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## A Snippet of Galois Theory

August 11, 2016 1 comment

The following hints at why the quintic equation cannot be solved using radicals. It follows the approach in the first part of Ian Stewart’s book “Galois Theory”. If time permits, a future post will summarise the approach in V. B. Alekseev’s book “Abel’s Theorem in Problems and Solutions”. Another candidate is Klein’s book “Lectures on the Icosahedron and the Solution of Equations of the Fifth Degree”.

To set the stage, consider the generic quadratic polynomial equation $x^2 + ax + b = 0$. Solving it is the same as factorising it: given the coefficients $a,b$ find the roots $t_1, t_2$ such that $x^2 + ax + b = (x-t_1)(x-t_2)$.

The first observation is the prominent role played by symmetric polynomials. Note $(x-t_1)(x-t_2) = x^2 - (t_1+t_2) x + t_1 t_2$. Here, $s_1 = t_1+t_2$ and $s_2 = t_1 t_2$ are called elementary symmetric polynomials. They are symmetric because swapping $t_1$ and $t_2$ does not change the product $(x-t_1)(x-t_2)$. This “annoying” ambiguity turns out to be a friend; the presence of symmetry forces certain relationships to be true, as will be seen presently.

Solving a quadratic equation is equivalent to the following. Given the values of the symmetric polynomials $s_1 = t_1 + t_2$ and $s_2 = t_1 t_2$ (up to sign, these are equivalent to the coefficients $a$ and $b$ in $x^2 + ax + b$), find the roots $t_1, t_2$.

Given $s_1,s_2$, we can readily evaluate any symmetric function of the roots that we wish; this is a property of elementary symmetric polynomials that is not too difficult to prove. For example, since $t_1^2 + t_2^2$ is symmetric, it must be capable of being written as a polynomial in $s_1,s_2$. Trial and error leads to finding that $s_1^2 - 2 s_2 = (t_1+t_2)^2 - 2 t_1 t_2 = t_1^2 + t_2^2$, verifying the claim in this instance.

The first insight is that this is not an all or nothing affair. Although $t_1 - t_2$ is not symmetric, it still behaves nicely when $t_1,t_2$ are swapped: $t_2 - t_1 = -(t_1 - t_2)$. [If we were trying to solve cubic or quartic equations, we would look for polynomial expressions of the roots that took on only two or three distinct values as the roots were permuted.] Moreover, since $(-1)^2 = 1$, this implies $(t_1-t_2)^2$ is symmetric, a fact that is also self-evident. Therefore, although we cannot express $t_1 - t_2$ as a polynomial function of $s_1,s_2$, we can do the next best thing: write $t_1-t_2$ as the square root of a polynomial in $s_1,s_2$.

Specifically, $(t_1-t_2)^2 = t_1^2 - 2 t_1 t_2 + t_2^2$, and from earlier, we saw that $t_1^2 + t_2^2 = s_1^2 - 2 s_2$, therefore $(t_1-t_2)^2 = s_1^2 - 4 s_2$.

At this point, the reader should pause to recognise that we have “rediscovered” the discriminant $b^2 - 4 a c$ of a quadratic $a x^2 + bx + c$. (In our case, $a = 1$, $b = -s_1$ and $c = s_2$, hence $b^2 - 4 a c = s_1^2 - 4 s_2$.)

It is now straightforward to derive the well-known formula for solving a general quadratic equation since we have reduced it to solving linear equations: we know that $t_1 + t_2 = s_1$ and $t_1 - t_2 = \sqrt{ s_1^2 - 4 s_2 }$.

Why is it impossible to derive a similar type of formula for solving a quintic equation?

It comes down to a property of the symmetry group of five elements: it behaves differently from the symmetry groups on two, three or four elements, which is why formulae exist for solving polynomials of degree less than 5 but not for those of degree 5. (Generic polynomials of higher degrees also have no closed-form solutions expressible in terms of radicals. Note that this does not mean a formula cannot be given whatsoever; for example, degree 5 polynomials can be solved using elliptic functions.)

The remainder of this note endeavours to shed some light on why there is this remarkable connection between symmetry groups and solutions of polynomial equations.

Solving the generic quintic equation amounts to inverting the map $(t_1,\cdots,t_5) \mapsto (s_1,\cdots,s_5)$ where $s_1 = t_1 + t_2 + t_3 + t_4 + t_5,\ \cdots\ , s_5 = t_1 t_2 t_3 t_4 t_5$ are the elementary symmetric polynomials. While it is possible to show that no formula exists for this inverse that uses only ordinary arithmetic operations (addition, subtraction, multiplication and division) and radicals (taking square roots, cube roots etc), we will prove a weaker but nonetheless insightful result. In his book, Stewart refers to this as showing no solution “by Ruffini radicals” exists. (Ruffini proved this result, without realising the stronger result does not follow from it.)

