A Question regarding root of Bring quintic not expressible with radicals

[Stephen Tashi]

Hi Stephen:
I don't understand the point you are making in the above. My original question is:

What would a proof that "r is not expressible in terms of radicals" look like?

The first step of such a proof is that the Galois theory shows the equation is not solvable in radicals.

My interpretation of what you meant by "r" was that "r" denoted an arbitrary root of $x^5 - x + 1 = 0$ or an arbitrary root of some arbitrary equation, but perhaps you mean "r" is the particular unique real root of that equation.

The mathematical question that I find interesting is "How do we (or Galois) show that the equation is not solvable in radicals?"

If you want to take Galois theory for granted, then I see what you are doing. It involves the interpretation of the phrase "not solvable in radicals". Your interpretation of "An equation p(x) = 0 is not solvable in radicals" is that "There exists at least one root r of p(x) = 0 such that r cannot be expressed as a function of the coefficients of the equation using only arithmetic operations and radicals". Another interpretation of "not solvable in radicals" would be "No root of p(x) = 0 can be expressed as a function...etc.".

When we say "Galois theory shows p(x) = 0 is not solvable in radicals" which of those interpretations is correct? At the moment, I don't know.

The problem for me is to interpret the abstract formulation of "solvable" in terms of "there exists a tower of field extensions such that...". Does "solvable" mean that there exists one root of p(x) = 0 that is contained in such a tower of field extensions? Or does "solvable" mean that there exists a single tower of field extensions that contains all the roots of p(x) = 0? Or does "solvable" mean for each root r of p(x) = 0 there exists a tower of field extensions that contains r - but possibly there are different towers for different roots?

 

[Buzz Bloom]

Hi Stephen:

perhaps you mean "r" is the particular unique real root of that equation.

Yes, that is what I meant by

This simple example of a Bring-Jerrard quintic equation has one real root, with an approximate value of
r = -1.16730397783.

When we say "Galois theory shows p(x) = 0 is not solvable in radicals" which of those interpretations is correct? At the moment, I don't know.

The meaning is that at least one root is not expressible in radicals.

And you're also right that this argument fails for higher degrees, because in this case q(x)=0q(x)=0 may not be solvable by radicals as it is for polynomials of degree four.

Here fresh_42 confirms that this is correct since for 6th order and higher degree equations my proof does not hold, and such equations may well have less that all roots failing to be expressible in radicals.

Hope this clarifies the issues we were discussing.

The problem for me is to interpret the abstract formulation of "solvable" in terms of "there exists a tower of field extensions such that...".

I had previously hoped to understand the process of using Galois theory to prove just the single example

x⁵ − x + 1 = 0.

is not solvable in radicals. However, I have given up on that.

I am wondering if you know a source that shows the application of Galois theory to demonstrate in detail that anyone 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

 

[fresh_42]

I had previously hoped to understand the process of using Galois theory to prove just the single example $x^5-x+1=0$ is not solvable in radicals. However, I have given up on that.

One could start with a solution $\sigma_0$. Then $x^5-x+1=(x-\sigma_0)(x^4+\sigma_0x^3+\sigma_0^2x^2+\sigma_0^3x+(\sigma_0^4-1))$. We now know that this polynomial in $\mathbb{Q}(\sigma_0)[x]$ can be factored into $(x-\sigma_1)(x-\sigma_2)(x-\sigma_3)(x-\sigma_4)$ where $\sigma_1,\sigma_2,\sigma_3,\sigma_4$ can be expressed by radicals of $\mathbb{Q}(\sigma_0)$.
Unfortunately the equations aren't as pleasant as they are for quadratic polynomials and it won't show us whether $\sigma_0$ can be expressed by radicals, too, or not.

