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In valued fields (of rank > 1), is the p-adic completion of the algebraic closure still algebraically closed?
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Let $K$ be a valued field and assume its valuation ring $mathcalO_K$ is $p$-adically separated. If we assume further that $K$ is algebraically closed and we take the $p$-adic completion, is the result algebraically closed?
I know many places where this is asserted for rank 1 valuations, and as far as I can tell the "usual" proof works basically the same way in the more general case. For example, I think the proof in these notes by Brian Conrad http://math.stanford.edu/~conrad/248APage/handouts/algclosurecomp.pdf works the same, as long as you replace the absolute values by valuations and replace things like $frac1j$ with something like $v(p^j)$. I've looked briefly through Bourbaki's Commutative Algebra and Zariski-Samuel's book and haven't seen this fact; those references seem so comprehensive I'm nervous that the result is not true and I am missing something.
Thanks very much for any help! Best would be a reference I can use, but I'd also be happy with someone reassuring me this result I want is correct. I'd also be happy to know if there is a more natural way to phrase it than how I've phrased it (in terms of a valuation ring being $p$-adically separated).
Here are a few additional comments. I think I have written up a proof of algebraic closedness, but I would still be reassured if someone can tell me this is a well-known fact.
Thanks @Lubin for your comment! I don't know what the most standard example of a valuation ring of rank > 1 is. The example I feel like I understand the best is the following. Let $k$ be a perfect valued field of characteristic $p$ and rank 1. (I picture something like $k = mathbbF_p((t^1/p^infty))$ with the $t$-adic valuation.) Then take the $p$-typical Witt vectors $W(k)$ and let $K = W(k)[1/p]$. I believe then you can put a rank 2 valuation on this field with value group $mathbbZ times mathbbR$ with the lexicographic ordering, where the first entry corresponds to the $p$-adic valuation and where the second entry corresponds to the $t$-adic valuation. The corresponding valuation ring is similar to an example in Remark 4.6 from these notes from a course taught by Bhargav Bhatt www-personal.umich.edu/~stevmatt/perfectoid2.pdf It includes elements like $p(1/[t])$ and $[t]$ but not $1/[t]$. I'm afraid I don't have an algebraically closed example if that's what you wanted... the best I could say would be take the algebraic closure of the example I just gave.
Because I said the valuation ring was $p$-adically separated, I'm pretty sure the $p$-adic completion is the same as the "usual" completion of a valued field... Explicitly I'd want the completion to consist of $p$-adic Cauchy sequences $(x_1, x_2, x_3, ldots)$, where for every $j > 0$, there exists an $N > 0$ such that if $n_1, n_2 > N$, we have $x_n_1 - x_n_2 in p^j mathcalO_K$.
abstract-algebra valuation-theory
asked Jul 21 at 15:14
CJD
23918
add a comment |Â
up vote
4
down vote
favorite
Let $K$ be a valued field and assume its valuation ring $mathcalO_K$ is $p$-adically separated. If we assume further that $K$ is algebraically closed and we take the $p$-adic completion, is the result algebraically closed?
I know many places where this is asserted for rank 1 valuations, and as far as I can tell the "usual" proof works basically the same way in the more general case. For example, I think the proof in these notes by Brian Conrad http://math.stanford.edu/~conrad/248APage/handouts/algclosurecomp.pdf works the same, as long as you replace the absolute values by valuations and replace things like $frac1j$ with something like $v(p^j)$. I've looked briefly through Bourbaki's Commutative Algebra and Zariski-Samuel's book and haven't seen this fact; those references seem so comprehensive I'm nervous that the result is not true and I am missing something.
Thanks very much for any help! Best would be a reference I can use, but I'd also be happy with someone reassuring me this result I want is correct. I'd also be happy to know if there is a more natural way to phrase it than how I've phrased it (in terms of a valuation ring being $p$-adically separated).
Here are a few additional comments. I think I have written up a proof of algebraic closedness, but I would still be reassured if someone can tell me this is a well-known fact.
Thanks @Lubin for your comment! I don't know what the most standard example of a valuation ring of rank > 1 is. The example I feel like I understand the best is the following. Let $k$ be a perfect valued field of characteristic $p$ and rank 1. (I picture something like $k = mathbbF_p((t^1/p^infty))$ with the $t$-adic valuation.) Then take the $p$-typical Witt vectors $W(k)$ and let $K = W(k)[1/p]$. I believe then you can put a rank 2 valuation on this field with value group $mathbbZ times mathbbR$ with the lexicographic ordering, where the first entry corresponds to the $p$-adic valuation and where the second entry corresponds to the $t$-adic valuation. The corresponding valuation ring is similar to an example in Remark 4.6 from these notes from a course taught by Bhargav Bhatt www-personal.umich.edu/~stevmatt/perfectoid2.pdf It includes elements like $p(1/[t])$ and $[t]$ but not $1/[t]$. I'm afraid I don't have an algebraically closed example if that's what you wanted... the best I could say would be take the algebraic closure of the example I just gave.
