While perusing Kluener's *Database of Number Fields*, I noticed that a lot of [the discriminants of 7T5](http://reh.math.uni-duesseldorf.de/cgi-klueners/groups3.pl?deg=7&t=5&zer=&bet=&teil=&noteil=&maxteil=&numprime=) came in pairs. After some doodling, I found four families. The first two are,


$$x^7 - x^6 + x^5 + (n - 1)x^4 - (n + 1) x^3 + 5x^2 - 2x - 1 = 0\tag{1}$$


$$y^7 - y^6 + y^5 - (n + 4)y^4 + (n + 2) y^3 + 5y^2 - 2y - 1 = 0\tag{2}$$


These have the same discriminant,


$$D\_1 = (n^2 + 3 n - 29)^2 (27 n^2 + 81 n + 661)^2$$


Eliminating $n$ between (1) and (2), one finds that $x,y$ are related by,


$$-x+x^2-y+x^3y+y^2-x^2y^2+xy^3$$


Another pair is given by,


$$x^7 - x^5 + (n - 1) x^4 - (n + 1)x^3 + 5x^2 - 3x - 1 = 0\tag{3}$$


$$y^7 - y^6 - y^5 + (n + 6)y^4 + (n + 4)y^3 - 7y^2 - 6y - 1 = 0\tag{4}$$


which has discriminant,


$$D\_2 = (27 n^4 + 274 n^3 - 967 n^2 - 6454 n - 26017)^2$$


Eliminating $n$ between (3) and (4) will give a different relation between its $x,y$. These are the easy families to find though, from looking at the database, I have a feeling there are more.


*Question*: Why do a lot of the discriminants/fields of 7T5 come in pairs? Is there anything special about 7T5, or these four families, or is this "pairing" simply an artifact of the search methods/assumptions used by Kluener in creating the database? 



Parametric septic fields $L(7) = L(3,2)$ with the same discriminant
*Every* septic number field $F$ with this Galois group
has a twin septic field $F'$ with the
same Galois closure $K$, the same Dedekind zeta functions
$\zeta\_F$ and $\zeta\_{F'}$, and the same discriminant.


These twins arise in the same way as twin sextic extensions with 
Galois groups $S\_6$ and $A\_6$: each of those groups has two
inequivalent actions on six objects. The 168-element group $G$
likewise has two inequivalent actions on seven objects
(namely the seven points and seven lines of the projective plane of order $2$);
and in this case the two actions have the further property that 
each $g \in G$ acts on the two $7$-element sets with the same
cycle structure (whereas in $S\_6$ and $A\_6$ the cycle structure 
can differ, e.g. the outer automorphism switches
$3$-cycles with double $3$-cycles). So if we let $H$ and $H'$ be
point stabilizers for the two actions (both isomorphic with $S\_4$)
then the fixed subfields $F = K^H$ and $F' = K^{H'}$ are septic fields
with the same zeta functions (the Euler products match up). The equality
of discriminants may be somewhat unpleasant to prove algebraically, 
but fortunately this is not necessary because it follows from the
functional equation!


