Why There Is No Quintic Formula
One of the most notable accomplishments of nineteenth-century algebra is the proof that there is no general solution by radicals for polynomial equations of degree five or higher. This result is known as the Abel–Ruffini theorem. Ruffini gave an important early argument, Abel published the first generally accepted proof in 1824, and Galois later explained exactly which polynomial equations are solvable by radicals.
Galois theory is extremely powerful, but it requires a substantial amount of abstract algebra and can obscure the topological idea behind a more elementary proof.
The aim of this post is to present that idea. The key notions parallel those of Galois theory, but we will work mainly with continuous paths, permutations, and nested radicals rather than field extensions. Strictly speaking, this is a proof sketch: a fully rigorous version must also keep track of the points at which expressions vanish or become singular. The presentation is heavily inspired by this video explanation and by Leo Goldmakher’s exposition of a proof due to Vladimir Arnold.
1. Polynomials and their roots
Let P \in \mathbb{C}[X] be a polynomial of degree n. Thus there exist a_0,a_1,\ldots,a_n\in\mathbb C, with a_n\ne0, such that P(X)=a_nX^n+a_{n-1}X^{n-1}+\cdots+a_1X+a_0.
We say that \alpha\in\mathbb C is a root of P if P(\alpha)=0. We are interested in recovering the roots from the coefficients. A natural first question is whether a root always exists.
Every non-constant complex polynomial has a complex root. This is the d’Alembert–Gauss theorem, more commonly called the Fundamental Theorem of Algebra.
There are many proofs. One short proof uses a basic result from complex analysis, Liouville’s theorem.
We first prove Liouville’s theorem. Let f be an entire function bounded by M. Cauchy’s integral formula for the derivative gives, for every r>0, f'(z_0)=\frac{1}{2\pi i}\oint_{|z-z_0|=r}\frac{f(z)}{(z-z_0)^2}\,dz. Consequently, |f'(z_0)|\leq M/r. Letting r\to\infty yields f'(z_0)=0. Since z_0 was arbitrary, f is constant.
Now let P have degree n\geq1 and suppose, for a contradiction, that it has no root. Then f(z)=1/P(z) is entire. Moreover, f(z)\to0 as |z|\to\infty. It is therefore bounded outside a sufficiently large disk; it is also bounded on the disk because it is continuous there. Liouville’s theorem says that f is constant, contradicting the fact that P is non-constant. Thus P has a root.
This only gives one root directly, but it implies that P has exactly n roots when multiplicities are counted. If \alpha is a root, then P(X)=(X-\alpha)Q(X), where Q has degree n-1. Repeating the argument gives all n roots.
2. From the roots to the coefficients
Going from the roots to the coefficients is straightforward. If P has roots \alpha_1,\ldots,\alpha_n, counted with multiplicity, then P(X)=a_n(X-\alpha_1)(X-\alpha_2)\cdots(X-\alpha_n).
Expanding this product gives Vieta’s formulas. They say that each coefficient is, up to its sign and the factor a_n, the sum of all products of a fixed number of distinct roots. More precisely, a_{n-k}=(-1)^k a_n\sum_{1\leq i_1<\cdots<i_k\leq n}\alpha_{i_1}\cdots\alpha_{i_k}.
We want to go in the opposite direction: given the coefficients, can we find a root using radicals? By a radical formula, we mean a finite expression involving the coefficients, addition, subtraction, multiplication, division, and extraction of roots of arbitrary positive integer orders.
We may divide by a_n and work with monic polynomials. Imagine, for the moment, that the desired formulas give a function f:\mathbb C^n\to\mathbb C^n sending (a_0,\ldots,a_{n-1}) to (\alpha_1,\ldots,\alpha_n): it takes the coefficients of a monic polynomial and returns its roots in some chosen order.
This construction should already feel slightly strange. The coefficients form an ordered tuple, whereas the roots naturally form an unordered set: every permutation of the roots gives the same polynomial. Nevertheless, let us suppose that f chooses a particular ordering. Since arithmetic operations and any locally chosen branch of a radical are continuous away from their singularities, continuously varying the coefficients should continuously vary the values returned by the formula. We will now see what goes wrong with this apparently reasonable function.
3. The quadratic case
Consider P(X)=X^2-1, whose roots are -1 and 1. Suppose, for instance, that our hypothetical function chooses f(-1,0)=(-1,1), where the input records the constant and linear coefficients of the monic polynomial.
