For example, the coefficent of ##x^2## is the function defined by ##f(a,b,c) = -a-b-c ## is invariant under permutation of ##a,b,c## in the sense that for each triple of real numbers ##f(a,b,c) = f(a,c,b) = f(b,a,c) = ## etc.

    By contrast, in the first example, where two equal roots ##r_1## are assumed , the coefficent ##g(r_1,r_2) = -2r_1 - r_2 ## is not invariant under permutations of ##r_1## and ##r-2##.
     
    Last edited: Dec 21, 2016
  9. Dec 21, 2016 #8

    fresh_42

    Staff: Mentor

    What do you think about the following alternative (again van der Waerden, which I try to translate as close as I can)?

    Given the situation ##\mathbb{K} \subseteq \mathbb{K}(\vartheta) \subseteq \mathbb{L}## and ##f(x) \in \mathbb{K}[x]## irreducible with ##f(\vartheta)=0## and ##\mathbb{L}## the splitting field of ##f(x)##.

    The relative (fixing ##\mathbb{K}##) isomorphisms of ##\mathbb{K}(\vartheta)## can be indicated by their transformations of ##\vartheta## into its conjugates ##\vartheta_1 , \ldots , \vartheta_n## in ##\mathbb{L}##. Each element ##\varphi(\vartheta)=\sum a_\lambda \vartheta^\lambda## is thus transformed into ##\varphi(\vartheta_\nu) = \sum a_\lambda \vartheta_\nu^\lambda## which allows us to speak of substitutions ##\vartheta \rightarrow \vartheta_\nu## instead.

    He also emphasizes that ##\vartheta## and ##\vartheta_\nu## are only auxiliaries to handle the isomorphisms, which by their nature are independent on the choice of ##\vartheta##. He proceeds in talking of substitutions instead of permutations which I always regarded as simply a bit of an old-fashioned way.
     
  10. Dec 21, 2016 #9

    Stephen Tashi

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    I don't (yet) see it as an "alternative" - in the sense of something having the same direct relation to the solution of equations as the example of how permuting roots affects (or doesn't affect) the coefficients of a polynomial equation.

    Ignoring that problem for a moment:

    I have (the usual!) comprehension problem with algebraic objects like ##\mathbb{K}(\vartheta)##. If we think of ##\vartheta## is "a symbol" or "an indeterminate" then we get a different structure than if we think of ##\vartheta## as something that can have more properties than a symbol. For example, if I "adjoin" the symbol ##w## to the field of rational numbers then "##1 + 3w^2##" is an element of the expanded structure and "##7##" is also an element of the expanded structure and ##7 \neq 1 + 3w^2##. - i.e. there is no doubt that "##7##" and "##1 + 3w^2##" are distinct elements. However if I "adjoin" ##\sqrt{2}## instead of the abstract symbol ##w## then we have ##7 = 1 + 3\sqrt{2}^2 ## and these are not distinct elements.

    Is there a notational convention that distinguishes between these types of "adjoining"? - perhaps "##\mathbb{K}[\vartheta]##" vs "##\mathbb{K}(\vartheta)##" ?


    What would "conjugates" mean in this context? ( One guess is that a conjugate of ##\vartheta## would be ##\omega_j \vartheta## where ##\omega_j## is one of the "n-th roots of unity", which still leaves open the question of whether "##\vartheta##" is an "indeterminate" or something with more properties than an "indeterminate".)

    I think I understand that, but it's not easy to say it precisely. The set of (field) automorphisms of ##\mathbb{K}(\vartheta)## that, when restricted to ##\mathbb{K}## are automorphisms on ##\mathbb{K}## can be put in 1-to-1 correspondence with set of 1-to-1 mappings from the set of conjugates of ##\vartheta## to itself. Does that convey the idea?
     
  11. Dec 21, 2016 #10

    fresh_42

    Staff: Mentor

    Just for short, I'm going to read it more carefully later.

    ##\mathbb{K}(\vartheta)## is the quotient field of the ring ##\mathbb{K}[\vartheta]##, whether ##\vartheta## fulfills an equation or not. If it doesn't, it's simply the same as the field of all rational polynomials with coefficients in ##\mathbb{K}## in one variable ##x=\vartheta##.

