Polynomial equations of degree larger than 4

calculus

1,270

Any polynomial over a field factors completely over its splitting field. For real polynomials, it turns out that adjoining $\sqrt{-1}$ to $\mathbb{R}$ gives the algebraic closure, and so every real polynomial factors over $\mathbb{C}$. So your question's assumption is incorrect.

On the other hand, there are equations using radicals that more or less give the roots of a quadratic, cubic or quartic polynomial over a field of characteristic zero. In some cases these formulae are somewhat silly because it is usually not easier to find the cube-root of a complex number than to find the roots of the original cubic polynomial! It is useless to say that we can use exponentials, trigonometric functions and their inverses, since they also cannot be simplified. If we eventually want to compute an approximation, we do not even want to use the cubic equation! There are root-finding algorithms that work for any arbitrary polynomial over the reals and make formulae using radicals completely redundant in the real world where we do not need infinite precision.

But it is still an interesting question whether there is a general formula for the roots of a quintic polynomial over a field $F$ that only uses radicals, in other words whether the roots are in some radical extension of $F$. It turns out that it is possible exactly when the Galois group of the Galois closure of the field is solvable ("solvable" arose from this very problem of trying to solve polynomials), in other words there is a chain of groups from that group to the trivial group such that the quotient between consecutive groups in the chain is cyclic. So everything reduces to the nature of the Galois group. See https://math.stackexchange.com/a/38901/21820 for an overview of the possible Galois group of an irreducible quintic. I know that for a quartic over the rationals there are easy ways to determine the Galois group just by looking at the coefficients. I do not know if there are deterministic tests for quintics.

1,270

Moti

Updated on October 18, 2022

Comments

Moti about 1 year

It is accepted that there are no general solutions for polynomial equations of degree higher than 4, unless they have some unique features. We know that if we could factor the polynomial into polynomial of degrees smaller than 5, we may find the roots for each such polynomial with degree smaller than 5.

Are there systematic tools for deciding whether polynomials with integer coefficients are irreducible over the rationals, and for factoring them over the rationals if they are not irreducible over the rationals?
- Matt Samuel about 8 years
  
  There are general solutions for fifth, sixth, and seventh degree equations, they just involve functions that can't be expressed in terms of radicals.
- Moti about 8 years
  
  I would like to understand it more - any suggestions where to look for?
- Matt Samuel about 8 years
  
  I'm not an expert but here's a link: en.wikipedia.org/wiki/…
- Moti about 8 years
  
  My question if we could by evaluation of the coefficients determine that the polynomial may be factored.
- Moti about 8 years
  
  Thanks. Not exciting solution.
- Gerry Myerson about 8 years
  
  Note that asking whether a polynomial can be factored is not the same as asking whether one can find its roots. Anyway, in principle, one can find the Galois group of any given polynomial, and then determine whether that group is a solvable group. If it is, we can solve the polynomial in radicals; if not, not. In practice, it's not so easy to find the Galois group. The question of how to find the Galois group has come up on this website before, so try to have a look around.
- Moti about 8 years
  
  I am asking what I think a clear question, where the focus is on factoring of polynomials. This relates to solving the equation but it is not the same. What I wonder if there are ways to evaluate the coefficients (real or complex for this matter - but limited to this field of numbers) and for this to determine that factoring is possible.
- user21820 about 8 years
  
  @Moti: As I said in my answer, your question is based on a wrong assumption. If you do not understand my answer, say so, rather than saying that your question is correct. Every polynomial over the reals can be factored completely over the complex numbers, by definition of factoring. If you want to ask about factoring into real polynomials, ask a different question, because it is totally unrelated to the difference between quartics and higher-degree polynomials.
- Moti about 8 years
  
  @user21820 What is the assumption? Read carefully my question and try to point where is the erroneous assumption.
- Gerry Myerson about 8 years
  
  OK, so you are asking about testing polynomials for irreducibility (over, I will assume, the rationals), and about factoring those that are reducible. This can't be any easier than testing integers for primality, and factoring integers known to be composite. But there are techniques that work for polynomials of not-too-high degree with not-too-large coefficients. You could start with the Wikipedia article, en.wikipedia.org/wiki/Factorization_of_polynomials
- user21820 about 8 years
  
  As I said, you clearly do not know what you are talking about. First you claimed "there are no general solutions for polynomial equations of degree higher than 4". That is utterly meaningless unless you are talking about expressing the roots using only radicals over the original field! Then you claim that it is impossible to factor some polynomials. Since you said nothing about radicals, this claim is false. My later paragraphs deal with the question about radical field extensions, which is the only situation where your first claim makes any sense at all.
- user21820 about 8 years
  