Now consider the continuous loop of polynomials P_t(X)=X^2-e^{2\pi it} for 0\leq t\leq1. Because of the ordering chosen at t=0, continuity forces the two outputs along this path to be \bigl(-e^{\pi it},e^{\pi it}\bigr). At t=1 the polynomial is again X^2-1, so the input to f is again (-1,0). However, following the roots continuously has changed the ordering from (-1,1) to (1,-1).
Thus f cannot be a globally continuous, single-valued function that orders the roots. If we follow a formula continuously around a closed loop, we may return to the same coefficients but a different value. In this sense, the formula must be multivalued. This behavior is called monodromy. This interactive polynomial-roots visualization makes the exchange of the roots surprisingly concrete.
There is no contradiction with the quadratic formula. The square root is itself multivalued over the complex numbers, and one turn around the origin exchanges its two values. This is exactly what allows the formula to exchange the two roots, and it is the meaning of the not-so-innocent \pm sign.
This observation is already quite strong. There can be no global root formula for quadratics using only addition, subtraction, multiplication, and division, or more generally only single-valued continuous operations. At least one multivalued operation is necessary.
4. The general case
The quadratic example exchanged two roots. In the general case, we can obtain any permutation of the roots by continuously varying the coefficients along a loop.
To see this, it is easier to work in the opposite direction. Start with n distinct roots in the complex plane and move them continuously without allowing any two of them to collide. We can exchange two roots by moving them around each other, and by composing such exchanges we can realize any permutation. Vieta’s formulas express the coefficients as continuous functions of the roots, so this motion of the roots produces a continuous path of coefficients. Since the final roots differ from the initial roots only by a permutation, the final polynomial is the same as the initial one. The coefficient path is therefore a loop, like the one we constructed above with P_t(X). Figure 1 illustrates this construction for three roots.
The core idea behind the proof of the Abel–Ruffini theorem is that a formula for the general polynomial must have enough “multivaluedness” to represent every permutation of the roots. These permutations become more complicated as the degree increases, which is precisely the feature we will use.
The question is whether the monodromy produced by a finite tower of radicals can realize all these permutations. To understand how we might answer it, let us first ask what information determines whether a single radical changes value.
Suppose r is a single-valued expression in the coefficients and that, along a closed coefficient path, r never vanishes. Its values trace a closed curve in \mathbb C\setminus\{0\}. If this curve winds k times around the origin, continuing an m-th root of r along the path multiplies its initial value by e^{2\pi ik/m}. Thus the winding number is the only information about the path that this root extraction sees.
A convenient way to guarantee that the m-th root returns to its initial value is therefore to construct a coefficient loop along which the winding number of r is zero. Notice that this condition does not depend on m: if the winding number is exactly zero, square roots, cube roots, and roots of every other order all return to their initial values.
How can we force the winding number to be zero while still permuting the polynomial’s roots? Suppose a loop \gamma induces a permutation \sigma of the roots. Traversing the loop backward, which we denote by \gamma^{-1}, induces \sigma^{-1} and reverses the winding number. The loop \gamma\gamma^{-1} therefore has winding number zero, but it also induces the trivial permutation \sigma\sigma^{-1}. This does not help.
Now introduce a second loop \delta, inducing a permutation \tau. Winding numbers add when paths are concatenated, so along \gamma\delta\gamma^{-1}\delta^{-1} the total winding number is k_\gamma+k_\delta-k_\gamma-k_\delta=0. This cancellation works for every single-valued nonzero expression r in the coefficients. The important point is that winding numbers are ordinary integers, so the order in which they are added does not matter, whereas permutations need not commute. The resulting permutation is \sigma\tau\sigma^{-1}\tau^{-1}, which can be nontrivial. This is how the commutator arises naturally: it consists of two cancelling pairs arranged in an order that need not cancel as a permutation. Figure 2 follows one such commutator step by step.
The corresponding motion of the roots, shown in Figure 2, consists of four steps:
- exchange \alpha_1 and \alpha_2;
- exchange \alpha_2 and \alpha_3;
- undo the first exchange;
- undo the second exchange.
More formally, we write [\sigma,\tau]=\sigma\tau\sigma^{-1}\tau^{-1} and call it the commutator of \sigma and \tau. If \sigma=(1\,2) and \tau=(2\,3), then, using the convention that the rightmost permutation acts first, [\sigma,\tau]=(1\,3\,2). This is a nontrivial 3-cycle. We have therefore found a coefficient loop along which every one-level radical returns to its initial value while the roots undergo a nontrivial permutation. From degree three onward, one layer of radicals cannot be enough. Notice that this phenomenon requires at least three roots: S_2 is abelian, so all of its commutators are trivial.