Because I said the valuation ring was $p$-adically separated, I'm pretty sure the $p$-adic completion is the same as the "usual" completion of a valued field... Explicitly I'd want the completion to consist of $p$-adic Cauchy sequences $(x_1, x_2, x_3, ldots)$, where for every $j > 0$, there exists an $N > 0$ such that if $n_1, n_2 > N$, we have $x_n_1 - x_n_2 in p^j mathcalO_K$.
abstract-algebra valuation-theory
asked Jul 21 at 15:14
CJD
23918
1
It’s been a while since I’ve thought about valuations of rank $>1$. Can you give us an example of one such field? And your mention of “$p$-adic completion†in this context disquiets me somewhat. Could you say exactly what that means (say, in opposition to completion with respect to the already-existing topological or uniform structure)?
– Lubin
Jul 22 at 2:43
Thanks for your further explanation. As you can see from my questions, I’m not likely to have the technical skills to help you at all. But I may ask for further clarification(s).
– Lubin
Jul 22 at 13:23
But do you really want the $p$-adic topology? Seems to me that the sequence $t^n$ is not $p$-adically convergent. Or do you care about that?
– Lubin
Jul 22 at 18:45
@Lubin I'm definitely not an expert on these rings, but my impression is that in my example, $ [t^n] $ should not be considered a Cauchy sequence for the valuation I described. I also don't think the valuation ring I described is $[t]$-adically separated... for example, I think $p$ is divisible by $[t]^n$ for all $n$.
– CJD
Jul 23 at 11:59
Yes, this only confirms that I’m totally confused, and unfortunately have nothing to contribute to the subject. Good luck, your question is definitely interesting.
– Lubin
Jul 23 at 14:01
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $K$ be a valued field and assume its valuation ring $mathcalO_K$ is $p$-adically separated. If we assume further that $K$ is algebraically closed and we take the $p$-adic completion, is the result algebraically closed?
I know many places where this is asserted for rank 1 valuations, and as far as I can tell the "usual" proof works basically the same way in the more general case. For example, I think the proof in these notes by Brian Conrad http://math.stanford.edu/~conrad/248APage/handouts/algclosurecomp.pdf works the same, as long as you replace the absolute values by valuations and replace things like $frac1j$ with something like $v(p^j)$. I've looked briefly through Bourbaki's Commutative Algebra and Zariski-Samuel's book and haven't seen this fact; those references seem so comprehensive I'm nervous that the result is not true and I am missing something.
Thanks very much for any help! Best would be a reference I can use, but I'd also be happy with someone reassuring me this result I want is correct. I'd also be happy to know if there is a more natural way to phrase it than how I've phrased it (in terms of a valuation ring being $p$-adically separated).
Here are a few additional comments. I think I have written up a proof of algebraic closedness, but I would still be reassured if someone can tell me this is a well-known fact.
Thanks @Lubin for your comment! I don't know what the most standard example of a valuation ring of rank > 1 is. The example I feel like I understand the best is the following. Let $k$ be a perfect valued field of characteristic $p$ and rank 1. (I picture something like $k = mathbbF_p((t^1/p^infty))$ with the $t$-adic valuation.) Then take the $p$-typical Witt vectors $W(k)$ and let $K = W(k)[1/p]$. I believe then you can put a rank 2 valuation on this field with value group $mathbbZ times mathbbR$ with the lexicographic ordering, where the first entry corresponds to the $p$-adic valuation and where the second entry corresponds to the $t$-adic valuation. The corresponding valuation ring is similar to an example in Remark 4.6 from these notes from a course taught by Bhargav Bhatt www-personal.umich.edu/~stevmatt/perfectoid2.pdf It includes elements like $p(1/[t])$ and $[t]$ but not $1/[t]$. I'm afraid I don't have an algebraically closed example if that's what you wanted... the best I could say would be take the algebraic closure of the example I just gave.
