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Generic polynomials, which are recalled below, play an important role in the constructive aspects of the inverse Galois problem.

Definition. Let $P(\mathbf{t},X)$ be a monic polynomial in $\mathbb{Q}(\textbf{t})[X]$ with $\textbf{t} = (t_1,\dots, t_n)$ and $X$ being indeterminates, and let $\mathbb{L}$ be the splitting field of $P(\textbf{t},X)$ over $\mathbb{Q}(\textbf{t})$. Suppose that:

  1. $\mathbb{L}/\mathbb{Q}(\textbf{t})$ is Galois with Galois group $G$, and that
  2. every $L/\mathbb{Q}$ with Galois group $G$ is the splitting field of a polynomial $P(\mathbf{a},X)$ for some $\textbf{a} = (a_1,\dots, a_n) \in \mathbb{Q}^n$.

We say that $P(\textbf{t},X)$ parametrizes $G$-extensions of $\mathbb{Q}$ and $P(\textbf{t},X)$ is a parametric polynomial. The parametric polynomial $P(\textbf{t},X)$ is generic if $P(\textbf{t},X)$ is parametric for $G$-extensions over any field containing $\mathbb{Q}$.

Much of the literature on such polynomials concerns their existence and their constructions.

Question. Is there is a useful criterion for determining whether a monic polynomial $P(\mathbf{t},X)$ in $\mathbb{Q}(\textbf{t})[X]$ is generic or not? In particular, I have encountered the following polynomial in my research $$ g_t(x) := x^3 + 147(t^2 + 13t + 49)x^2 + 147(t^2 + 13t + 49)(33t^2 + 637t + 2401)x + 49(t^2 + 13t + 49)(881t^4 + 38122t^3 + 525819t^2 + 3058874t + 5764801), $$ and I would like to know whether this is a generic polynomial for $\mathbb{Z}/3\mathbb{Z}$-extensions. (For my purposes, it would be enough to know that this is a parametric polynomial for $\mathbb{Z}/3\mathbb{Z}$-extensions.) It is easy to verify condition (1) in the definition using Magma, but I do not know of a way to test condition (2).

Any references and/or suggestions are greatly appreciated!

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    $\begingroup$ If you have not done so already, I would recommend looking at Serre's "Topics in Galois Theory". In Section 1.1, he proves that the polynomial $x^3 - tx^2 + (t - 3)x + 1$ is generic for $\mathbb{Z}/3\mathbb{Z}$. Maybe his proof can be adapted to your case. $\endgroup$ Commented Apr 30, 2020 at 16:40
  • $\begingroup$ @DanielLoughran Thank you for the comment! Yes I have looked at Serre's notes, and the crux of his arguement seems to be that for $x^3 + tx^2 + (t-3)x + 1$, the function $t = (x^3 - 3x + 1)/(x^2 - x)$ is $\mathbb{Z}/3\mathbb{Z}$-invariant for the action of $\mathbb{Z}/3\mathbb{Z}$ on $\mathbb{P}^1$ given by $\sigma x = 1/(1-x)$. I don't see how to adapt his arguement to my setting as the coefficients in my polynomial are non-linear. $\endgroup$ Commented Apr 30, 2020 at 19:33

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I believe there is no good way to determine in general if a polynomial $P(\mathbf{t},X)$ is generic. In fact, given a number field $K$ and a univariate polynomial $P(t,x) \in \mathbb{Q}[t,x]$, the problem of determining whether this is some $t \in \mathbb{Q}$ for which $P(t,x)$ has a root in $K$ is quite hard (and in some cases boils down to determining the $\mathbb{Q}$-points of a curve of genus greater than $1$).

The polynomial that you give is a generic $\mathbb{Z}/3\mathbb{Z}$-extension of $\mathbb{Q}(t)$ however. If you let $f(t,x) = x^{3} - tx^{2} + (t-3)x + 1$, be the example of a generic $\mathbb{Z}/3\mathbb{Z}$ extension given in the comments by Daniel Loughran, one can verify (using Magma for example) that your polynomial and $f(\frac{49}{t} + 8, x)$ define the same degree $3$ extension of $\mathbb{Q}(t)$. This proves the claim, expect possible for the cyclic cubic extension obtained from $f(0,x) = x^{3} - 3x + 1$. But this cubic extension is also obtained from $f(-20,x)$.

EDIT: The OP asked for some info about how I found this. I did a fair amount of stumbling around to come up with $f(49/t + 8,x)$. First, I noticed that the discriminant of the maximal order of $\mathbb{Q}(t)[x]/(g_{t}(x))$ was $(t^{2} + 13t + 49)^{2}$, while the discriminant of the maximal order of $\mathbb{Q}(t)[x]/(f(t,x))$ was $(t^{2} - 3t + 9)^{2}$, and replacing $t$ with $t+8$ gives two polynomials with discriminants that are in the same square class. However, these do not define isomorphic extensions - specializing for lots of values of $t$ reveals that in most cases the discriminant of the ring of integers for $g_{t}(x)$ and that for $f(t+8,x)$ differ by a factor of $49$. This suggests that for a fixed $t$, if $K_{1}$ is the number field defined by $f(t+8,x)$ and $K_{2}$ is that defined by $g_{t}(x)$, then $K_{2} \subseteq K_{1}(e^{2 \pi i / 7} + e^{-2 \pi i /7})$ [note that $\mathbb{Q}(e^{2 \pi i /7} + e^{-2 \pi i / 7})$ is the unique cyclic extension with discriminant $49$]. One can verify that this is true. I suspected that there was a linear fractional transformation in $t$ that would take $f(t+8,x)$ to a polynomial defining one of the other cubic subfields of $K_{1}(e^{2 \pi i / 7} + e^{-2 \pi i /7})$. Looking at several specific cubic fields and searching for the values of $t$ (there are many) that result in $f(t+8,x)$ and $g_{t}(x)$ defining them leads to the suggestion that $t \to 49/t$ would work.

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  • $\begingroup$ Thank you very much for your answer! Could you say a word about how you found that my above polynomial and $f(49/t + 8,x)$ define the same degree 3 extension of $\mathbb{Q}(t)$? $\endgroup$ Commented May 1, 2020 at 0:38

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