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The question is: for any finite group, $G$, and any finite set of primes (of $\mathbb{Z}$), $P$, is there a number field $K$, such that there is a regular $G$-Galois extension of $\mathbb{P}^1_K$, and such that $K$ is unramified over all the primes in $P$ as an extension of $\mathbb{Q}$. Supposedly, of course, the answer is yes because conjecturally there is a $G$-Galois extension of $\mathbb{P}^1_{\mathbb{Q}}$. To clarify: by a regular extension, I mean one that descends from a geometric extension (meaning that if you base change to $\mathbb{C}$ you get a cover of the same degree). Basically this means that I don't allow the action of $G$ to come from an extension of scalars. Also: you should notice that when I talk about a $G$-Galois cover of a $K$-curve, I mean that the cover itself is defined over $K$ and that even over $Spec(K)$ it is $G$-Galois (meaning the $G$-action is defined over $K$).

It is a little known fact that for any finite set of primes $P$ and any finite group $G$, there is a number field $K$ unramified over $P$ and having a $G$-Galois extension (in fact the extension constructed will itself be unramified over $P$.) If you follow the proof carefully, you see that it also proves that for any group, $G$, and any such finite set of primes $P$, there is a $G$-Galois extension of some curve $C$, which descends such that it's still Galois to $K$; where $K$ is unramified over the primes in $P$ as an extension of $\mathbb{Q}$. I'm wondering if this can be extended to $C$ being $\mathbb{P}^1_K$

The proof goes like this:

Imbed $G$ in some $S_m$, and embed this $S_m$ in an $S_n$ such that n is coprime with all the primes in $P$. Start with $\mathbb{Q}[X_1, ..., X_n]$, and mod out by the obvious action of $S_n$. You get a $\mathbb{Q}[\sigma_1, ..., \sigma_n]$ (where the $\sigma_i$'s are the elementary symmetric polynomials). Look at the impositions on the $(a_1, ..., a_n)$ in $\mathbb{Z}^n$: $a_i$ is divisible by all the $p \in P$, for $i=1, ..., n-1$, and $a_n \equiv 1$ modulo $\displaystyle\prod_{p \in P} p$. This is gives a $\displaystyle\prod_{p \in P} p$-adically open set, in which we can find an $(a_1, ..., a_n)$ that would make Hilbert irreducibility work. Meaning that over the $\mathbb{Q}$-rational point $(a_1, ..., a_n)$ the fiber is connected. So we get that the fiber is $Spec$ of the splitting field of $t^n-a_1t^{n-1}+...+(-1)^na_n$. Call this extension of $\mathbb{Q}$, $L$.

Now, in the original proof, one just looks at $L^G$, and gets that since $L$ is unramified over $\mathbb{Q}$ (this can be seen by the impositions on the $a_i$'s), then obviously $L^G$ does also.

Let's take a different route. Instead of plugging in all of the $a_i$'s, we can plug in all but one. For example plug in $a_1, ..., a_{n-1}$, and thus get an $S_n$ cover of $\mathbb{A}^1_{\mathbb{Q}}$ (all defined over $\mathbb{Q}$) given by $t^n-a_1t^{n-1}+...+(-1)^nx$ (where $x$ is my new name for $X_n$). Let's think of this as an $S_n$ cover of $\mathbb{P}^1_{\mathbb{Q}}$. We don't know the genus of the cover. Let's call this cover $D$. If you mod out $D$ by $G$: $D \rightarrow D/G$, we get a $G$-Galois cover of $D/G$, and all is defined over $L$. So $D/G$ is the $C$ I was talking about. My question is: can we be clever about our choice of $(a_1, ..., a_n)$ so that $D/G$ would be a $\mathbb{P}^1$?

Of course completely different approaches are also welcome!

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Not sure whether this is of help after such a long time, but anyway:

The answer is yes, and this even works over some number field in which all the primes in $P$ are completely split. Namely, denote by $K^P$ the maximal algebraic extension of $\mathbb{Q}$ in which all $p\in P$ split completely. As noted by Pop on p.3 here: https://www.math.upenn.edu/~pop/Research/files-Res/LF_6Oct2013.pdf , the field $K^P$ is a "large" field, meaning that every irreducible curve over $K^P$ with a smooth $K^P$-point has infinitely many such points (I suppose for this particular field, the "large" property actually follows from some variant of Krasner's lemma). The regular inverse Galois problem is known to have a positive answer over all large fields (due to e.g. Pop, Harbater, Jarden etc.), so there is a regular $G$-Galois extension of $\mathbb{P}^1(K^P)$. But this extension must of course be defined over some number field $F\subset K^P$, giving the answer.

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  • $\begingroup$ What did you have in mind when you wrote "I suppose for this particular field, the "large" property actually follows from some variant of Krasner's lemma"? $\endgroup$ – Arno Fehm May 25 '20 at 9:09
  • $\begingroup$ I guess I just meant Hensel Lifting. Say you have a simple Q_p-point (x0,y0) (in a plane curve model, for simplicity) whose coordinates are also algebraic numbers, then lift those to points (x,y) p-adically close and such that x has a prescribed minimal polynomial equal to the one of x0 mod p. Then x (and thus y) are still algebraic. Weak approximation lets you do this for finitely many p at the same time, yielding K^P-points. $\endgroup$ – Joachim König May 26 '20 at 15:24
  • $\begingroup$ Thanks for the explanation, but I don't think this works, even for one p. This only produces points in $\mathbb{Q}_p\cap\overline{\mathbb{Q}}$, not in $\mathbb{Q}^{\{p\}}$. $\endgroup$ – Arno Fehm May 26 '20 at 15:45
  • $\begingroup$ Fair enough; coming back after a long time I failed to check properly what the field actually was. So thinking about it, I guess I DID mean Krasner's lemma, because you need to ensure that the minimal polynomial in the previous comment splits completely over Q_p. Krasner's lemma guarantees that, if you know that for the starting point (x_0,y_0) and then stay p-adically sufficiently close to it, this property gets preserved. $\endgroup$ – Joachim König May 29 '20 at 2:38
  • $\begingroup$ Could you give more details? Suppose we know that $(0,0)$ is a smooth point on the curve. How do I find even one further point in $\mathbb{Q}^{\{p\}}$? You seem to say that Krasner guarantees points in $\mathbb{Q}^{\{p\}}$ of the form $(p^n,y)$ for large enough $n$. $\endgroup$ – Arno Fehm May 29 '20 at 12:19

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