6 minor clarification

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 $P$, P$and any finite group$G$, there is a$G$-Galois extension of a number field$K$, such that$K$over$\mathbb{Q}$is unramified over$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$? Of course completely different approaches are also welcome! 5 added 10 characters in body The question is: for any finite group,$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}}$. To clarify: by a regular extension, I mean one that descends from a geometric extension (meaning that it is the descent of the cover if you base changed change to$\mathbb{C}$). \mathbb{C}$ 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 $P$, there is a $G$-Galois extension of a number field $K$, such that $K$ over $\mathbb{Q}$ is unramified over $P$ (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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# 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?

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