• Dummit showed that given the five roots {$x_1, x_2, x_3, x_4, x_5$} of the quintic, then the expression,

$x_1^2(x_2 x_5+x_3 x_4)+x_2^2(x_1x_3+x_4x_5)+x_3^2(x_1x_5+x_2x_4)+x_4^2(x_1x_2+x_3 x_5)+x_5^2(x_1 x_4+x_2x_3)$

is the root of a $(5-2)!=3!=6$-deg eqn.

  • Boswell and Glasser gave an analogous one for the sextic {$y_1, y_2, y_3, y_4,y_5, y_6$} as,


which is the root of a 10th-deg eqn.

  • Question: What is a corresponding expression for the septic {$z_1, z_2, z_3, z_4,z_5, z_6, z_7$}?

(I know it will be a root of a $(7-2)! = 5! = 120$-deg equation, but I want to know what combination of the $z_i$ will be suitable.)

  • $\begingroup$ Hold on, how is the sextic resolvent you cite analogous to the quintic resolvent? By your reasoning it should have degree $(6 - 2)! = 24$, not $10$. $\endgroup$ – Zhen Lin Apr 23 '12 at 15:10
  • $\begingroup$ Boswell and Glasser's decic resolvent is analogous in the sense that if it has a rational root, then the sextic is solvable. Kindly see Theorem 2 of their paper at arxiv.org/pdf/math-ph/0504001v1.pdf. The formula $(p-2)!$ applies only to prime p. $\endgroup$ – Tito Piezas III Apr 23 '12 at 15:59

Let $k$ be a field, let $x_1, \ldots, x_n$ be indeterminates, let $F = k(x_1, \ldots, x_n)$, let $G$ be a subgroup of $S_n$, and let $L$ be the subfield of $L$ fixed by the evident action of $G$, and let $K$ be the subfield fixed by the entire group $S_n$. The field extension $F \mid L$ is a finite Galois extension with Galois group $G$ (by construction!) and each $x_i$ is a root of the equation $$(T - x_1) \cdots (T - x_n) = T^n - e_1 T^{n-1} + \cdots + (-1)^n e_n = 0$$ where $e_1, \ldots, e_n$ are the elementary symmetric polynomials in $x_1, \ldots, x_n$. Since the $e_1, \ldots, e_n$ are fixed by $S_n$, they are elements of $K$. We think of $x_1, \ldots, x_n$ as being the zeros of a generic $n$-th degree polynomial with coefficients in $k$, but strictly speaking they are not. However, they do have minimal polynomials with coefficients in $K$: we think of $e_1, \ldots, e_n$ as being generic coefficients.

Now, since $F \mid K$ is a Galois extension, the subextension $L \mid K$ must be separable. The primitive element theorem then implies there is a $\rho$ in $L$ such that $L = K (\rho)$. Such a $\rho$ is called a generic resolvent for the group $G$. If $G$ is a soluble group, then Kummer theory implies that $x_1, \ldots, x_n$ can be expressed in terms of sums, products, and iterated radicals of elements of $L$, and by definition every element of $L$ can be written as a polynomial in $\rho$ with coefficients in $K$. Standard Galois theory tells us that $\rho$ will satisfy an equation of degree $\left. {n!} \middle/ |{G}| \right.$; the minimal polynomial of $\rho$ over $K$ is called the generic resolvent polynomial for $G$.

Now, if we specialise to a concrete polynomial $P$ over $k$ with Galois group $G$, this means that we will have a formula for the zeros of $P$ in terms of the coefficients of $P$ and $\rho$, which will be an element of $k$ depending only on the coefficients of $P$. We can tell whether $P$ has Galois group $G$ by looking at the resolvent polynomial: if it has no zeros in $k$, then $P$ cannot have a Galois group $G$; otherwise, the Galois group of $P$ is a subgroup of $G$.

For example, let $n = 3$, $G = \{ \textrm{id} \}$, and suppose $k$ contains a primitive third root of unity $\omega$. (In particular, $\operatorname{char} k \notin \{ 2, 3 \}$.) This is a slightly degenerate case since $F = L$, but it was the first one considered historically. The Lagrange resolvent is defined as $$\alpha = x_1 + \omega x_2 + \omega^2 x_3$$ Define also $$\beta = x_1 + \omega^2 x_2 + \omega x_3$$ Clearly, the Galois orbit of $\alpha$ is $\{ \alpha, \omega \alpha, \omega^2 \alpha, \beta, \omega \beta, \omega^2 \beta \}$, and these are all distinct, so $\alpha$ is of degree $6$ over $K$ and therefore $L = K(\alpha)$ as required. We know $1 + \omega + \omega^2 = 0$, so \begin{align} \alpha + \beta + e_1 & = 3 x_1 \\ \omega^2 \alpha + \omega \beta + e_1 & = 3 x_2 \\ \omega \alpha + \omega^2 \beta + e_1 & = 3 x_3 \end{align} Thus, as long as we can find $\alpha$ and $\beta$ in terms of $e_1, e_2, e_3$, we will be able to find $x_1, x_2, x_3$ as well. Now, notice that the action of the subgroup $A_3$ partitions the Galois orbit of $\alpha$ into two sets $\{ \alpha, \omega \alpha, \omega^2 \alpha \}$ and $\{ \beta, \omega \beta, \omega^2 \beta \}$, so $\alpha^3$ and $\beta^3$ must each be of degree $2$ over $K$, and the minimal polynomial of $\alpha$ and $\beta$ is therefore $$T^6 - (\alpha^3 + \beta^3) T^3 + \alpha^3 \beta^3$$ Notice that both $\alpha^3 + \beta^3$ and $\alpha \beta$ are invariant under the action of $S_3$ so must indeed be in $K$. In fact, $$\alpha^3 + \beta^3 = 2 {e_1}^3 - 9 {e_1} {e_2} + 27 e_3$$ $$\alpha \beta = {e_1}^2 - 3 e_2$$ and so we have the general solution of the cubic equation, at least when $\operatorname{char} k \notin \{ 2, 3 \}$.

Your example of the Dummit resolvent $\rho$ is invariant under the action of the subgroup $$F_{20} = \langle (1 \; 2 \; 3 \; 4 \; 5), (2 \; 3 \; 4 \; 5) \rangle$$ The group $F_{20}$ has $20$ elements, so $\rho$ satisfies an equation of degree $5! / 20 = 6$ over the elementary symmetric polynomials $e_1, \ldots, e_5$. The group $F_{20}$ is soluble, so there is an expression for $x_1, \ldots, x_5$ in terms of sums, products, and iterated radicals of elements of $L = K (\rho)$. Thus, if a quintic polynomial $P$ over $mathbb{Q}$ has Galois group $F_{20}$, then once we have found what $\rho$ is when specialised to $P$, we will be find the zeros of $P$.

You should now be able to invent your own resolvents.


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