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edited Apr 14 2010 at 6:14 |
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!
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edited Apr 14 2010 at 4:09 |
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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edited Apr 14 2010 at 2:31 |
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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edited Apr 14 2010 at 2:23 |
For any finite group G, and any finite set of primes P, does Does there exist a number fieldK, and unramified over a predetermined finite G-Galois cover set of the projective line (all defined over K)primes of Q, such that K the inverse regular problem is unramified over P as an extension of Qcorrect 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 . The 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 (regular) inverse Galois problem meaning that it is equivalent the descent of the cover base changed to there existing, for any finite group $G$, \mathbb{C}$). 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 extension (cover of a $K$-curve, I mean that the cover itself is defined over $\mathbb{Q}$) of K$ and that even over $\mathbb{P}^1_{\mathbb{Q}}$.Spec(K)$ it is $G$-Galois (meaning the $G$-action is defined over $K$).
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edited Apr 14 2010 at 1:36 |
Supposedly, of course, the answer is yes. The (regular) inverse Galois problem is equivalent to there existing, for any finite group $G$, a $G$-Galois extension (defined over $\mathbb{Q}$) of $\mathbb{P}^1_{\mathbb{Q}}$.
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 $P$. 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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asked Apr 14 2010 at 1:26 |
For any finite group G, and any finite set of primes P, does there exist a number field K, and a finite G-Galois cover of the projective line (all defined over K), such that K is unramified over P as an extension of Q?
Supposedly, of course, the answer is yes. The (regular) inverse Galois problem is equivalent to there existing, for any finite group $G$, a $G$-Galois extension (defined over $\mathbb{Q}$) of $\mathbb{P}^1_{\mathbb{Q}}$.
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 $P$. 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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