    If there is an equation, like ##f(x) = x^2-2## with ##f(w)=0## in the example, then ##\mathbb{K}(w) \cong \mathbb{K}[x] / \left( \mathbb{K}[x]\cdot f(x)\right) = \mathbb{K}[x]/(x^2-2)## a quotient or factor ring. So in general, if we have the polynomials ##\mathbb{K}[x]## then ##\mathbb{K}(\vartheta)## is always a field of the form ##\mathbb{K}[x] / \left(\mathbb{K}[x]\cdot f(x)\right)##. (If ##f(x)=0## which corresponds to the case of a "symbolic" (better: transcendental of degree ##1##) extension, we have to require it to be the quotient field of this ring, because ##f(x)=0## doesn't define a maximal ideal anymore, but the construction is given by this isomorphism of rings.
     
  12. Dec 21, 2016 #11

    fresh_42

    Staff: Mentor

    A remark on the difference between ##\mathbb{K}(\vartheta)## and ##\mathbb{K}[\vartheta]##.
    ##\mathbb{K}(\vartheta)## usually denotes the quotient field of the integral domain (ring) ##\mathbb{K}[\vartheta]##.
    In the case of complex numbers this means ##\mathbb{C}=\mathbb{R}(i)## and ##\mathbb{R}[ i ] = \mathbb{R}[x]/(x^2+1) = \mathbb{R}(i)## in this case, because ##\frac{1}{i} = -i##. The reason for this coincidence is the maximality of ##(x^2+1)## as ideal in the ring ##\mathbb{R}[x]##. As mentioned above, this is no longer true for a transcendental extension and the zero ideal.
    It means the same as in the complex number field: its fellow roots, as ##-i## is the other root to ##i## in ##x^2+1##.
    I guess so. Although if elements of ##Aut_\mathbb{K}(\mathbb{L})## are restricted to ##\mathbb{K}##, they'll be the identity (by definition) - only this one automorphism. Since all coefficients are thus fixed, such a field automorphism has to leave ##f(x)\in \mathbb{K}[x]## unchanged. In the splitting field - which might be larger than ##\mathbb{K}(\vartheta)##, where ##f(x)## can be written as ##f(x)=c_0\cdot (x-\vartheta_1)\cdot \ldots \cdot (x-\vartheta_n) = \sum c_\nu(\vartheta_1, \ldots , \vartheta_n)x^{\nu}##, the only possibility is therefore to switch between the ##\vartheta_\nu \, : \,##
    $$f(x) = \varphi(f(x))= \sum c_0 \cdot (x-\varphi(\vartheta_1))\cdot \ldots \cdot (x-\varphi(\vartheta_n))=\sum c_\nu(\varphi(\vartheta_1), \ldots , \varphi(\vartheta_n))x^{\nu}$$
    and the permutation drops out for free.
     
  13. Dec 22, 2016 #12

    Stephen Tashi

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    Thank you clarifying that notation!

    However, we are digressing from the question in the original post unless we can explain the connection between solving equations and the abstract algebra of field extensions.

    Perhaps we need something like the discussion that begins on page 41 of this PDF:
    http://pages.uoregon.edu/koch/Galois.pdf , but I don't fully comprehend it yet.
     
  14. Dec 22, 2016 #13

    fresh_42

    Staff: Mentor

    I find Koch's presentation pretty detailed and good, although he avoids to give a precise meaning to "solvable by radicals", simply by defining it (p.56) as what can be proven in algebraic terms. This appears a little bit like cheating, in the sense that it leaves open the question between the algebraic formulation and the language theoretical formulation (using an alphabet). Nevertheless, his chapter 9 is quite illuminating.
     
  15. Dec 25, 2016 #14

    Stephen Tashi

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    I agree. It's cheating! It appears to be very difficult to make a connection between the definition of "solvable by radials" in the sense of "solvable with a limited repertoire of constants and operations" and the definitions used in contemporary algebra.

    The best attempt I've found online is an exposition of Abel's work: http://fermatslasttheorem.blogspot.com/2008_08_24_archive.html and (to me) the connection is still hazy.
     
