On Elkies' $\text{9T32}$ nonic and a shared property with j-function formulas

Question

I. First Set

Before going to Elkies' nonic, we start with something a bit simpler. There is a list of j-function formulas in this MSE post. For example, for prime levels $p = 5,7,13,$ we have,

$$j=\frac{(x^2+10x+5)^3}x$$ $$j=\frac{(x^2+5x+1)^3(x^2+13x+49)}x$$ $$j=\frac{(x^4+7x^3+20x^2+19x+1)^3(x^2+5x+13)}x$$

Assume $j$ as any number, hence a free parameter. Then these equations have groups $\text{PGL}(2,5), \text{PGL}(2,7), \text{PGL}(2,13),$ respectively, so generally is not solvable in radicals. However, $x$ is solvable if $j$ is the $j$-function OR if we do this,

$$\frac{(x^2+10x+5)^3}x =\frac{(n^2+10n+5)^3}n$$

for any non-zero $n$. For example, let $n=2$,

$$\frac{(x^2+10x+5)^3}x =\frac{29^3}2$$

Its irreducible $5$th-deg factor has Frobenius group $F(5)=4\times5=20$ hence is now solvable. Likewise for its higher siblings which have $F(7)=6\times7=42$ and $F(13)=12\times13=156$.

Edit (May 3): For an example where the result is non-solvable, let

$$j = \frac{(x + 432)(x^2 + 80x - 3888)^3}{7^7 x^3}$$

which has familiar discriminant (without its numerical factors) as $D = -(j-1728)^3\, j^4$. Then,

$$\frac{(x + 432)(x^2 + 80x - 3888)^3}{x^3} = \frac{(n + 432)(n^2 + 80n - 3888)^3}{n^3}$$

does not yield a solvable sextic factor.

Q: So the procedure doesn't work on just any rational function. What condition should be satisfied such that the procedure yields a solvable factor?

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II. Second Set

Searching the literature for similar "formulas", the paper Galois Number Fields with Small Root Discriminant by J. Jones and D. Roberts was a gold mine. They had degrees for $p \leq 10$ and even for $p=17$ which I mentioned in an old MO post. But this equating of,

$$P(j,x) = P(j,n)$$

was not there. A small sample for deg 9,

$$j = \frac{(x^3 + 4x^2 + 10x + 6)^3}{(4x^2 + 13x + 32)}$$

Again, assume $j$ as a free parameter. This equation in Magma notation is $\text{9T32}$ and has order $9\times12\times14 = 1512$ (unsolvable). However,

$$\frac{(x^3 + 4x^2 + 10x + 6)^3}{(4x^2 + 13x + 32)} = \frac{(n^3 + 4n^2 + 10n + 6)^3}{(4n^2 + 13n + 32)}$$

(after removing the linear factor) is $\text{8T36}$ and has order $12\times14 = 168$ (solvable). Elkies' nonic is also $\text{9T32}$ but a bit trickier,

$$j^2+(9x^4 - 42x^3 - 675x^2 - 1485x - 441)j - (x^3 - 9x^2 - 69x - 123)^3 =0$$ $$j^2+(9n^4 - 42n^3 - 675n^2 - 1485n - 441)j - (n^3 - 9n^2 - 69n - 123)^3 =0$$

Eliminating $j$ between the two using resultants and getting rid of two linear factors yields a $16$-deg equation with order $12\times14\times16 = 2688$ (and solvable according to Magma).

Note: For consistency, I still used the variable $j$ for the second set, but they are not to be understood as level-9 formulas for the j-function.

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III. Question

So why is it for these two sets (and some others) that eliminating the common variable $j$ between $P(j,x) = P(j,n) = 0$ suddenly yields a parametric equation with a solvable Galois group?

Answer 1

This has relatively little to do with the j-invariant itself. If you take any rational function $f(x)\in k(x)$ and let $G:=Gal(f(x) - t/k(t))$ (also referred to as the monodromy group of $f$), then by elementary Galois theory, $Gal(f(x)-f(y) / k(y))$ is a point stabilizer in $G$ (in the usual action on the roots). In your cases $G$ is $P\Gamma L_2(q)$ acting on $q+1$ points ($q=5,7,8,13$) which has solvable point stabilizer.

(Edit: Of course the reason why this observation "wasn't there" in the linked Jones/Roberts paper is that this is comparatively the "trivial case", working for every $f$. More interesting are the (exceptional) cases where a group is a monodromy group of two essentially different rational functions in the same Galois closure, which then gives shrinking of $Gal(f(x)-g(y)/k(y))$ in the analogous way - the sextic above together with the right quintic (relates to the $j$-invariant) is one such pair.)