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So I've just been over a bit of Galois theory, and I'm trying to understand what the implications are for a polynomial's Galois group to not be solvable. My book says this means that there is no formula for finding the roots of the polynomial using just the field operations and the extraction of roots.

So does this mean that there is no one single formula, like the quadratic formula, which one can apply mindlessly to unsolvable polynomials in order to find their solutions? Or does it mean the stronger statement that no matter how long you spend messing around with the poly using the field operations and the extraction of roots you will NEVER be able to figure out its solutions?

I've been under the impression that the roots of a polynomial with integer coefficients, that is to say algebraic numbers, could always be written in the form of some finite combination (sum/product/difference/quotient) of integers under roots of varying degree. This is correct right? Or are there algebraic numbers which must be written in terms of say an infinite series?

And assuming there aren't such algebraic integers, how does one generally go about finding the roots of unsolvable polynomials with integer coefficients? Just brute force and cleverness?


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As for solving ‘unsolvable’ polynomial equations, there is this vast subject called numerical analysis. – Zhen Lin Jan 6 '12 at 1:39
There are two concepts here: (1) unsolvable which means there is no formula to express roots in terms of radicals, and (2) irreducible which means that the roots live in an algebraic extension of the field of coefficients. – user2468 Jan 6 '12 at 1:46
"Or are there algebraic numbers which must be written in terms of say an infinite series?" - yes, something like that. The roots of a quintic can be expressed using either theta functions or hypergeometric functions, for instance. Look up the Bring radical. – J. M. Jan 6 '12 at 2:08
See my answer to this question:… – deoxygerbe Jan 6 '12 at 3:01

If the polynomial $p(x)$ is irreducible over $K$ and has unsolvable Galois group over $K$, then no expression involving the coefficients, the elements of $K$, addition, multiplication, division, subtraction, and extraction of roots, will equal the roots of $p(x)$.

In particular, there can be no general formula à la quadratic formula for the roots.

Since there are monic polynomials with integer coefficients that have unsolvable Galois group, there are certainly algebraic numbers/integers that cannot be given as an expression involving rationals, addition, multiplication, division, subtraction, and extraction of roots.

As to "finding roots of irreducible polynomials with integer coefficients", it really depends on what you mean by "finding". There are a lot of good methods to approximate roots (both real and complex) to any desired degree of accuracy (e.g., Newton's method). But generally, we don't write down 'expressions' that give them "exactly" in some sense.

(In a sense, what does it mean that we can "find" the roots of $x^2-2$? We write them out as $\sqrt{2}$ and $-\sqrt{2}$, but all we've done is give a name to those roots, without actually "exhibiting" them; we can compute $\sqrt{2}$ to any degree of accuracy we wish, but we can never write it down exactly, except by cheating and writing things like "$\sqrt{2}$").

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I like the last parenthesized paragraph; there's really not much difference in using $\sqrt{}$ to represent the roots of a quadratic, trigonometric functions to represent the roots of a cubic, or theta functions to represent the roots of an arbitrary-order polynomial. They're all just notation. – J. M. Jan 6 '12 at 2:57

One implication is that finding eigenvalues is a non-linear problem for which there is no finite algorithm, only numerical approximations. (Actually, finding eigenvalues and finding roots of polynomials are equivalent problems, because of the companion matrices.)

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"because of..." shouldn't this be "because of Cayley-Hamilton theorem"? – user2468 Jan 6 '12 at 1:48
@J.D.: No, it's companion matrices, not Cayley-Hamilton. Cayley-Hamilton tells you the matrix satisfies its characteristic polynomial; it doesn't tell you anything about the real/complex roots of the polynomial. The point of the parenthetical comment is that every polynomial is the characteristic/minimal polynomial of some matrix, so if you know how to find the eigenvalues of any matrix, then you know how to find the roots of any polynomial (take the companion matrix of that polynomial); the converse is the fact that the eigenvalues are the zeros of the characteristic/minimal polynomial. – Arturo Magidin Jan 6 '12 at 1:51
@J.D., not really. I meant, that if you knew how to find eigenvalues then you'd know how to find roots of polynomials by finding the eigenvalues of companion matrices. Conversely, if you knew how to find roots, then you could find eigenvalues by finding the roots of $\det(A-\lambda I)$. – lhf Jan 6 '12 at 1:52
It has to be noted, however, that in practice, one finds the roots of a polynomial by constructing an appropriate companion matrix and applying eigenvalue algorithms to it, but one almost never finds the eigenvalues of a matrix by constructing its characteristic polynomial. – J. M. Jan 6 '12 at 2:07

The stronger statement is true.

You are right to wonder about the possibility that every individual polynomial might have a root despite there not existing a general formula, but it turns out that there are, in fact, polynomials whose roots cannot be written in terms of field operations and taking roots.

There are lots of other ways of representing numbers than infinite series. One useful method for an representing a real algebraic number $\alpha$ is in the following way:

$\alpha$ is the unique root of the polynomial $f(x)$ lying in the interval $(a,b)$

Mathematica, for example, can do computations with a similar form.

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To be precise: Mathematica Root[] objects store the minimal polynomial and an index (and I have to admit I remain confused by how it indexes its roots after all these years). Maple's RootOf() objects behave similarly. – J. M. Jan 6 '12 at 3:05

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