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 $F(13)=12\times13=156$.

So what is the reason they now have solvable groups?

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.

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?

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?

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.

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?

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 $n$. 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 $F(13)=12\times13=156$.

So what is the reason they now have solvable groups?

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.

III. Question

So why is it that 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?

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 $F(13)=12\times13=156$.

So what is the reason they now have solvable groups?

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.

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?

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,

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

which, for random $j$, 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 $n$. 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 $F(13)=12\times13=156$.

So what is the reason they now have solvable groups?

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,

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

which is $\text{9T32}$ and has order $9\times12\times14 = 1512$ (unsolvable) while,

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

which 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).

III. Question

So why is it that 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?

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 $n$. 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 $F(13)=12\times13=156$.

So what is the reason they now have solvable groups?

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.

III. Question

So why is it that 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?