# Does there exist a number field, unramified over a predetermined finite set of primes of Q, such that the inverse regular Galois problem is correct for that number field?

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!

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.
• 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 $\mathbb{Q}^{\{p\}}$. – Arno Fehm May 26 '20 at 15:45
• 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 $n$. – Arno Fehm May 29 '20 at 12:19