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I always wondered if the fact that the quartic can be solved by a cubic can be generalized to other even degrees $n$, namely if there is an ordering of the roots $x_i$ of form $x_1x_2+x_3x_4+\dots+x_{n-1}x_{n} = y$ such that $y$ is reduced to an algebraic number of deg $n-1$. It seems it can be if its Galois group is of a certain kind.

Call the resolvent polynomial formed by $y$ as $P_{n}(y)$. (The labeling of the transitive groups below is from Kluener's database of number fields.)

The 3-deg $P_4(y)$ has a 3th deg factor for the general quartic.

The 15-deg $P_6(y)$ has a 5th deg factor if $6T12, 6T14$.

The 105-deg $P_8(y)$ has a 7th deg factor if $8T25, 8T36, 8T37, 8T48$ (the first two are solvable groups).

The 945-deg $P_{10}(y)$ has an 9th deg factor if (none?),

The 10395-deg $P_{12}(y)$ has an 11th deg factor if $12T179, \text{(and?)}$

For example, given,

x^12-2x^11+17x^10-28x^9+138x^8-157x^7+549x^6-382x^5+1099x^4-77x^3+1016x^2+9x+1912 = 0

(I've left it unformatted for easier copy-paste) which has group $L(2,11)$ and discriminant $D_{12} = 1831^6$. I use Mathematica, and its root ordering is,

$$x_1, x_2 = -0.822\mp0.744 i$$

$$x_3, x_4 = -0.801\mp2.308 i$$

$$x_5, x_6 = -0.452\mp2.271 i$$

$$x_7, x_8 = 0.800\mp1.760 i$$

$$x_9, x_{10} = 0.949\mp1.080 i$$

$$x_{11}, x_{12} = 1.324\mp2.123 i$$

Define,

$$y = x_1x_7+x_2x_8+x_3x_6+x_4x_5+x_9x_{10}+x_{11}x_{12}=15.60720071\dots$$

then $y$ is a root of the monic 11-deg,

$$14155013839 + 4895349263 y + 3076830095 y^2 + 520581160 y^3 - 43394407 y^4 - 3261079 y^5 + 376457 y^6 + 33595 y^7 - 42 y^8 - 162 y^9 - 17 y^{10} + y^{11} = 0$$

which has discriminant $D_{11} = 23^6(1831^4)$.

Questions:

  1. What other transitive groups for 12-deg eqns will have a $P_{12}(y)$ that has an 11-deg factor?

  2. Does it follow that there are 14-deg eqns such that $P_{14}(y)$ has an 13-deg factor? (I suspect $14T30, 14T39$ which have groups $PSL(2,13), PGL(2,13)$, respectively.)

P.S. My thanks to the Spearman, Watanabe, Williams paper PSL(2,5) Sextic Fields with a Power Basis which was the clue.

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2 Answers 2

up vote 8 down vote accepted

There are no other examples of degree $12$ besides the one you found. Moreover, there are no examples of degrees $10$ or $14$, and in fact the only additional degrees up to $32$ in which there exist examples are $16$, $28$, and $32$, where the groups in the latter cases are precisely (in Klueners/GAP/MAGMA notation) 16T447, 16T777, 16T1079, 16T80, 16T1329, 16T1508, 16T1653, 16T1654, 16T1753, 16T1840, 16T1906, 28T165, 32T35272, 32T397065, 32T2795174. Of these, only the first three degree-$16$ groups and the first two degree-$32$ groups are solvable; those degree-$16$ groups are the unique solvable transitive degree-$16$ groups of orders $240$, $480$, and $960$, and the solvable degree-$32$ groups are the unique transitive degree-$32$ groups of orders $992$ and $4960$. The biggest nonsolvable degree-$16$ group on this list is $\text{AGL}_4(2)$, the degree-$28$ group is $\text{P}\Gamma\text{L}_2(8)$, and the nonsolvable degree-$32$ group is $\text{AGL}_5(2)$.

More generally, let $f(X)$ be a separable irreducible polynomial over a field $K$, and suppose that $n:=\text{deg}(f)$ is even. Suppose in addition that, if $\{C_1,C_2,\dots,C_{n/2}\}$ and $\{D_1,D_2,\dots,D_{n/2}\}$ are any two distinct partitions of the roots of $f(X)$ into $n/2$ two-element sets, then $$ \sum_{i=1}^{n/2} \prod_{x\in C_i} x \ne \sum_{i=1}^{n/2}\prod_{x\in D_i} x. $$ (Note that this condition will hold for any "randomly selected" polynomial having any prescribed Galois group.) Under these hypotheses, if $G$ denotes the Galois group of $f(X)$ over $K$, then the following are equivalent:

  1. there is an ordering $x_1,\dots,x_n$ of the roots of $f(X)$ such that $x_1x_2+x_3x_4+\dots+x_{n-1}x_n$ has degree $n-1$ over $K$.
  2. there is an index-$(n-1)$ subgroup $H$ of $G$, and a pair of distinct roots $(x_1,x_2)$ of $f(X)$, such that every element of $H$ maps $\{x_1,x_2\}$ to either itself or to a disjoint set of roots, but every subgroup of $G$ which properly contains $H$ must contain an element which maps $\{x_1,x_2\}$ to a set having exactly one element in common with $\{x_1,x_2\}$.

This can be used to test your condition for specific values $n$, and perhaps group-theoretic results can be used to prove theorems for infinitely many $n$. For instance, I will show later that $G=\text{AGL}_d(2)$ has the required property for $n=2^d$.