Because I said the valuation ring was $p$-adically separated, I'm pretty sure the $p$-adic completion is the same as the "usual" completion of a valued field... Explicitly I'd want the completion to consist of $p$-adic Cauchy sequences $(x_1, x_2, x_3, ldots)$, where for every $j > 0$, there exists an $N > 0$ such that if $n_1, n_2 > N$, we have $x_n_1 - x_n_2 in p^j mathcalO_K$.
abstract-algebra valuation-theory
asked Jul 21 at 15:14
CJD
23918
Let $K$ be a valued field and assume its valuation ring $mathcalO_K$ is $p$-adically separated. If we assume further that $K$ is algebraically closed and we take the $p$-adic completion, is the result algebraically closed?
I know many places where this is asserted for rank 1 valuations, and as far as I can tell the "usual" proof works basically the same way in the more general case. For example, I think the proof in these notes by Brian Conrad http://math.stanford.edu/~conrad/248APage/handouts/algclosurecomp.pdf works the same, as long as you replace the absolute values by valuations and replace things like $frac1j$ with something like $v(p^j)$. I've looked briefly through Bourbaki's Commutative Algebra and Zariski-Samuel's book and haven't seen this fact; those references seem so comprehensive I'm nervous that the result is not true and I am missing something.
Thanks very much for any help! Best would be a reference I can use, but I'd also be happy with someone reassuring me this result I want is correct. I'd also be happy to know if there is a more natural way to phrase it than how I've phrased it (in terms of a valuation ring being $p$-adically separated).
Here are a few additional comments. I think I have written up a proof of algebraic closedness, but I would still be reassured if someone can tell me this is a well-known fact.
Thanks @Lubin for your comment! I don't know what the most standard example of a valuation ring of rank > 1 is. The example I feel like I understand the best is the following. Let $k$ be a perfect valued field of characteristic $p$ and rank 1. (I picture something like $k = mathbbF_p((t^1/p^infty))$ with the $t$-adic valuation.) Then take the $p$-typical Witt vectors $W(k)$ and let $K = W(k)[1/p]$. I believe then you can put a rank 2 valuation on this field with value group $mathbbZ times mathbbR$ with the lexicographic ordering, where the first entry corresponds to the $p$-adic valuation and where the second entry corresponds to the $t$-adic valuation. The corresponding valuation ring is similar to an example in Remark 4.6 from these notes from a course taught by Bhargav Bhatt www-personal.umich.edu/~stevmatt/perfectoid2.pdf It includes elements like $p(1/[t])$ and $[t]$ but not $1/[t]$. I'm afraid I don't have an algebraically closed example if that's what you wanted... the best I could say would be take the algebraic closure of the example I just gave.
Because I said the valuation ring was $p$-adically separated, I'm pretty sure the $p$-adic completion is the same as the "usual" completion of a valued field... Explicitly I'd want the completion to consist of $p$-adic Cauchy sequences $(x_1, x_2, x_3, ldots)$, where for every $j > 0$, there exists an $N > 0$ such that if $n_1, n_2 > N$, we have $x_n_1 - x_n_2 in p^j mathcalO_K$.
abstract-algebra valuation-theory
asked Jul 21 at 15:14
CJD
23918
edited Jul 22 at 12:41
asked Jul 21 at 15:14
CJD
23918
asked Jul 21 at 15:14
CJD
23918
asked Jul 21 at 15:14
CJD
23918
23918
1
It’s been a while since I’ve thought about valuations of rank $>1$. Can you give us an example of one such field? And your mention of “$p$-adic completion†in this context disquiets me somewhat. Could you say exactly what that means (say, in opposition to completion with respect to the already-existing topological or uniform structure)?
– Lubin
Jul 22 at 2:43
Thanks for your further explanation. As you can see from my questions, I’m not likely to have the technical skills to help you at all. But I may ask for further clarification(s).
– Lubin
Jul 22 at 13:23
But do you really want the $p$-adic topology? Seems to me that the sequence $t^n$ is not $p$-adically convergent. Or do you care about that?
– Lubin
Jul 22 at 18:45
@Lubin I'm definitely not an expert on these rings, but my impression is that in my example, $ [t^n] $ should not be considered a Cauchy sequence for the valuation I described. I also don't think the valuation ring I described is $[t]$-adically separated... for example, I think $p$ is divisible by $[t]^n$ for all $n$.
– CJD
Jul 23 at 11:59
Yes, this only confirms that I’m totally confused, and unfortunately have nothing to contribute to the subject. Good luck, your question is definitely interesting.
– Lubin
Jul 23 at 14:01
add a comment |Â
1
It’s been a while since I’ve thought about valuations of rank $>1$. Can you give us an example of one such field? And your mention of “$p$-adic completion†in this context disquiets me somewhat. Could you say exactly what that means (say, in opposition to completion with respect to the already-existing topological or uniform structure)?
