The collection of all degree-$n$ polynomials in the variable $w$ (call this set $\mathbb{C}[w]_n$) can be identifies with $\mathbb{C}^{n+1}$ by the bijection $F:\mathbb{C}^{n+1}\to\mathbb{C}[w]_n$ defined by $$F:(a_0,a_1,\ldots,a_n)\mapsto w=p(z)=a_0+a_1z+\cdots+a_nz^n.$$ Let $\mathcal{A}_n\subset\mathbb{C}^{n+1}$ denote the set of polynomials (using the above identification) defined by saying that $p(z)\in\mathcal{A}_n$ if and only if each branch of the inverse $z=p^{-1}(w)$ is expressible as an explicit formula using finitely many algebraic operations (ie addition/subtraction, multiplication/division, and roots). That is, something like $$z=p^{-1}(w)=\sqrt{w+\sqrt{2/w}}.$$ Note that for any polynomial $p(z)\in\mathbb{C}^{n+1}$, either each branch of $p^{-1}$ is so expressible, or none is (since every branch can be reached as an analytic continuation of any other).

My Question: What is the structure of $\mathcal{A}_n$ in $\mathbb{C}^{n+1}$? Is it an algebraic variety? What is its dimension? What can we say about its topology?

  • $\begingroup$ are you basically asking about the polynomials that are solvable in radicals ? $\endgroup$
    – mercio
    Commented Aug 4, 2018 at 20:08
  • $\begingroup$ @mercio, yes, that is right. $\endgroup$ Commented Aug 5, 2018 at 22:59

1 Answer 1


So you are asking if we can tell when an extension $\Bbb C(w = p(z)) \subset \Bbb C(z)$ is solvable by radicals, so you want to study the Galois group of that extension and see if it is a solvable group. The answer to your question is that yes, this can be given by algebraic equations.

This is a pretty geometric question, because there is a correspondance between extensions of $\Bbb C(w)$ of dimension $n$ and $n$-sheeted branched coverings of the Riemann sphere, and the Galois group in the algebraic side corresponds to the group of deck transformations in the geometric side.

If you choose a finite subset $\{w_1 ; \ldots ; w_k\}$ of the Riemann sphere and fix some non-intersecting branch cuts between $w_0$ and each $w_i$, then you can make a new Riemann surface by building an $n$-sheeted covering branched at those points, by appropriately choosing permutations of the branches $\alpha_i$ when looping around each of the $w_i$. Given a $k$-uple of elements of $S_n$ $\alpha = (\alpha_1, \ldots, \alpha_k)$, the corresponding covering is connected if and only if the subgroup $G_\alpha$ of $S_n$ generated by the $\alpha_i$ is transitive. If so the covering defines a Riemann surface, and the meromorphic functions on it are a field extension of $\Bbb C(w)$ with Galois group $G_\alpha$.
Finally, two $k$-uples $\alpha^1,\alpha^2$ give isomorphic coverings if they are the same after reordering the sheets, that is, if there is a $\beta \in S_n$ such that $\alpha_i^1 = \beta \alpha_i^2 \beta^{-1}$ for $i \in \{1 \cdots k\}$, so we will only be interested in equivalence classes with regard to this. Let us denote this relation with $\sim$.

So to understand the Galois group of polynomial maps we simply have to look at their possible branching data.

Because polynomials have an $n$-uple pole at infinity (around which a loop makes an $n$-cycle permutation of the sheets), there is an $n-1$-uple branch point above $w_n = \infty$, and there are another $n-1$ branch points (counted with multiplicity) above the critical values of $p$, that is the $w_i = p(z_i)$ where the $z_i$ are the critical points of $p$, the points where $p'(z)$ vanishes.
Generically, those $n-1$ critical values are all distinct and the loops induce simple transpositions there, so the choice of the branching data amounts to the choice of $n-1$ transpositions the composition of which give an $n$-cycle.
Let $T_n \subset S_n$ be the subset of transpositions, and let $G_n = \{ (\alpha_1,\ldots,\alpha_{n-1}) \in T_n^{n-1} \mid \alpha_1\cdots \alpha_{n-1} = (123\cdots n)\} / \sim$, the set of possible polynomial branching data on $n-1$ distinct branch points.

In the non-generic case, some critical values have higher multiplicity, and you can obtain their data by starting with a generic one and composing some of the transpositions together. This is also the only way that the group can possibly be smaller than $S_n$.

