Given a field $F$ of characteristic $0$, we know that a general polynomial $p(x) \in F \left[ x \right]$ is not solvable by radicals if $n \geq 5$. However, I was wondering what restrictions can one place on the coefficients of $p(x)$ to get it to be solvable.

For example, suppose $$ p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 $$ where $a_i = a_0^i$.

With the above restriction on the coefficients (for example), can one get relatively simple formulas for the roots or at least prove the associated Galois group is not solvable?

Edit/Clarification. The way I am thinking about this problem is by analogy with how one proves the general polynomial of degree $n$ over a field $F$ is not solvable by radicals for $n \geq 5$. This is done using the following theorem due to Galois

Theorem: Let $F$ be a field. Then $p(x) \in F \left[ x \right]$ is solvable if and only if the Galois group of $p(x)$ is a solvable group.

Once one has this correspondence, we show that the Galois group of a polynomial in $F[x]$ with degree $n \geq 5$ is $S_n$. This is achieved by recognizing that the Galois group of $t^n + a_1 t^{n-1} + \cdots + a_n$ over $F(a_1, \cdots , a_n)$, the field of rational functions in $n$ variables, is $S_n$. (We of course can always make a polynomial monic with out changing the solvability.) However $S_n$ is not a solvable group for $n \geq 5$. Hence, such polynomials cannot be solvable by radicals.

However, the proofs of the various lemmas and theorems that eventually produce the above theorem assume that the coefficients are arbitrary. So my question is

Are there special restrictions one can put on the coefficients that make the polynomial solvable? More explicitly, when we restrict ourselves from an arbitrary polynomial in $F[x]$ to a polynomial that has coefficients which are related to one another in some simple way (e.g. $a_i = a_0^i$), how does the Galois group change? In particular, is the Galois group of this new polynomial some solvable subgroup of $S_n$?. Moreover, can we find the associated formula for the roots with ease? (For the case of $a_i = a_0^i$, I imagine the formula for the roots is some relatively 'simple' function of $a_0$)

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    $\begingroup$ Note that replacing $x$ by $x/a_0$ reduces this to $x^n+x^{n-1}+\dots + x + a_0$. So the problem is reduced to whether or not this is soluble by radicals. $\endgroup$ – user98602 May 13 '14 at 19:33
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    $\begingroup$ An irreducible polynomial equation is solvable iff its Galois group is solvable — if this doesn't answer you question, what is you question exactly? $\endgroup$ – Grigory M May 13 '14 at 20:47
  • $\begingroup$ @MikeMiller I had realized this shortly after I post it, but that only applies to this example. I am also looking for other possible structure of the coefficients that allowed solvability. $\endgroup$ – o0BlueBeast0o May 13 '14 at 21:34
  • $\begingroup$ You should be explicit about what other possible structure there can be on the coefficients. $\endgroup$ – user98602 May 13 '14 at 21:36
  • $\begingroup$ @GrigoryM Yes I am aware of this. The fact that you stated about the Galois Group is a way one can prove the insolvability of the general polynomial of degree $n$ (where the coefficients are assumed to be entirely arbitrary). Once one has this criteria, it is easy to show that the Galois group of the polynomial for $n \geq 5$ contains $S_5$ which is not a solvable group. This is achieved by recognizing that the galois group of $t^n + a_1 t^{n-1} + \cdots + a_n$ over $F(a_1, \cdots , a_n)$ (the field of rational functions in $n$ variables) is $S_n$. I will edit the post and include more detail. $\endgroup$ – o0BlueBeast0o May 13 '14 at 21:42

Consider the case in which $f(x)$ is an irreducible quintic with rational coefficients.

A well-known paper by D.S. Dummit, Solving Solvable Quintics, approaches the problem by constructing a sixth degree rational polynomial $g(x)$ from the coefficients of $f(x)$. The result is that $f(x)$ is solvable if and only if $g(x)$ has a rational root. The Rational Root Thm. then gives us a way to test for rational roots of $g(x)$.

In particular any quintic is easily transformed into an equivalent depressed monic quintic, with a leading coefficient 1 and zero coefficient of $x^4$. Dummit then gives the explicit sixth degree polynomial $g(x)$ that corresponds to this depressed quintic $f(x)$.

Once a rational root of $g(x)$ is found, Dummit gives formulas that allow us to express the roots of $f(x)$ "analogous to Cardano’s formulas for the general cubic and quartic polynomials".


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