Indeed, if our hypothetical function f were built with only one level of radicals, zero winding would force f to return to its original value, even though following the roots continuously requires its output to be permuted. This is why the cubic formula needs nested radicals and contains a square root inside a cube root. For instance, Cardano’s formula has terms of the form \sqrt[3]{-\frac q2+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}. Along a commutator loop the innermost single-valued quantity has zero winding, but the continued value of its square root need not trace a loop with zero winding. An outer radical can therefore still change branch.
We call this a two-level nested radical: one radical occurs inside another. In general, the radical depth of an expression is the largest number of root extractions nested along any one branch of the expression. Coefficients and single-valued functions of them have depth zero; addition, subtraction, multiplication, and division take the maximum depth of their inputs; and taking any m-th root increases the depth by one.
This explains why neither the particular orders of the radicals nor the number of arithmetic operations is important. A root extraction of any order introduces one possible branch change, controlled by a winding number. Arithmetic operations introduce no new branches: if their inputs return to their initial values, so does their output. For example, adding ten separate radicals still has depth one, whereas taking a square root of their sum raises the depth to two. Every finite radical formula can be viewed as such an expression tree, so it is enough to argue using its maximum radical depth rather than assuming that the whole formula is one simple chain of nested roots.
The next question is therefore whether increasing this depth can eventually represent every permutation.
5. The final argument
Since we have built a permutation that one layer of radicals cannot represent, the natural next step is to look for one that two nested layers cannot represent.
Why is an ordinary commutator not already enough? Along a commutator loop, every first-level radical returns to its initial value, but the path traced by that radical may itself wind around the origin. If it appears inside a second radical, this new winding can change the outer radical’s branch. This is precisely the extra freedom provided by nesting.
The key point is that, because all first-level radicals return to their initial values along a commutator loop, every expression built from them using arithmetic operations also returns to its initial value. In particular, the expression inside a second-level radical traces a closed curve, so it has its own winding number. We can now cancel this new winding number by the same construction as before.
Take two commutator loops \Gamma and \Delta. Along each of them, all depth-one expressions close. Now form \Gamma\Delta\Gamma^{-1}\Delta^{-1}. For the radicand of any depth-two root, the windings accumulated along \Gamma and \Delta are cancelled by those along the reverse paths. Its total winding number is zero, so every second-level radical returns to its initial value. This new loop is a commutator of commutators, or a double commutator.
The same reasoning continues inductively. A loop built from d successive levels of commutators makes every expression of radical depth at most d return to its initial value. At each stage, the preceding stage ensures that all inner expressions trace closed curves; the next commutator cancels the winding numbers of those curves. The only remaining question is whether the permutation of the polynomial’s roots can stay nontrivial through arbitrarily many such stages.
In small degrees, these repeated commutators eventually become trivial. In S_3, every double commutator is trivial. In S_4, nontrivial double commutators exist, but every triple commutator is trivial. This matches what the cubic and quartic formulas suggest: finitely many levels suffice in degrees three and four.
It is useful to summarize the process with a little notation. Let S_n^{(0)}=S_n,\qquad S_n^{(j+1)}=[S_n^{(j)},S_n^{(j)}], where the bracket denotes the subgroup generated by commutators. The terms S_n^{(k)} are called the derived series of S_n. The continuation of an expression with d nested layers of radicals cannot detect a root permutation lying sufficiently deep in this series.
For S_3 and S_4, this series eventually reaches \{e\}, so the commutator obstruction disappears after finitely many levels. For S_5, however, the first step gives A_5 and every subsequent step gives A_5 again. Thus, whatever finite radical depth d a proposed formula has, there remains a nontrivial permutation in S_5^{(d)} that the formula cannot detect. This is the essential reason the argument stops radical formulas in degree five but not in degrees at most four.
Because the sign map is multiplicative, \operatorname{sgn}([\sigma,\tau])=1, so every commutator in S_n is even. Conversely, for n\geq3, every 3-cycle is a commutator of two transpositions; for example, [(1\,2),(2\,3)]=(1\,3\,2). Since the 3-cycles generate A_n, it follows that [S_n,S_n]=A_n.
For S_3, the subgroup A_3 is cyclic and therefore abelian. Hence [A_3,A_3]=\{e\}, giving S_3\supset A_3\supset\{e\}.