  If your question is what Gerry just guessed, then it is completely unclear from your question, in which case as he said it would be at least as hard as factoring integers, which almost certainly has no formula to determine factorizability (we don't even know whether it is in P or not). Sorry Gerry I didn't intend to direct my comment to you.
- Moti about 8 years
  
  Actually Gerry provided a great respond. With integers we have tools for factoring in a systematic way EVERY integer by just checking for all primes smaller than the integer. It seems that basically the trouble with polynomials that PRIMES can not be well defined.
- Gerry Myerson about 8 years
  
  There are systematic ways to factor polynomials (over, I again assume, the rationals). It always reduces to a finite problem, and exhaustive search is available as a method. But exhaustive search is an impractical method for factoring integers (once they get beyond a few digits), and it is an impractical method for factoring polynomials (once the degree and/or the coefficients are not small).
- Moti about 8 years
  
  If I understand correctly what you say, in case the coefficients are rational (not real or complex or...) it is possible to factor a polynomial. Is this factoring to the level of other polynomials with rational coefficients or actually allow to get the roots? It seems that you understood perfectly what I was looking for.
- Gerry Myerson about 8 years
  
  Have you tried reading the Wikipedia page? Yes, given a polynomial with rational coefficients, there is a systematic way to decide whether it is irreducible over the rationals, and a systematic way to factor it into polynomials that are irreducible over the rationals. For polynomials of degree 5 or greater, there is, in general, no closed form for their roots in terms of radicals and field operations. By the way, if you want to be certain that I see a comment, you have to put @Gerry into it.
- Moti about 8 years
  
  @Gerry If you make your comment an answer I will accept it. You need to be rewarded for your help here. Thanks!
- Gerry Myerson about 8 years
  
  Since the question is on hold, I can't post an answer. What you can do is edit ithe statement of the question so as to clarify (people shouldn't have to wade through a long thread of comments to understand the question), and then post to the meta site, noting that you have edited the question, and asking for it to be reopened. If it gets reopened, then you can post an answer yourself, if you wish, based on what you have learned from the discussion.
- Moti about 8 years
  
  @Gerry The problem is that when you ask persons that know TOO much it hard for them to understand simpler question. Since I looked at the site it is clear that the basic assumption I made with regard to solutions could mislead, but anyone that knows the specifics could relate to the fact that the up to degree 4 (included) we have a solution which is straight forward - meaning an equation for the roots. It turns out that there are also solutions (not as elegant) for higher degrees. I do not know how I could edit - I mentioned the degree 4 and mentioned factoring. What else?
- Gerry Myerson about 8 years
  
  If I have worked out what you want, your question is, are there systematic tools for deciding whether polynomials with integer coefficients are irreducible over the rationals, and for factoring them over the rationals if they are not irreducible over the rationals? If that's the case, you could just erase everything that's in your question now, and replace it with the question I just wrote. Or, you could state the question as I have written it, and then put in a note saying you are including the original wording for historical reasons, and then put in the current wording.
- Moti about 8 years
  
  @Gery modified the question, though it is not reflecting exactly on my question, but is good enough.
Matt Samuel about 8 years

Finding the cube root of a complex number is certainly not nearly as hard as solving a cubic equation.
obataku about 8 years

finding cube roots is trivial by de Moivre's theorem: $z^3=re^{it}$ gives $z\in\{\sqrt[3]{r}e^{it/3},\sqrt[3]{r}e^{i(t+2\pi)/3},\sqrt[3]{r}e^{i(t+4\pi)/3}\}$
user21820 about 8 years

@MattSamuel: It is. Tell me the exact value of $\sqrt[3]{1+2i}$ using only arithmetic.
Matt Samuel about 8 years

No. But I have no problem using trigonometry. The point is there's nothing to solve, it's just plugging something in.
user21820 about 8 years

@oldrinb: Tell me the exact value of $t$ for $(1+2i)$. It is no point saying you know the cube roots when I too can write $\exp(\frac{1}{3}\ln(1+2i))$ (multi-valued) to express them.
user21820 about 8 years

@MattSamuel: That's why you have to read my answer carefully. I said that if you only want an approximation...
user21820 about 8 years

@MattSamuel: And I have no problem using a root-finding algorithm to find roots of arbitrary polynomials (not just cubics) instead of asking for a formula that is artificially restricted to only use radicals.
Moti about 8 years

I am asking what I think a clear question, where the focus is on factoring of polynomials. This relates to solving the equation but it is not the same. What I wonder if there are ways to evaluate the coefficients (real or complex for this matter - but limited to this field of numbers) and for this to determine that factoring is possible.