To see this equivalence, first assume that the first condition holds; then we can let $H$ be the set of elements in $G$ which fix $y:=x_1x_2+x_3x_4+\dots+x_{n-1}x_n$. This $H$ will be an index-$(n-1)$ subgroup of $G$, and every element of $H$ permutes the collection of 2-element sets $\{\{x_1,x_2\}, \{x_3,x_4\}, ..., \{x_{n-1},x_n\}\}$. Thus, each element of $H$ maps $\{x_1,x_2\}$ to some $\{x_i,x_{i+1}\}$ with $i$ odd. Moreover, since $f(X)$ is irreducible, $G$ acts transitively on $\{x_1,\dots,x_n\}$, so $[G:G_{x_1}]=n$; but since $[G:H]=n-1$ is coprime to $n$, we have $[G:G_{x_1}\cap H]=n(n-1)$ and thus $[H:G_{x_1}\cap H]=n$, so $H$ is transitive. Any subgroup $J$ of $G$ which properly contains $H$ must contain an element $j$ which does not fix $y$, so this element must not permute the collection $\{\{x_1,x_2\}, \{x_3,x_4\}, ..., \{x_{n-1},x_n\}\}$; by multiplying $j$ on both sides by suitable elements of $H$, we obtain an element of $J$ which maps $\{x_1,x_2\}$ to a set having one element in common with $\{x_1,x_2\}$.

Conversely, assume that the second condition holds. As above, $H$ is transitive. It follows that the images of $\{x_1,x_2\}$ under $H$ consist of $n/2$ pairwise disjoint two-element sets, which we can write as $\{x_1,x_2\}, \{x_3,x_4\}, ..., \{x_{n-1},x_n\}$. Then $y:=x_1x_2+x_3x_4+...+x_{n-1}x_n$ is fixed by $H$, but not by any larger subgroup of $G$, so $[K(y):K]=n-1$. This completes the proof of the equivalence.

Added later: here is a proof that $G:=\text{AGL}_d(2)$ has the required property for $n=2^d$. The group $G$ acts on the $\mathbf{F}_2$-vector space $V:=(\mathbf{F}_2)^d$, and $\text{GL}_d(2)$ consists of the elements fixing $0$. Pick some nonzero $c\in V$, and let $J$ be the stabilizer of $c$ in $G$. Let $H$ be the subgroup of $G$ generated by $J$ and the translations $x\mapsto x+u$. I claim that $H$ and $\{0,c\}$ have the required properties. Note that $\text{GL}_d(2)$ acts transitively on $V\setminus\{0\}$, so $[\text{GL}_d(2):J]=2^d-1$. Since the group of translations has order $2^d$, and is normalized by $\text{GL}_d(2)$ (and hence by $J$), we have $\#H=2^d\cdot\#J$ so $[G:H]=2^d-1$. Every translation maps $\{0,c\}$ to a set $\{b,b+c\}$, and any such set either equals $\{0,c\}$ or is disjoint from $\{0,c\}$. Finally, let $K$ be a subgroup of $G$ which properly contains $H$. Since $K$ contains all the translations, and every element of $G$ is a translation times an element of $\text{GL}_d(2)$, it follows that $K$ contains an element of $\text{GL}_d(2)$ which is not in $H$. Any such element maps $c$ to some $b\notin\{0,c\}$, and hence maps $\{0,c\}$ to the set $\{0,b\}$ which has exactly one element in common with $\{0,c\}$.

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That is fascinating! I guess just like quartics can be solved as $x = \sqrt{u_1}+\sqrt{u_2}+\sqrt{u_3}$ where the $u_i$ are roots of a cubic, and some octics (such as this one) can be solved as $x = \sqrt{u_1}+\sqrt{u_2}+\dots+\sqrt{u_7}$ where the $u_i$ are the roots of a solvable 7th deg, then it seems there are 16 and 32-deg eqns that can be solved as $x = \sqrt{u_1}+\sqrt{u_2}+\dots+\sqrt{u_{15}}$ and $x = \sqrt{u_1}+\sqrt{u_2}+\dots+\sqrt{u_{31}}$, respectively. I would love to see such a 32-deg, but I guess that's out of reach for now. –  Tito Piezas III Nov 19 '13 at 0:12

The common feature in all of your examples is that the nonabelian composition factors of the transitive permutation groups of degree $2n$ in question have permutation representations of degree $2n-1$. This is the case for 6T12 and 6T14 ($A_5$), 8T37 and 8T48 ($L(2,7)$), and 12T179 ($L(2,11)$).

There are no transitive groups of degree 10 with composition factors having a representation of degree 9, which might explain why you find no examples there (although $A_5$ and $A_6$ arise here, which have smaller degree representations).

Note that $L(2,13)$ has no permutation representation of degree 13, so I would not expect this to work in those cases.

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Why did it work for 6T14 ($PGL(2,5)$), but not for 8T43 ($PGL(2,7)$)? (I tested it for the latter, and $P_8(y)$ only had a 14-deg factor.) –  Tito Piezas III Nov 16 '13 at 21:50
    
I imagine that is because ${\rm PGL}(2,5) \cong S_5$ has a permutation representation of degree 5, but ${\rm PGL}(2,7)$ has none of degree 7 (but ${\rm PSL}(2,7)$ does). That suggests that it is not just the composition factor that matters, but the automorphisms that are present in the transitive group. Do you have a polynomial for 12T272 = $M_{11}$? According to my theory, that should work too. –  Derek Holt Nov 17 '13 at 8:19
    
Although $M_{11}$ has a degree-$11$ permutation representation, still that group doesn't yield the required types of polynomials, since there are no degree-$11$ elements of the form $x_1x_2+x_3x_4+...+x_{11}x_{12}$. There are degree-$11$ elements given by other expressions in the $x_i$'s, but not by an expression of the required form. –  Michael Zieve Nov 18 '13 at 4:49
    
@Michael Zieve: Yes I see that now! Thanks! –  Derek Holt Nov 18 '13 at 10:59

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