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*](http://www.math.uni-duesseldorf.de/~klueners/minimum/).) 


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](http://people.math.carleton.ca/~williams/papers/pdf/283.pdf)* which was the clue.



Reducing 12th degree eqns (12T179) to an 11th degree eqn
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 $\{0,c\}$.