The type of formulae we will permit are of the same form as that used to solve the quadratic equation. We start off with known values for $t_1+\cdots+t_5,\ \cdots\ , t_1 \cdots t_5$; indeed, these are given by $s_1,\cdots,s_5$ which are the coefficients of the quintic equation under consideration. At each step, we wish to determine more information about the roots $t_i$ by finding a rational function of the $t_i$ whose nth power is a rational function of the already known quantities, for some integer $n \geq 2$. At the first step, this means finding a rational function of the $t_i$ whose nth power is equal to a rational function of $s_1,\cdots,s_5$. Taking an nth root then gives us a new quantity that can be used in subsequent steps, and so forth. (Ruffini’s oversight was not realising that a priori it might be useful to compute nth roots of other quantities; the above description only permits an nth root to be computed if the answer is expressible as a rational function of the roots. For example,  we are not allowed to take the nth root of $s_1$ because the nth root of $s_1$ cannot be written as a rational function of the roots $t_1,\cdots,t_5$. Abel was able to prove that computing nth roots of other quantities did not help, thereby deducing the stronger result from Ruffini’s result.)

Consider step 1. A rational function $g(t_1,\cdots,t_5)$ is sought such that, for some integer $n \geq 2$, the $n$th power of $g$ is a rational function of the elementary symmetric polynomials $s_1,\cdots,s_5$, that is, $g^n(t_1,\cdots,t_5) = h(s_1,\cdots,s_5)$ where $g,h$ are rational. Without loss of generality, we may assume $n$ is a prime number. (If we wanted a sixth root, we could first take a cube root then, in step 2, take a square root.)

To emphasise the dependence on the roots, we have $g^n(t_1,\cdots,t_5) = f(t_1+t_2+t_3+t_4+t_5,\cdots,t_1 t_2 t_3 t_4 t_5)$. Can $f,g$ be chosen relatively freely, or does this constraint impose severe restrictions on the possible choices of the rational functions $f,g$?

Here comes the crux of the matter. Despite the seemingly unwieldy equation $g^n(t_1,\cdots,t_5) = f(t_1+t_2+t_3+t_4+t_5,\cdots,t_1 t_2 t_3 t_4 t_5)$, the symmetry of the right-hand side (irrespective of what $f$ actually is) forces $g$ and $n$ to be essentially unique. Permuting the roots does not change the right-hand side, therefore permuting the roots cannot change the left-hand side either. It might change $g(t_1,\cdots,t_5)$, but it will not change $g^n(t_1,\cdots,t_5)$. [Actually, if it did not change $g(t_1,\cdots,t_5)$ then $g$ would be directly expressible as a rational function of the $s_1,\cdots,s_5$ and hence we can ignore this case, which would correspond to $n=1$.]

Recall momentarily the quadratic case, where $g(t_1,t_2) = t_1 - t_2$. Swapping $t_1,t_2$ caused the sign of $g$ to change, but that was it. Here we have, for example, that $g^n(t_1,t_2,t_3,t_4,t_5) = g^n(t_2,t_1,t_3,t_4,t_5)$, therefore, $\left(\frac{g(t_2,t_1,t_3,t_4,t_5)}{g(t_1,t_2,t_3,t_4,t_5)}\right)^n = 1$, or in other words, there exists an nth root of unity $\omega$ such that $g(t_2,t_1,t_3,t_4,t_5) = \omega\,g(t_1,t_2,t_3,t_4,t_5)$. [To be clear, $\omega$ is a complex number satisfying $\omega^n = 1$.]

Different permutations of the roots will lead to possibly different values of $\omega$. Sure, some $\omega$ might equal 1, but there will be at least one permutation such that the corresponding $\omega$ does not equal one. (If this were not true then $g$ would be symmetric and hence directly expressible as a rational function of the $s_1,\cdots,s_5$, hence we could ignore this step as no radical would need computing.)

The mapping from the permutation of the roots to the value of $\omega$ is a (nontrivial) group homomorphism. It cannot be arbitrary, and in particular, we will see that it forces $n$ to equal 2.

Denote the set of all permutations of $t_1,\cdots,t_5$ by $S_5$. This set naturally has a group structure: multiplication corresponds to applying one permutation after another, so to speak. Precisely, it is a symmetric group.

Denote by $W_n$ the set of all $n$th roots of unity; $\omega$ must lie in $W_n$. This set also has a standard group structure given by ordinary multiplication: if you multiply two $n$th roots of unity you get another $n$th root of unity. In fact, it is what is known as a cyclic group (and is isomorphic to $Z/nZ$, the abelian group of integers modulo $n$).

