Using higher-order Bring radicals to solve arbitrary polynomials - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T10:57:15Z http://mathoverflow.net/feeds/question/61409 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/61409/using-higher-order-bring-radicals-to-solve-arbitrary-polynomials Using higher-order Bring radicals to solve arbitrary polynomials Daniel Miller 2011-04-12T15:07:14Z 2011-10-06T13:30:34Z <p>It is well known that there is no general formula for the solution of the quintic. Of course, what this really means is that there is no general formula that only involves addition, subtraction, multiplication, division, and the extraction of $n$-th roots. Indeed, if one is allowed to use the Bring radical, that is, solutions of the equation $x^5+x+a=0$, then it is indeed possible to solve any quintic. It would seem that if one introduced higher order Bring radicals, it would be possible to solve polynomials of higher degree. More precisely, define a Bring radical of order $n$ to be a continuous function $B_n(t)$ such that $B_n(t)$ is a solution to an $n$-th degree polynomial, one of whose coefficients is $t$. (Of course, I am being rather vague, most of these Bring radicals are only continuously definable on some proper subset of $\mathbb{C}$) It is trivial that any $n$-th degree polynomial may be solved by means of some $n$-th order Bring radical. However, it is not at all apparent that for some fixed $n$, there exist a finite collection $B_n^1,B_n^2,\cdots,B_n^k$ such that <em>any</em> $n$-th degree polynomial may be solved using the $B_n^i$. So my question is: </p> <blockquote> <p>Is it the case that for any $n$, there is a finite collection of Bring radicals that may be used to solve any $n$-th degree polynomial? </p> </blockquote> <p>Another question, to which the answer is most likely negative, is whether there exists a finite collection $B_{r_1}^1,B_{r_2}^2,\cdots$ of Bring radicals such that any polynomial of any degree is solvable using the $B_{r_i}^j$.</p> <p><strong>Edit</strong>: My definition of a higher-order Bring radical was rather narrow. I'd also be interested in any answer that involved Bring radicals $B_n(t)$ that were solutions to a polynomial of the form $x^n+p_{n-1}(t)x^{n-1}+\cdots+p_{1}(t)x+p_{0}(t)$. The general idea concerns whether or not one can solve <em>all</em> $n$-th degree polynomials just be adjoining the roots of some finite family of them polynomials, whose coefficients depend smoothly on $t$.</p> http://mathoverflow.net/questions/61409/using-higher-order-bring-radicals-to-solve-arbitrary-polynomials/61547#61547 Answer by Dylan Thurston for Using higher-order Bring radicals to solve arbitrary polynomials Dylan Thurston 2011-04-13T14:12:55Z 2011-04-13T14:12:55Z <p>The answer to the second question is "no". For a family of polynomials $p_t$ depending polynomially on a complex parameter, as in the polynomials satisfied by your $B_r(t)$, define its <em>Galois group</em> to be the group of permutations of the roots you see by moving around the branch points. (Assume that the roots of $p_t$ are distinct for generic values of $t$ to make this work well. This is the same as the Galois group of `$\mathcal{C}(t)(\text{roots of }p_t)$ over $\mathcal{C}(t)$.) Then there are families of polynomials exhibiting an arbitrary finite group as its Galois group. Any finite family of $B^j(t)$ will only exhibit finitely many groups $G_j$, and any tower of roots of the $B^j$ will give only groups whose composition factors are among the $G_j$. Since there are infinitely many finite simple groups, you cannot acheive all possible finite groups this way.</p> <p>However, I suspect the question was stronger than you meant to ask, since even the original conjecture that all polynomials are solvable by radicals wouldn't fit in to the framework of your second question.</p> http://mathoverflow.net/questions/61409/using-higher-order-bring-radicals-to-solve-arbitrary-polynomials/61558#61558 Answer by David Speyer for Using higher-order Bring radicals to solve arbitrary polynomials David Speyer 2011-04-13T15:51:49Z 2011-04-13T16:03:00Z <p>If I understand it correctly, <a href="http://www.nsc.ru/EMIS/journals/SC/1997/2/pdf/smf_sem-cong_2_1-11.pdf" rel="nofollow">this paper of Abhyankhar</a> says that what you want is not achievable.