  16. Dec 25, 2016 #15

    fresh_42

    Staff: Mentor

    Yes, but I guess it is pretty much straight forward, so most (all?) authors avoid the translation. I mean "solvable by radicals" as being a formal expression over an alphabet ##\{+,-,* , \frac{*}{*}, \sqrt[\text{*}\,]{*},a \in \mathbb{F}\}## is only an expression with field operations and adjunction of roots, i.e. every word over this alphabet is contained in some Galois extension of ##\mathbb{F}##. The other direction is clear by the definition of a field extension. So the only critical part is that our symbol for roots isn't unique and we have to choose exactly one to be meant (as mentioned by van der Waerden).
    (I think Abel lived before formal languages became a matter of mathematics, but I'm not sure.)
     
  17. Dec 25, 2016 #16
    Hi @fresh_42:

    I am unsure about what
    means. I assume F means a field. Does the "F" font you use mean any specific field? What does the "a" represent? I am guessing it represents any member of the field F, so that
    a∈F​
    represents the syntax to be used to say that a particular expression indicated by something that substitutes for "a" in "a∈F" is a member of the field F. Is that correct?

    Regards,
    Buzz
     
  18. Dec 25, 2016 #17

    fresh_42

    Staff: Mentor

    Yes. ##\mathbb{F}## was meant to be the (any) field over which the expression "solvable by radicals" is defined. It is usually abbreviated by a bold ##F## (field) or a ##K## (Körper) or in similarity to ##\mathbb{Q}, \mathbb{R}, \mathbb{C}## as ##\mathbb{F}## or ##\mathbb{K}##. I was talking about a possible definition of this expression as a word in a formal language. Therefore all elements of the field are needed to be part of the alphabet as well as the arithmetic operators and the radicals = roots.
     
  19. Dec 25, 2016 #18
    Hi to all of you who have participated in this thread:
    I much appreciate all your contributions.

    When I tried to look up various concepts used in various posts on the internet, mostly Wikipedia, I confess I got lost. In each article I read about a concept, it was explained in terms of other concepts I had to look up elsewhere. When the nesting of these concepts reached about seven, I gave up. At my advanced age, it was too much too much for me to try to keep these interrelationships of all these concepts in my head.

    My original question in post #1 was:
    It has taken me a while to figure out something like a sketch of a proof about how this question relates to proofs about the radical solvability of quintics and higher order equations. It goes like this:
    If a fifth order equation has any root expressible in a radical form, then all the roots will be similarly expressible. This is because the single such root r can be used to reduce the quintic equation to a quartic equation with coefficients in the rational field extended by expressions using radical operations involving r. The solution of the quartic in terms of these these coefficients will then also be radical.
    Therefore, if one proves that a particular quintic has a radical solution, this means that all of its roots have a radical expression.
    This "proof" may fail to work for sixth and higher order equations.​
    If the above is incorrect, I would much appreciate an explanation regarding my error(s).

    Regards,
    Buzz
     
  20. Dec 25, 2016 #19

    fresh_42

    Staff: Mentor

    Yes, it's correct what you've said. Polynomial equations up to order four are solvable by radicals, i.e. can be expressed in the form discussed above.

    Now in general if one has a polynomial of degree ##n##, say ##p(x) = a_0 + a_1x+\ldots+a_nx^n \in \mathbb{F}[x]## and a root ##r## in some field extension, say ##r \in \mathbb{G}## then ##(x-r)## divides ##p(x)##, i.e. ##p(x)=(x-r)q(x)## and ##q(x)## is of degree ##n-1## and ##q(x)## can be written in terms of the basis field and ##r##, i.e. ##q(x) \in \mathbb{G}[x]=\mathbb{F}(a)[x]##. (Often ##a=r##.)

    In the special case ##n=5##, we would get ##\deg q(x)=4## which is solvable by radicals wherever ##q(x)## lives in. So in summation we could solve ##p(x)=0## with the radicals from the solution of ##q(x)=0## and those needed for the coefficients of ##q(x)## in ##\mathbb{G}##. Then everything can be written by radicals.

    And you're also right that this argument fails for higher degrees, because in this case ##q(x)=0## may not be solvable by radicals as it is for polynomials of degree four.
     
  21. Dec 25, 2016 #20
    Hi @fresh_42:

    Thank you very much for your confirmation that my thinking was OK.

    I am wondering if you know a source that shows the application of Galois theory to demonstrate in detail that any one particular chosen example of a quintic equation has no radical solution. I am unable to understand Galois theory well enough to apply the general theory to make such a demonstration on my own, but I think I might be able to understand some one else's detailed description of such a single example demonstration.

    Regards,
    Buzz
     
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