For S_4, let V_4=\{e,(1\,2)(3\,4),(1\,3)(2\,4),(1\,4)(2\,3)\}. The quotient A_4/V_4 is cyclic of order 3, so [A_4,A_4]\subseteq V_4. On the other hand, [(1\,2\,3),(1\,2\,4)]=(1\,2)(3\,4), and conjugating this identity inside A_4 gives the other double transpositions. Thus [A_4,A_4]=V_4. Since V_4 is abelian, [V_4,V_4]=\{e\}, and therefore S_4\supset A_4\supset V_4\supset\{e\}.
For S_5, we again have [S_5,S_5]=A_5. Now take \sigma=(1\,2\,4) and \tau=(2\,3\,5). Both lie in A_5, and [\sigma,\tau]=(2\,4\,3) is a 3-cycle. Every 3-cycle in A_5 is obtained from this one by conjugation by an element of A_5: if a relabeling permutation is odd, compose it with the transposition of the two points fixed by the target 3-cycle. Therefore every 3-cycle belongs to [A_5,A_5]. Since 3-cycles generate A_5, we obtain [A_5,A_5]=A_5. In other words, A_5 is perfect, and the derived series remains equal to A_5 forever.
No finite number of radical layers can therefore account for all the possible permutations of the roots of a general quintic. If there were such a formula, we could follow it continuously along a loop of coefficients that induces a nontrivial permutation in S_5^{(d)}. The formula would return to its initial value by construction of the permutation, but the roots would not.
This is basically it, of course, the same conclusion holds for every degree n\geq5 since we can just look at the restruction of such formula to the first 5 roots for instance.
The word finite is essential. Abel–Ruffini concerns finite expressions, whereas an infinite nesting is a limit. For x^5-x-1=0, the iteration x_{j+1}=\sqrt[5]{1+x_j} converges to a solution. Thus the solutions can be represented by corresponding convergent infinite nested radicals of the form x=\sqrt[5]{1+\sqrt[5]{1+\sqrt[5]{1+\cdots}}}. This does not contradict Abel–Ruffini: a convergent infinite limit is not a finite formula by radicals.
6. Some final remarks
We have suppressed some technical details. The roots must remain distinct, so the coefficient paths live in the complement of the discriminant locus. This space is open and path-connected, and its loops produce the permutations used above. One must also choose the paths generically so that denominators do not vanish and the quantities under radicals avoid zero when winding numbers are used. These issues can be handled, but doing so carefully requires more topology than this post aims to develop.
The argument we described says that there is no radical formula for the general quintic. It does not say that every quintic is unsolvable by radicals. This is because we assumed that the same formula continued to give us the roots throughout the loops in the coefficient space, so the formula must work for any polynomial. If we restrain to a particular family of quintics, it may be possible to find a radical formula for that family. For example, consider X^5-t where t \in \mathbb{C}^\star, then the same loop argument when t goes around the origin only allow for a cyclic permutation of the roots. Those permutations can actually be represnted by radicals. In fact, the roots are given by t^{1/5}\zeta^k for k=0,1,2,3,4 where \zeta is a primitive fifth root of unity, therefore the formula .
Galois theory is more powerful here in the sense that it gives the precise criterion: a polynomial is solvable by radicals if and only if its Galois group is solvable. For example, X^5-X-1 is not solvable by radicals over \mathbb Q, whereas X^5-1 is; the latter’s roots are the fifth roots of unity.
However, the argument developed here has some advantages over the classical Galois theory one. The proof shows that any finite expression made from arithmetic operations, radicals, and globally single-valued functions such as the exponential, sine, and cosine cannot solve the general quintic equation since they do not address the “multivaluedness” issue.
Going beyond radicals
If radicals are insufficient, what additional operation can solve a general quintic? By algebraic changes of variables, a general quintic can be reduced to the Bring–Jerrard form z^5+z+t=0. The Bring radical is defined as a chosen local inverse of the map z\mapsto-(z^5+z); in other words, \operatorname{BR}(t) is a chosen solution of z^5+z+t=0. This function can be represented locally by a hypergeometric series and then analytically continued to obtain its other branches. Once the Bring radical is admitted as a new operation, the reduced quintic—and hence the general quintic—can be solved.
There is also a classical solution, due to Hermite and developed further by Kronecker and Brioschi, using elliptic modular functions. The point is not simply that these functions are more complicated than radicals. Their associated modular equation has the non-abelian A_5 monodromy needed to permute the five roots, whereas the monodromy produced by any finite tower of radicals is solvable. Adding such a function therefore supplies exactly the kind of multivaluedness that the preceding proof shows radicals cannot provide.
For more detail, see Leo Goldmakher’s Arnold’s Elementary Proof of the Insolvability of the Quintic and Paul Ramond’s The Abel–Ruffini Theorem: Complex but Not Complicated.