What we can do is gather the equations that $x^4+\sigma_0x^3+\sigma_0^2x^2+\sigma_0^3x+(\sigma_0^4-1))=(x-\sigma_1)(x-\sigma_2)(x-\sigma_3)(x-\sigma_4)$ gives us:

• $\sigma_1\sigma_2\sigma_3\sigma_4=\sigma_0^4-1=(\sigma_0+i)(\sigma_0-i)(\sigma_0+1)(\sigma_0-1)$
  
• $\sigma_1+\sigma_2+\sigma_3+\sigma_4=-\sigma_0$
  
• $\sigma_1\sigma_2+\sigma_1\sigma_3+\ldots+\sigma_3\sigma_4=\sigma_0^2$
  
• $\sigma_1\sigma_2\sigma_3+\sigma_1\sigma_2\sigma_4+\sigma_1\sigma_3\sigma_4+\sigma_2\sigma_3\sigma_4=-\sigma_0^3$

Now $\mathcal{Sym}(\{\sigma_i\}) \twoheadrightarrow Aut_\mathbb{Q}(\mathbb{Q}(\sigma_0,\ldots,\sigma_4))$, i.e. every permutation of the $\sigma_i$ induces an automorphism of $\mathbb{Q}(\sigma_0,\ldots,\sigma_4)$, and the problem is reduced to the question of the kernel of this homomorphism. If it's injective or at least not bigger than $\mathbb{Z}_2$ we're done. If not we should look for a normal series which would guide us the way the solutions can be found.

 

[Buzz Bloom]

kernel of this homomorphism.

Hi @fresh42:
I appreciate your effort to educate me. I am stuck right away on the above.

I am guessing that to actually do the necessary work, one has to write down the kernel of this homomorphism.
First: what is the definition of the underlined term.
From https://en.wikipedia.org/wiki/Kernel_(algebra):

The definition of kernel takes various forms in various contexts. But in all of them, the kernel of a homomorphism is trivial (in a sense relevant to that context) if and only if the homomorphism is injective. The fundamental theorem on homomorphisms (or first isomorphism theorem) is a theorem, again taking various forms, that applies to the quotient algebra defined by the kernel.​

I confess this does not help me at all. So, maybe you can tell me, is the homomorphism injective for the particular quintic, therefore leading to a "trivial" case?
From https://en.wikipedia.org/wiki/Injective_function

In mathematics, an injective function or injection or one-to-one function is a https://www.physicsforums.com/javascript:void(0) that preserves distinctness: it never maps distinct elements of its domain to the same element of its codomain.​

I confess this does not help me at all. Furthermore, when I keep digging down through more and more definitions, I get lost.

So for now, perhaps you can tell me whether this homomorphism is injective for the particular quintic equation I started with. If so, then perhaps you can also write down the trivial kernel. Then I may be able to ask another question.

Regards,
Buzz

 

[Stephen Tashi]

How do we know when we need the n-th power of a number? (The equation might not be as simple as $x^n - A = 0$)

If we have added some radicals to the list of constants that we are permitted to use then how do we know when a different radical can't be equal to one of the permitted expressions?

In these cases we would have to define a basis of this field, e.g. over $\mathbb{Q}$, and see whether our number can be expressed in this basis or not, i.e. we'll have to test on linear dependency.

I agree, but the poorly explained aspect of Galois theory is how "test on linear dependency" has something to do with groups and field automorphisms.

A typical "test on linear dependency" on vectors involves trying to solve a linear vector equation, which is interpreted as a set of simultaneous linear equations.

Your post #33 indicates that the simultaneous equations we are trying to solve in Galois theory are non-linear. This suggests that that the theory of groups and field automorphisms is useful in reasoning about the solutions of special types of non-linear equations.

The exposition of Galois theory would be clearer if it first explained how group theory is useful in solving certain special types of simultaneous multivariate algebraic equations. It is not necessary to begin by explaining how these equations arise from reasoning about the roots of polynomial equations.

 

[fresh_42]

Your post #33 indicates that the simultaneous equations we are trying to solve in Galois theory are non-linear. This suggests that that the theory of groups and field automorphisms is useful in reasoning about the solutions of special types of non-linear equations.

