Why the fifth order polynomial roots can,t be obtained exactly?

[eljose]

this is a question that only mathematicians know about why the roots of the Polynomial:

\sum_{n=0}^{k}a_{n}x^{n} with k>4 and integer can,t be obtained by elementary algebra (addition,substraction,and so on)..

 

[matt grime]

It's called Galois theory. You're glossing over some of the technicalities: some people might not consider extracting radicals as elementary algebra for instance. And it is *the general poly of degree greater than 5* and for the very easy to state reason that A_n is simple if n>4. (Harder to understand, but there are very many good texts on this by people such as Artin, or Stewart. Try getting hold of the latter one if you are interested in finding out why.)

 

[Hurkyl]

Galois theory, incidentally, is about symmetries.

The basic idea (I think) is this: (keep in mind I am suppressing a whole lot of details -- this is only the coarsest of sketches)

Basic arithemtic operations have only the trivial symmetry. (Given a and b, there's only one choice for a+b)

Taking n-th roots has a cyclic symmetry. (e.g. there are three choices for the cube root of a number, and it doesn't matter which one we choose)

So any expression you can build out of basic arithmetic operations and n-th roots can only have a symmetry that is "built" with cyclic symmetries. (In some sense)


When you take the root of a polynomial, you also have some symmetries in your choice of root.


So, the question is can you "build" the symmetry group corresponding to your polynomial out of cyclic symmetries?


The question is answered by the fact that A_5 is irreducible -- it cannot be "built" out of other symmetry groups, and we can find polynomials whose corresponding symmetry groups contain A_5.

 

[eljose]

Um..but if you have the differential operator:

L[y]=\sum_{n=0}^{k}a_{n}D^{n}y=0 you could propose the "ansatz" as a solution:

\sum_{n=0}^{\infty}b_{n}x^{n} this series is analytic for every x, if we substitute it into de differential equation we would get the coefficients (n!)b_{n}=(r)^{n} with r a root of the system..

By the way if you use group teory...wouldn,t be a form of representing the Group R(n) (the group of the equation of grade n) as the sum product or something in the form: R(n)=R(a)*R(b) with a+b=n

 

[Hurkyl]

There are lots of ways to solve higher degree equations -- you just can't (generally) do it with the basic arithmetic operations, and n-th roots.

Incidentally, differential operators and infinite sums are not basic arithmetic operations.



The symmetry group associated to the general n-th degree polynomial is (isomorphic to) the symmetric group on n-symbols: S_n. This group consists of all permutations on the symbols 1, 2, 3, ..., n.

S_n contains A_n, the group of all even permutations.

matt grime pointed out that for n>4, A_n is a simple group -- that means that it cannot be "built" from smaller groups. (At least in the way relevant to Galois theory)


I feel the need to point out that the "building" operation is not simply the direct product of groups -- it is a trickier notion. (But I think that the direct product would just be a special case)


Incidentally, if you want to know the notation...

If E is the (smallest) extension of Q that contains the roots of your polynomial f, then we call E the splitting field of f. The symmetry group I mentioned is:

Gal(E / Q)

which is called the "Galois group of E over Q".

There exist polynomials of degree 5 for which this Galois group contains A_5, and is thus not solvable by radicals. (And the general polynomial with symbolic coefficients has Galois group isomorphic to S_n, as I've said)

Other polynomials of degree 5 have smaller Galois groups, and we can solve those polynomials with radicals.

For example, I think the equation x^5 - 2 = 0 has a Galois group isomorphic to Z_5: the cyclic group with 5 elements.

 

[eljose]

-Algebra was always too hard for me to understand... sorry..i hoped that somehow S_{m}=S_{a}xS_{b} with a+b=m which according to you is not true..but i think that for "solvable" roots you have the same problem...of course you can get that the roots of a Polynomial are for example \sqrt(3) \sqrt(1/4)+3 the problem is that you can,t calculate them exactly, you can calculate with any degree of accuracy the square root of 2 4 or 3/8 or even the number \pi or e but you can,t know exactly how they are,aother question i have is if would be considered "exact" if somehow we managed to obtain the root r in the form of r=g(a) where g is a known function for example exp(x), sin(x),tan(x),...or a combination of them or their derivatives, integrals...

 

[matt grime]

Galois theory has nothing to do with 'exactness' of an trunctation of a decimal representation of anything.

Surely it's rather clear that S_n is not S_a x S_b, they don't even have the same order (n! is not a!b! in general ).

 

[robert Ihnot]

eljose: problem is that you can,t calculate them exactly, you can calculate with any degree of accuracy the square root of 2 4 or 3/8 or even the number or e but you can,t know exactly how they are,

Things like pi and square root 2 are know exactly. Pi is the ratio of the circumference to the diameter, and X^2-2 = 0, has two exact answers.
!/3 is exactly known, even though the decimal representation stretches on forever.

There are, however, differences in the numbers examined above, particularly pi, which is not the root of any finite polynomial with integral coefficients.

Galois theory, Abel's Theorem, is concern with answers in terms of a "finite number of additions, subtractions, multiplications, divisions, and extraction of roots." You are well aware of this with regards to the quadratic equation in particular, but third and fourth power equations as well.

 

[matt grime]

Incidentally, when we say things like 'can't be built using algebraic operations' we should say that we are using the coefficients of the polynomial as the building blocks.Example: C_2 is solvable, hence the general quadratic has roots that exrpessible algebraically in terms of the coefficients.

more explicitly: if x^2+Bx+C is the quadratic with roots r and s, then s+s=-B, and rs=C, note that (r+s)^2- 4rs = (r-s)^2, hence we can recover r and s from B and C using only field operations and radicals.

In generaly a degree n poly is

p(x)=\sum_{r} (-1)^r \sigma_r x^{n-r}

where \sigma_r is the r'th elementary symmetric polynomial in the roots. If the roots have solvable galois group then this states that we can extract any of them (or a power) by manipulating the symmetric polys. In general this cannot be done because the general galois group will not be solvable.

 

[eljose]

and for a n-th degree Polynomial P(x)=y what would happen with its inverse function?..i mean x=f(y) so P[f(x)]=x by the way..could you write down the general expression for \sigma_{r} ? and could exist a transform that map our unsolvable P(x) polynomial into another Polynomial of the same or higher degree that is solvable?..thanks.

 

[matt grime]

The inverse of a polynomial is not (normally) a polynomial.

I am not going to feed you any more than that. It's called google.

And lastly forget the transforms. Galois theory isn't about properties preserved under transforms. It's about algebra, nothing to do with analysis. Anything that introduces the real numbers is not, in general, going to play well with algebra. Remeber, algebraists don't even like infinite sums, never mind integrals. (This last bit is not serious, though many of us hate integrals.)