– Lubin
Jul 22 at 2:43
Thanks for your further explanation. As you can see from my questions, I’m not likely to have the technical skills to help you at all. But I may ask for further clarification(s).
– Lubin
Jul 22 at 13:23
But do you really want the $p$-adic topology? Seems to me that the sequence $t^n$ is not $p$-adically convergent. Or do you care about that?
– Lubin
Jul 22 at 18:45
@Lubin I'm definitely not an expert on these rings, but my impression is that in my example, $ [t^n] $ should not be considered a Cauchy sequence for the valuation I described. I also don't think the valuation ring I described is $[t]$-adically separated... for example, I think $p$ is divisible by $[t]^n$ for all $n$.
– CJD
Jul 23 at 11:59
Yes, this only confirms that I’m totally confused, and unfortunately have nothing to contribute to the subject. Good luck, your question is definitely interesting.
– Lubin
Jul 23 at 14:01
1
1
It’s been a while since I’ve thought about valuations of rank $>1$. Can you give us an example of one such field? And your mention of “$p$-adic completion†in this context disquiets me somewhat. Could you say exactly what that means (say, in opposition to completion with respect to the already-existing topological or uniform structure)?
– Lubin
Jul 22 at 2:43
It’s been a while since I’ve thought about valuations of rank $>1$. Can you give us an example of one such field? And your mention of “$p$-adic completion†in this context disquiets me somewhat. Could you say exactly what that means (say, in opposition to completion with respect to the already-existing topological or uniform structure)?
– Lubin
Jul 22 at 2:43
Thanks for your further explanation. As you can see from my questions, I’m not likely to have the technical skills to help you at all. But I may ask for further clarification(s).
– Lubin
Jul 22 at 13:23
Thanks for your further explanation. As you can see from my questions, I’m not likely to have the technical skills to help you at all. But I may ask for further clarification(s).
– Lubin
Jul 22 at 13:23
But do you really want the $p$-adic topology? Seems to me that the sequence $t^n$ is not $p$-adically convergent. Or do you care about that?
– Lubin
Jul 22 at 18:45
But do you really want the $p$-adic topology? Seems to me that the sequence $t^n$ is not $p$-adically convergent. Or do you care about that?
– Lubin
Jul 22 at 18:45
@Lubin I'm definitely not an expert on these rings, but my impression is that in my example, $ [t^n] $ should not be considered a Cauchy sequence for the valuation I described. I also don't think the valuation ring I described is $[t]$-adically separated... for example, I think $p$ is divisible by $[t]^n$ for all $n$.
– CJD
Jul 23 at 11:59
@Lubin I'm definitely not an expert on these rings, but my impression is that in my example, $ [t^n] $ should not be considered a Cauchy sequence for the valuation I described. I also don't think the valuation ring I described is $[t]$-adically separated... for example, I think $p$ is divisible by $[t]^n$ for all $n$.
– CJD
Jul 23 at 11:59
Yes, this only confirms that I’m totally confused, and unfortunately have nothing to contribute to the subject. Good luck, your question is definitely interesting.
– Lubin
Jul 23 at 14:01
Yes, this only confirms that I’m totally confused, and unfortunately have nothing to contribute to the subject. Good luck, your question is definitely interesting.
– Lubin
Jul 23 at 14:01
add a comment |Â
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1
It’s been a while since I’ve thought about valuations of rank $>1$. Can you give us an example of one such field? And your mention of “$p$-adic completion†in this context disquiets me somewhat. Could you say exactly what that means (say, in opposition to completion with respect to the already-existing topological or uniform structure)?
– Lubin
Jul 22 at 2:43
Thanks for your further explanation. As you can see from my questions, I’m not likely to have the technical skills to help you at all. But I may ask for further clarification(s).
– Lubin
Jul 22 at 13:23
But do you really want the $p$-adic topology? Seems to me that the sequence $t^n$ is not $p$-adically convergent. Or do you care about that?
– Lubin
Jul 22 at 18:45
@Lubin I'm definitely not an expert on these rings, but my impression is that in my example, $ [t^n] $ should not be considered a Cauchy sequence for the valuation I described. I also don't think the valuation ring I described is $[t]$-adically separated... for example, I think $p$ is divisible by $[t]^n$ for all $n$.
– CJD
Jul 23 at 11:59
Yes, this only confirms that I’m totally confused, and unfortunately have nothing to contribute to the subject. Good luck, your question is definitely interesting.
– Lubin
Jul 23 at 14:01