Yes, you're right. It is basically of the same difficulty as finding a relation in a group, or to show there is none. This is - as far as I know - not a trivial issue. I'm still thinking about @Buzz Bloom 's quest to solve it. At least the isomorphism $\mathbb{Q}(\sigma_i) \cong \mathbb{Q}[x]/(x^5-x+1)$ allows us to only consider polynomials and no fractions. I think the automorphism group is the whole $\mathcal{Sym}(5)$, but I have no idea so far, how to prove the $\sigma_i$ actually behave like variables, or indeterminates, i.e. have no non-trivial relation $f(\sigma_0,\ldots,\sigma_4)=0\,.$

 

[mathwonk]

one year after teaching graduate algebra many times without getting to galois theory, i wanted to understand it so much that I began my course with this topic. It turned out one cannot do that, since one needs to know about polynomial rings, vector space dimension, and groups, including the concept of normal sub groups and simple groups. After developing this prerequisite material, I did address the problem of which polynomials can be solved by radicals, in detail, in section 843-2 of my algebra course notes. I tried to survey this in the first 7 pages or so of these notes, and then spent the rest of the 56 or so pages filling in the details. See if reading the first 4, or first 7 pages of these free notes helps: If necessary you may want to refer back to the notes from 843-1 for background on groups.

http://alpha.math.uga.edu/%7Eroy/843-2.pdf

If you are still interested in why X^5 - X + 1 has no root expressible over Q by radicals, note that if it did, then taking the negative of that root would give such a root of X^5 -X -1, and this is a standard example of an equation with non solvable Galois group, going back to the book of Van der Waerden and reproduced in some more modern books like Dummit and Foote. Also see these free notes, example 3.33. The technique is to reduce the equation modulo the primes 2,3, and use the fact that those reduced Galois groups are isomorphic to subgroups of this one, and in a way that preserves the cycle type of the permutations of the roots.

http://citeseerx.ist.psu.edu/viewdoc/download?rep=rep1&type=pdf&doi=10.1.1.211.2314

pages 16 and 17 of those notes are especially relevant to your question, and explain how to check in many cases, using a computer, whether a galois group is isomorphic to S(p) or A(p).

 

[Stephen Tashi]

It turned out one cannot do that, since one needs to know about polynomial rings, vector space dimension, and groups, including the concept of normal sub groups and simple groups.

Even after those topics in abstract algebra are understood there is still a big pedagogical problem of explaining the connection between the definition of "solving an equation by radials" in the sense it is understood in secondary school algebra and the re-definition of that concept which is used in abstract algebra. I wonder if the solution to this pedagogical problem is to employ even more abstraction by applying mathematical ideas from the study of abstract formal languages -i.e. formal rules for manipulating strings of symbols. (Something like what Ritt did for the concept of "closed form solutions" https://en.wikipedia.org/wiki/Joseph_Ritt .)

For example, considering Abel's proof, as expounded on a blog: http://fermatslasttheorem.blogspot.com/2008/09/abels-impossibility-proof-radicals-of.html , we find that it uses language that could be rephrased as properties of string manipulations in some formal computer language. It uses ideas such as "nested" , "at the deepest level" , "of the form" etc.

e.g.

If the solution to the general quintic is expressible as nested radicals then for all radicals of the form $R^{1/m}$, $m = 2$.

e.g. ( http://fermatslasttheorem.blogspot.com/2008/10/abels-impossibility-proof.html)

If a solution to this equation exists it can be expressed as follows (see Theorem 5 here):

$y = p + R^{1/m} + p_2 R^{2/m} + p_3 R^{3/m} + ...+ p_{m-1} R^{(m-1)/m}$

where $m$ is a prime number and $R,p_1,p_2..$ .are functions of this same form finitely nested at the deepest level each $p,p_i, R$ is a function of the coefficients of the general quintic equation.

From the point of view of intellectual honesty, to formally present the connection between the elementary algebra sense of solution by radicals and the abstract algebra definition of the concept, it is necessary to formalize all the language involving concepts of string manipulations (e.g. "nested" , "of the same form"). To do this in the typical algebra course would be a kind of mathematical culture shock because the study of formal languages is traditionally the domain of the Computer Science department.

Traditional graduate education of algebraists can proceed satisfactorily by hand-waving and citing the typical examples (e.g. quadratic and cubic equations) to connect the elementary algebra concept of solution by radicals to its abstract re-definition. I'm not advocating a revision to the standard curriculum. However, for the benefit of people who are curious about the connection between the two concepts and also understand formal presentations, it would be nice to have a formal presentation of string manipulation aspects of the topic.