</p> <p>Here is what Abhyankar shows. I warn you that I haven't read the paper, and am just skimming the introduction. Let $K$ be the algebraic closure of $k(X,Y)$. So an element $z$ of $K$ should be thought of as a root of a polynomial $z^n + a_{n-1}(X,Y) z^n + \cdots + a_{0}(X,Y)$ whose coefficients depend on two parameters $x$ and $y$. </p> <p>Let $L^1$ be the subfield of $K$ generated by all solutions to polynomials of the form $z^n + a_{n-1}(T) z^{n-1} + \cdots + a_{0}(T)$ where $T \in k(X,Y)$ and the $a_i$ are polynomials. This is pretty close to the class of $z$'s you refer to in your update, except that you ask for $T$ and the $a_i$ to be smooth and Abhyankar takes them to be rational functions. Let $L^2$ be the subfield of $K$ generated by solutions to polynomials of the form $z^n + a_{n-1}(T) z^{n-1} + \cdots + a_{0}(T)$ where $T \in L^1$ and the $a_i$ are polynomials. Inductively, make $L^3$, $L^4$, etcetera. Abhyankar shows that $\bigcup L^i$ is strictly smaller than $K$.</p> <hr> <p>In fact, the meat of Abyhankar's argument is about formal power series, not polynomials. Here I had a little trouble following Abyhankar, so I hope I am summarizing him correctly. Let $\hat{A}$ be the ring of formal power series $k[[X,Y]]$ and let <code>$\hat{K} =\mathrm{Frac} \ \hat{A}$</code>. Let $\hat{L}^1$ be the field generated by all roots of polynomials $\sum a_i(T) z^i$ where $a_i$ are formal power series and $T$ is in the maximal ideal of $\hat{A}$. Let $\hat{B}^1$ be the integral closure of $\hat{A}$ in $\hat{L}^1$. Let $\hat{L}^2$ be the field generated by all roots of polynomials $\sum a_i(T) z^i$ where $a_i$ are formal power series and $T$ is in the maximal ideal of $\hat{B}^1$. Let $\hat{B}^2$ be the integral closure of $\hat{A}^1$ in $\hat{L}^2$. In this manner, define $\hat{L}^j$ for all $j$. Then $\bigcup \hat{L}^j$ is smaller than the algebraic closure of $\hat{K}$. </p> <p>The essence of the proof is that (1) $\hat{L}^{j+1}$ is solvable<sup>1</sup> over $\hat{L}^j$ and (2) the splitting field of $z^6+Xz+Y$ is not solvable over <code>$\mathrm{Frac} \ k[[X,Y]]$.</code> If you've never thought about these issues before, part (1) may surprise you. To get some intuition for this, note that the root of $x^2 - (1+t)$ is degree $2$ over <code>$\mathrm{Frac} \ k[t]$</code> but this polynomial factors as $\left( x - \sum \binom{1/2}{j} t^j \right)\left( x + \sum \binom{1/2}{j} t^j \right)$ over <code>$\mathrm{Frac} \ k[[t]]$</code>. So passing to power series can make Galois groups smaller.</p> <p>You might enjoy thinking about the topology in part (2). Here is as far as I have gotten. The polynomial $z^6+Xz+Y$ has a multiple root if and only if $5^5 X^6 + 6^6 Y^5=0$. This is a real $2$-fold in $\mathbb{C}^2$, with a singularity at $(0,0)$. Its intersection with a sphere around the origin is a torus knot of type $(6,5)$. The roots of $z^6+Yz+X$ form a $6$-sheeted cover of this torus-knot complement. Abyhankar's claim is that the monodromy of this cover is not solvable.</p> <hr> <p>So, Abhyankar is disproving your hope not just for polynomials but for formal power series. The map "take the Taylor series" is a map from the ring of smooth functions to the ring of formal power series. (This map has a kernel, containing things like $e^{-1/x^2}$.) I think (but have not checked in detail) that this map should mean that Abhyankar's result also works for smooth functions.</p> <p><sup>1</sup> I think that, at $j=0$, this statement is only right if we take $k$ algebraically closed, although I don't see where Abhyankar says this. But I'm perfectly willing to do that. Note that $\hat{L}^1$ contains the algebraic closure of $k$, so the issue of whether or not $k$ is algebraically closed goes away once we get one step up the tower. Also, I think that when $k=\mathbb{C}$, we should be able to strengthen solvable to abelian.</p>