For every such branching data, the covering will be a Riemann surface of genus $0$, so another Riemann sphere, and by deciding that the preimage of $w_n = \infty$ should be $z_n = \infty$ the covering becomes a polynomial map. Two polynomial maps give the same branching data if and only if they differ by pre-composition with an automorphism of the Riemann sphere, and since polynomials fix $\infty$, they have to differ by a map of the form $z_1 = az_2+b$ from some nonzero $a \in \Bbb C$. Then, if we restrict ourselves to polynomials with $a_n=1$ and $a_{n-1}=0$, then for $n \ge 3$ there are exactly $n$ possible polynomials $p(z), p(\zeta_n z), p(\zeta^2 z)$ etc (for $n=2$ there is only one because if $p(x) = x^2+a_0$, $p(-x) = p(x)$).

Let $g_n$ be the cardinal of $G_n$. Geometrically, the space of polynomials (up to precomposition) is a $g_n$-sheeted branched covering of $Sym^{n-1}(\Bbb C)$, and we again have an isomorphism of groups between the (very large !!) Galois group and the deck transformations of that covering. This group, as a subgroup of permutations of $G_n$, is generated by the $n-2$ operations you get by replacing $(\alpha_i, \alpha_{i+1})$ with $(\alpha_i\alpha_{i+1}\alpha_i^{-1},\alpha_i)$ (think of this as moving two critical values in order to swap them). For $G_n$, it is transitive, but that won't be the case when you start merging elements of $\alpha$, then the number of orbits tells you the number of components you should expect.

Now let us go to the algebraic side.

Given a degree $n$ polynomial $p(z) = \sum_{i=0}^n a_i z^i$, if you denote $z_1 \ldots z_{n-1}$ the critical points, and $w_1,\ldots,w_{n-1}$ the critical values, then the polynomial expression $q(w) = \prod_{i=1}^{n-1} (w - w_i) = \prod_{i=1}^{n-1} (w - p(z_i))$ is symmetric in the $z_i$.
Call $b_i$ the coefficients of $q$ ($q(w) = w^{n-1} + \sum_{i=0}^{n-2} b_i w^i$).
Given that $p'(z) = na_n \prod (z - z_i)$, it is possible to express $b_i$ as polynomial expressions in terms of $p'(z)/na_n$, which are themselves algebraic in the $a_i$.

When you assume $a_{n-1}=0$ and $a_n=1$, they become polynomial in the $a_i$, and the field extension $K_n = \Bbb C(q_0,\ldots,q_{n-2}) \subset L_n = \Bbb C(a_0,a_1, \ldots, a_{n-2})$ has finite degree $d_n$. From what we know of the geometric side, we should have $d_n = n g_n$ for $n \ge 3$ and for $n = 2$, $d_n = g_n$.
(it is possible to compute $d_n = n^{n-2}$, so $g_n = n^{n-3}$ for $n \ge 3$)

The places where this cover can branch is when two critical values coincide, which means when the discriminant $\Delta$ of $q$ vanishes. Meanwhile, the only chance for your polynomial to not have Galois group $S_n$ is when you merge some critical values together in order to push all of the generators of the deck transformation into a subgroup of your choice. And this gives a polynomial equation $\Delta_n(a_0, \ldots, a_{n-2}) = 0$ that has to vanish for your polynomial to be non-generic.

There is a lot of information encoded in the discriminant $\Delta_n(a_0, \ldots,a_{n-2})$. It doesn't actually depend on $a_0$. The number of components and how they intersect themselves or each other into a number of singularity loci and so on all can all be described combinatorially on the geometric side.

The discriminant of $p'$ has to divide it three times (because if two critical points coincide, then two critical values will also coincide, and do so by making a $3$-cycle holonomy).
Its other irreducible factors have exponent $2$ (there is one for each geometrically different way to merge two commuting critical values).

At this point I should illustrate with what happens for small $n$

I don't know how to easily get nice systems of equations for the lower dimensional pieces, nor do I know how to interpret the various multiplicites you see in the geometric/combinatoric side into algebraic statements (at least in the case $n=5$, I did see lots of factorisations happening)

degree $2$ : $p(z) = z^2 + a_0$

$G_2$ only has one element, $[(12)]$, so $g_2=1,d_2=1$ and so $K_2 = L_2$.
There is only one possible geometry here. Given a critical value $w_1$, there is only one $2$-sheeted covering branched at $w_1$ and $\infty$, and there is only one possible polynomial, $p(z) = z^2 + w_1$.

degree $3$ : $p(z) = z^3 + a_1z + a_0$

Again $G_3$ is the singleton $[(12),(23)]$, so we have $g_3=1$ and $d_3=3$.