As mentioned above, the fact that $g^n(t_1,\cdots,t_5)$ is symmetric induces a function, call it $\phi$, from $S_5$ to $W_n$. Precisely, if $\sigma \in S_5$ is a permutation of the $t_1,\cdots,t_5$ then $\phi(\sigma) = \frac{g(\sigma(t_1,\cdots,t_5))}{g(t_1,\cdots,t_5)}$; recall from earlier that this ratio will be an element of $W_n$. Crucially, $\phi$ has structure: $\phi(\sigma_1 \sigma_2) = \phi(\sigma_1) \phi(\sigma_2)$ because $\frac{g(\sigma_1(\sigma_2(t_1,\cdots,t_5)))}{g(t_1,\cdots,t_5)} = \frac{g(\sigma_1(\sigma_2(t_1,\cdots,t_5)))}{g(\sigma_2(t_1,\cdots,t_5))} \frac{g(\sigma_2(t_1,\cdots,t_5))}{g(t_1,\cdots,t_5)}$. That is, $\phi$ is compatible with the group structure and hence is a group homomorphism. Furthermore, from earlier observations, we know $\phi$ is nontrivial: there exists at least one permutation $\sigma$ such that $\phi(\sigma) \neq 1$.

There is only a single possible nontrivial group homomorphism from $S_5$ to a cyclic group of prime order!

While $S_5$ is not a very small group (it has 120 elements), its group structure is what prevents more such group homomorphisms from existing. [Since $\phi \colon S_5 \rightarrow W_n$ is a nontrivial homomorphism then the kernel of $\phi$ must be a proper normal subgroup of $S_5$. This proper normal subgroup must also contain every commutator because $W_n$ is abelian. It can be shown the only such subgroup is what is known as the alternating subgroup $A_5$, implying $n=2$.]

At this juncture, the group structure of $S_5$ has forced $g$ to be (up to a constant) the square-root of the discriminant: $g(t_1,\cdots,t_5) = \prod_{i < j} (t_i - t_j)$. [I have omitted the proof of this as it is not the main focus; I mentioned it only because it helps keep the outline as concrete as possible, and indeed, it follows closely the situation encountered at the start when the quadratic polynomial was being considered.] Going into Step 2, we know not only the values of the elementary symmetric polynomials $t_1+t_2+t_3+t_4+t_5,\ \cdots\ , t_1t_2t_3t_4t_5$ but also the value of $\prod_{i < j} (t_i - t_j)$. It turns out though that we get blocked at Step 2 from going any further. The reasoning is similar to above, but with $S_5$ replaced by a smaller group.

Precisely, at Step 2, we want to find a rational function $g$ and a prime number $n \geq 2$ such that $g^n(t_1,\cdots,t_5)$ is equal to a rational function of not only the $s_1,\cdots,s_n$ but also the new quantity we found in Step 1, namely $\prod_{i < j} (t_i - t_j)$. This time it is no longer true that $g^n(t_1,\cdots,t_5)$ must be symmetric in $t_1,\cdots,t_5$ because the new quantity $\prod_{i < j} (t_i - t_j)$ is not symmetric. However, if we only look at what are known as even permutations then $\prod_{i < j} (t_i - t_j)$ will remain invariant. The alternating subgroup $A_5$ of $S_5$ consists precisely of the even permutations. In particular, we therefore know that if $\sigma$ is in $A_5$ then $g^n(\sigma(t_1,\cdots,t_5)) = g^n(t_1,\cdots,t_5)$. This is the same as in Step 1 except $S_5$ has been replaced by $A_5$. [We did not need to know that Step 1 produced $\prod_{i < j} (t_i - t_j)$. A simpler group-theoretic argument mentioned parenthetically above implies that the kernel of $\phi$ in Step 1 is $A_5$ and hence that whatever new function of the roots is determined in Step 1, it will be invariant with respect to permutations belonging to $A_5$.]

Exactly as in Step 1, the invariance of $g^n(t_1,\cdots,t_5)$ forces the existence of a nontrivial group homomorphism, this time from $A_5$ to a cyclic group of order $n$. And once again, the group structure of $A_5$ limits the possibilities.

There are no nontrivial group homomorphisms from $A_5$ to a cyclic group of prime order.

The above (which is a relatively straightforward result in group theory) immediately implies that there is no formula of the kind specified earlier for solving a quintic polynomial.

Galois theory goes well beyond the basic argument sketched here. It can be used to conclude that a generic quintic (or any higher-degree polynomial for that matter) has no solution in terms of radicals (rather than use Abel’s result to strengthen Ruffini’s result, as alluded to earlier, a cleaner proof is possible). It can also be used to conclude that there exist particular quintic polynomials whose roots cannot be expressed using radicals; this does not follow from the generic case because the generic case sought a single formula for all quintic polynomials, whereas a priori there might exist a different formula for each quintic equation. (Certainly, some quintic equations can be solved using radicals.)