Some quick computation shows that
$b_1 = -2a_0, b_0 = a_0^2 + 4a_1^3/27 : K_3 = \Bbb C(a_0, a_1^3)$.

$\Delta_3(a_0,a_1) = b_1^2 - 4b_0 = 4a_1^3/27$, so it only contains the cube of the discriminant of $p'$, as expected.

The Galois group of $p$ is $S_3$ unless $a_1=0$ and then it is a cyclic group of order $3$.

degree $4$ : $p(z) = z^4 + a_2z^2 + a_1z + a_0$

Now things start getting a bit more interesting : $G_4 = \{[(12),(23),(34)],[(12),(23),(24)],[(12),(23),(14)],[(12),(34),(13)]\}$.

Merging two critical values gives two orbits, that of $[(123),(34)]$ and that of $[(12)(34),(13)]$, which reflects the factorisation $\Delta_4 = (-27a_1^2-8a_2^3)^3 a_1^2/2^{16}$.

If the first factor vanishes the Galois group is still $S_4$, but if $a_1$ vanishes then the Galois group of $p$ is the diedral group of order $8$. If both vanish ($a_1=a_2=0$) then you get a cyclic group of order $4$

Also note that the curve $(-27a_1^2-8a_2^3)=0$ is pinched at $a_1=a_2=0$.

degree $5$ : $p(z) = z^5 + a_3z^3 + a_2z^2 + a_1z + a_0$.

The number of branching data is increasing quickly, since now $G_5$ has $25$ elements (in a single orbit) So $L_5$ should be an extension of degree $125$ of $K_5$ (I didn't check that).

Moving down a dimension, all the possible ways to merge two critical values fall (again, and forever) in two orbits, which reflects the factorisation (up to a multiplicative constant)

$\Delta_5 = (-16a_3^6+224a_3^4a_1-88a_3^3a_2^2-1040a_3^2a_1^2+360a_3a_2^2a_1+135a_2^4+1600a_1^3)^2 \times \\ (-81a_3^4a_1+27a_3^3a_2^2+360a_3^2a_1^2-540a_3a_2^2a_1+135a_2^4-400a_1^3)^3$

The surface $S_1$ given by the first factor corresponds to the orbit of $[(12),(24),(23)(45)]$, while the surface $S_2$ corresponds to the orbit of $[(12),(23),(345)]$.

Polynomials falling on any smooth point of the $\Delta_5=0$ surfaces still have Galois Group $S_5$

Going down a dimension, we now get five orbits, so five curves. (by the way, the curves will always have genus $0$ because they are branched coverings of the Riemann sphere with only two branch points).
Hopefully, I paired the curves with the orbits properly :

$[(12), (2345)]$ : where $S_1$ and $S_2$ intersect and $S_2$ is singular.
It is parametrized with $p(z) = z^5-10t^2z^3+20t^3z^2-15t^4z+a_0$.
The Galois group is $S_5$.

$[(13), (345)(12)]$ : where $S_1$ and $S_2$ intersect and $S_1$ is singular.
It is parametrized with $p(z) = z^5-15t^2z^3+10t^3z^2+60t^4z+a_0$.
The Galois group is $S_5$.

$[(123), (345)]$ : where $S_2$ intersects itself.
It is parametrized with $p(z) = z^5+10tz^3+45t^2z+a_0$.
The Galois group is $A_5$.

$[(124), (45)(23)]$ : where $S_1$ and $S_2$ intersect transversally.
It is parametrized with $p(z) = z^5+30t^2z^3+100t^3z^2+105t^4z+a_0$.
The Galois group is $A_5$.

$[(23)(14), (45)(13)]$, where $S_1$ intersects itself.
It is parametrized with $p(z) = z^5+5tz^3+5t^2z+a_0$.
The Galois group is the diedral group of order $10$, so there you have your solvable degree $5$ polynomials.

Down another dimension, you end up at the point $(a_1=a_2=a_3=0)$, at the intersection of every curve, the Galois group there is cyclic of order $5$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .