Mixing
Ring of integers of $mathbb{C}_p$
$begingroup$
Sorry, maybe this is a really stupid question. Let $mathbb{C}_p$ be the completion of the algebraic closure $overline{mathbb{Q}_p}$ of the field of $p$-adic numbers. We know that there exists a way to extend the $p$-adic valuation to $mathbb{C}_p$, and in this way it is possible to define the ring of integers $mathcal{O}_{mathbb{C}_p}$ given by elements of valuation greater or equal than $0$. It is also possible to define the maximal ideal of this valuation ring given by elements of valuation strictly greater than $0$. Now, is this ideal finitely generated?
commutative-algebra p-adic-number-theory
asked Dec 28 '18 at 17:08
Zariski93Zariski93
876
$endgroup$
add a comment |
$begingroup$
Sorry, maybe this is a really stupid question. Let $mathbb{C}_p$ be the completion of the algebraic closure $overline{mathbb{Q}_p}$ of the field of $p$-adic numbers. We know that there exists a way to extend the $p$-adic valuation to $mathbb{C}_p$, and in this way it is possible to define the ring of integers $mathcal{O}_{mathbb{C}_p}$ given by elements of valuation greater or equal than $0$. It is also possible to define the maximal ideal of this valuation ring given by elements of valuation strictly greater than $0$. Now, is this ideal finitely generated?
commutative-algebra p-adic-number-theory
asked Dec 28 '18 at 17:08
Zariski93Zariski93
876
$endgroup$
5
$begingroup$
Hint: elements of the maximal ideal can have valuations arbitrarily close to $0$.
$endgroup$
– Wojowu
Dec 28 '18 at 17:12
$begingroup$
Let $K^{unr} = mathbf{Q}_p(zeta_{p^infty-1})$ the maximal unramified extension of $mathbf{Q}_p$ (so that $(p)$ is the unique maximal ideal of both rings of integers). Since $O_{K^{unr}}/(p)$ is algebraically closed then for any finite extension $L/K$ of degree $n$ you'll have $|pi_L|_L= |p|_L^{1/n}$. Thus only finite extensions of $K$ are local fields and any infinite extension of $K$ will be dense in $mathbb{C}_p$.
$endgroup$
– reuns
Dec 28 '18 at 18:47
1
$begingroup$
@reuns, I do not believe that any infinite extension of $K=K^{unr}$ is dense in $Bbb C_p$. What about the maximal tame extension of $K$?
$endgroup$
– Lubin
Dec 29 '18 at 22:21
$begingroup$
@Lubin Thank you. You are saying that $p nmid n = [K^{unr}(a):K^{unr}]$ implies that $|a|= p^{l/n}$ so $|a| ne |p^{1/p}|$ and $|p^{1/p}-a| ge |p^{1/p}|$ ? Whence with $L$ the maximal tame extension of $K^{unr}$ then $inf_{a in L} |p^{1/p} - a| = |p^{1/p}|$ and my last sentence was a big mistake and $mathbb{C}_p$ is very different to $mathbb{C}$ in this regard.
$endgroup$
– reuns
Dec 30 '18 at 17:11
1
$begingroup$
Right you are. I’m pretty sure that even if your valuation group was all of $Bbb Q$, that would not be enough to guarantee that the completion was all of $Bbb C$. Wild ramification is truly a wild phenomenon!
$endgroup$
– Lubin
Dec 30 '18 at 18:15
add a comment |
$begingroup$
Sorry, maybe this is a really stupid question. Let $mathbb{C}_p$ be the completion of the algebraic closure $overline{mathbb{Q}_p}$ of the field of $p$-adic numbers. We know that there exists a way to extend the $p$-adic valuation to $mathbb{C}_p$, and in this way it is possible to define the ring of integers $mathcal{O}_{mathbb{C}_p}$ given by elements of valuation greater or equal than $0$. It is also possible to define the maximal ideal of this valuation ring given by elements of valuation strictly greater than $0$. Now, is this ideal finitely generated?
commutative-algebra p-adic-number-theory
asked Dec 28 '18 at 17:08
Zariski93Zariski93
876
$endgroup$
Sorry, maybe this is a really stupid question. Let $mathbb{C}_p$ be the completion of the algebraic closure $overline{mathbb{Q}_p}$ of the field of $p$-adic numbers. We know that there exists a way to extend the $p$-adic valuation to $mathbb{C}_p$, and in this way it is possible to define the ring of integers $mathcal{O}_{mathbb{C}_p}$ given by elements of valuation greater or equal than $0$. It is also possible to define the maximal ideal of this valuation ring given by elements of valuation strictly greater than $0$. Now, is this ideal finitely generated?
commutative-algebra p-adic-number-theory
commutative-algebra p-adic-number-theory
asked Dec 28 '18 at 17:08
Zariski93Zariski93
876
asked Dec 28 '18 at 17:08
Zariski93Zariski93
876
asked Dec 28 '18 at 17:08
Zariski93Zariski93
876
asked Dec 28 '18 at 17:08
Zariski93Zariski93
876
asked Dec 28 '18 at 17:08
Zariski93Zariski93
876
876
5
$begingroup$
Hint: elements of the maximal ideal can have valuations arbitrarily close to $0$.
$endgroup$
– Wojowu
Dec 28 '18 at 17:12
$begingroup$
Let $K^{unr} = mathbf{Q}_p(zeta_{p^infty-1})$ the maximal unramified extension of $mathbf{Q}_p$ (so that $(p)$ is the unique maximal ideal of both rings of integers). Since $O_{K^{unr}}/(p)$ is algebraically closed then for any finite extension $L/K$ of degree $n$ you'll have $|pi_L|_L= |p|_L^{1/n}$. Thus only finite extensions of $K$ are local fields and any infinite extension of $K$ will be dense in $mathbb{C}_p$.
$endgroup$
– reuns
Dec 28 '18 at 18:47
1
$begingroup$
@reuns, I do not believe that any infinite extension of $K=K^{unr}$ is dense in $Bbb C_p$. What about the maximal tame extension of $K$?
$endgroup$
– Lubin
Dec 29 '18 at 22:21
$begingroup$
@Lubin Thank you. You are saying that $p nmid n = [K^{unr}(a):K^{unr}]$ implies that $|a|= p^{l/n}$ so $|a| ne |p^{1/p}|$ and $|p^{1/p}-a| ge |p^{1/p}|$ ? Whence with $L$ the maximal tame extension of $K^{unr}$ then $inf_{a in L} |p^{1/p} - a| = |p^{1/p}|$ and my last sentence was a big mistake and $mathbb{C}_p$ is very different to $mathbb{C}$ in this regard.
$endgroup$
– reuns
Dec 30 '18 at 17:11
1
$begingroup$
Right you are. I’m pretty sure that even if your valuation group was all of $Bbb Q$, that would not be enough to guarantee that the completion was all of $Bbb C$. Wild ramification is truly a wild phenomenon!
$endgroup$
– Lubin
Dec 30 '18 at 18:15
add a comment |
5
$begingroup$
Hint: elements of the maximal ideal can have valuations arbitrarily close to $0$.
$endgroup$
– Wojowu
Dec 28 '18 at 17:12
$begingroup$
Let $K^{unr} = mathbf{Q}_p(zeta_{p^infty-1})$ the maximal unramified extension of $mathbf{Q}_p$ (so that $(p)$ is the unique maximal ideal of both rings of integers). Since $O_{K^{unr}}/(p)$ is algebraically closed then for any finite extension $L/K$ of degree $n$ you'll have $|pi_L|_L= |p|_L^{1/n}$. Thus only finite extensions of $K$ are local fields and any infinite extension of $K$ will be dense in $mathbb{C}_p$.
$endgroup$
– reuns
Dec 28 '18 at 18:47
1
$begingroup$
@reuns, I do not believe that any infinite extension of $K=K^{unr}$ is dense in $Bbb C_p$. What about the maximal tame extension of $K$?
$endgroup$
– Lubin
Dec 29 '18 at 22:21
$begingroup$
@Lubin Thank you. You are saying that $p nmid n = [K^{unr}(a):K^{unr}]$ implies that $|a|= p^{l/n}$ so $|a| ne |p^{1/p}|$ and $|p^{1/p}-a| ge |p^{1/p}|$ ? Whence with $L$ the maximal tame extension of $K^{unr}$ then $inf_{a in L} |p^{1/p} - a| = |p^{1/p}|$ and my last sentence was a big mistake and $mathbb{C}_p$ is very different to $mathbb{C}$ in this regard.
$endgroup$
– reuns
Dec 30 '18 at 17:11
1
$begingroup$
Right you are. I’m pretty sure that even if your valuation group was all of $Bbb Q$, that would not be enough to guarantee that the completion was all of $Bbb C$. Wild ramification is truly a wild phenomenon!
$endgroup$
– Lubin
Dec 30 '18 at 18:15
5
5
$begingroup$
Hint: elements of the maximal ideal can have valuations arbitrarily close to $0$.
$endgroup$
– Wojowu
Dec 28 '18 at 17:12
$begingroup$
Hint: elements of the maximal ideal can have valuations arbitrarily close to $0$.
$endgroup$
– Wojowu
Dec 28 '18 at 17:12
$begingroup$
Let $K^{unr} = mathbf{Q}_p(zeta_{p^infty-1})$ the maximal unramified extension of $mathbf{Q}_p$ (so that $(p)$ is the unique maximal ideal of both rings of integers). Since $O_{K^{unr}}/(p)$ is algebraically closed then for any finite extension $L/K$ of degree $n$ you'll have $|pi_L|_L= |p|_L^{1/n}$. Thus only finite extensions of $K$ are local fields and any infinite extension of $K$ will be dense in $mathbb{C}_p$.
$endgroup$
– reuns
Dec 28 '18 at 18:47
$begingroup$
Let $K^{unr} = mathbf{Q}_p(zeta_{p^infty-1})$ the maximal unramified extension of $mathbf{Q}_p$ (so that $(p)$ is the unique maximal ideal of both rings of integers). Since $O_{K^{unr}}/(p)$ is algebraically closed then for any finite extension $L/K$ of degree $n$ you'll have $|pi_L|_L= |p|_L^{1/n}$. Thus only finite extensions of $K$ are local fields and any infinite extension of $K$ will be dense in $mathbb{C}_p$.
$endgroup$
– reuns
Dec 28 '18 at 18:47
1
1
$begingroup$
@reuns, I do not believe that any infinite extension of $K=K^{unr}$ is dense in $Bbb C_p$. What about the maximal tame extension of $K$?
$endgroup$
– Lubin
Dec 29 '18 at 22:21
$begingroup$
@reuns, I do not believe that any infinite extension of $K=K^{unr}$ is dense in $Bbb C_p$. What about the maximal tame extension of $K$?
$endgroup$
– Lubin
Dec 29 '18 at 22:21
$begingroup$
@Lubin Thank you. You are saying that $p nmid n = [K^{unr}(a):K^{unr}]$ implies that $|a|= p^{l/n}$ so $|a| ne |p^{1/p}|$ and $|p^{1/p}-a| ge |p^{1/p}|$ ? Whence with $L$ the maximal tame extension of $K^{unr}$ then $inf_{a in L} |p^{1/p} - a| = |p^{1/p}|$ and my last sentence was a big mistake and $mathbb{C}_p$ is very different to $mathbb{C}$ in this regard.
$endgroup$
– reuns
Dec 30 '18 at 17:11
$begingroup$
@Lubin Thank you. You are saying that $p nmid n = [K^{unr}(a):K^{unr}]$ implies that $|a|= p^{l/n}$ so $|a| ne |p^{1/p}|$ and $|p^{1/p}-a| ge |p^{1/p}|$ ? Whence with $L$ the maximal tame extension of $K^{unr}$ then $inf_{a in L} |p^{1/p} - a| = |p^{1/p}|$ and my last sentence was a big mistake and $mathbb{C}_p$ is very different to $mathbb{C}$ in this regard.
$endgroup$
– reuns
Dec 30 '18 at 17:11
1
1
$begingroup$
Right you are. I’m pretty sure that even if your valuation group was all of $Bbb Q$, that would not be enough to guarantee that the completion was all of $Bbb C$. Wild ramification is truly a wild phenomenon!
$endgroup$
– Lubin
Dec 30 '18 at 18:15
$begingroup$
Right you are. I’m pretty sure that even if your valuation group was all of $Bbb Q$, that would not be enough to guarantee that the completion was all of $Bbb C$. Wild ramification is truly a wild phenomenon!
$endgroup$
– Lubin
Dec 30 '18 at 18:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No it isn't - indeed, suppose it were, say $mathfrak{m}_{mathbb{C}_p} = (x_1,ldots,x_n)$. Then $v(x) geq min v(x_i) > 0$ for any $xinmathfrak{m}_{mathbb{C}_p}$. However, this is a contradiction as there exist elements of $mathfrak{m}_{mathbb{C}_p}$ with valuation strictly less than $min v(x_i)$ (for example, one of the $sqrt{x_i}$).
edited Jan 21 at 12:08
Pierre-Yves Gaillard
13.4k23184
answered Dec 28 '18 at 17:17
ODFODF
1,486510
$endgroup$
add a comment |
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$begingroup$
No it isn't - indeed, suppose it were, say $mathfrak{m}_{mathbb{C}_p} = (x_1,ldots,x_n)$. Then $v(x) geq min v(x_i) > 0$ for any $xinmathfrak{m}_{mathbb{C}_p}$. However, this is a contradiction as there exist elements of $mathfrak{m}_{mathbb{C}_p}$ with valuation strictly less than $min v(x_i)$ (for example, one of the $sqrt{x_i}$).
edited Jan 21 at 12:08
Pierre-Yves Gaillard
13.4k23184
answered Dec 28 '18 at 17:17
ODFODF
1,486510
$endgroup$
add a comment |
$begingroup$
No it isn't - indeed, suppose it were, say $mathfrak{m}_{mathbb{C}_p} = (x_1,ldots,x_n)$. Then $v(x) geq min v(x_i) > 0$ for any $xinmathfrak{m}_{mathbb{C}_p}$. However, this is a contradiction as there exist elements of $mathfrak{m}_{mathbb{C}_p}$ with valuation strictly less than $min v(x_i)$ (for example, one of the $sqrt{x_i}$).
edited Jan 21 at 12:08
Pierre-Yves Gaillard
13.4k23184
answered Dec 28 '18 at 17:17
ODFODF
1,486510
$endgroup$
add a comment |
$begingroup$
No it isn't - indeed, suppose it were, say $mathfrak{m}_{mathbb{C}_p} = (x_1,ldots,x_n)$. Then $v(x) geq min v(x_i) > 0$ for any $xinmathfrak{m}_{mathbb{C}_p}$. However, this is a contradiction as there exist elements of $mathfrak{m}_{mathbb{C}_p}$ with valuation strictly less than $min v(x_i)$ (for example, one of the $sqrt{x_i}$).
edited Jan 21 at 12:08
Pierre-Yves Gaillard
13.4k23184
answered Dec 28 '18 at 17:17
ODFODF
1,486510
$endgroup$
No it isn't - indeed, suppose it were, say $mathfrak{m}_{mathbb{C}_p} = (x_1,ldots,x_n)$. Then $v(x) geq min v(x_i) > 0$ for any $xinmathfrak{m}_{mathbb{C}_p}$. However, this is a contradiction as there exist elements of $mathfrak{m}_{mathbb{C}_p}$ with valuation strictly less than $min v(x_i)$ (for example, one of the $sqrt{x_i}$).
edited Jan 21 at 12:08
Pierre-Yves Gaillard
13.4k23184
answered Dec 28 '18 at 17:17
ODFODF
1,486510
edited Jan 21 at 12:08
Pierre-Yves Gaillard
13.4k23184
edited Jan 21 at 12:08
Pierre-Yves Gaillard
13.4k23184
edited Jan 21 at 12:08
Pierre-Yves Gaillard
13.4k23184
13.4k23184
answered Dec 28 '18 at 17:17
ODFODF
1,486510
answered Dec 28 '18 at 17:17
ODFODF
1,486510
answered Dec 28 '18 at 17:17
ODFODF
1,486510
1,486510
add a comment |
add a comment |
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5
$begingroup$
Hint: elements of the maximal ideal can have valuations arbitrarily close to $0$.
$endgroup$
– Wojowu
Dec 28 '18 at 17:12
$begingroup$
Let $K^{unr} = mathbf{Q}_p(zeta_{p^infty-1})$ the maximal unramified extension of $mathbf{Q}_p$ (so that $(p)$ is the unique maximal ideal of both rings of integers). Since $O_{K^{unr}}/(p)$ is algebraically closed then for any finite extension $L/K$ of degree $n$ you'll have $|pi_L|_L= |p|_L^{1/n}$. Thus only finite extensions of $K$ are local fields and any infinite extension of $K$ will be dense in $mathbb{C}_p$.
$endgroup$
– reuns
Dec 28 '18 at 18:47
1
$begingroup$
@reuns, I do not believe that any infinite extension of $K=K^{unr}$ is dense in $Bbb C_p$. What about the maximal tame extension of $K$?
$endgroup$
– Lubin
Dec 29 '18 at 22:21
$begingroup$
@Lubin Thank you. You are saying that $p nmid n = [K^{unr}(a):K^{unr}]$ implies that $|a|= p^{l/n}$ so $|a| ne |p^{1/p}|$ and $|p^{1/p}-a| ge |p^{1/p}|$ ? Whence with $L$ the maximal tame extension of $K^{unr}$ then $inf_{a in L} |p^{1/p} - a| = |p^{1/p}|$ and my last sentence was a big mistake and $mathbb{C}_p$ is very different to $mathbb{C}$ in this regard.
$endgroup$
– reuns
Dec 30 '18 at 17:11
1
$begingroup$
Right you are. I’m pretty sure that even if your valuation group was all of $Bbb Q$, that would not be enough to guarantee that the completion was all of $Bbb C$. Wild ramification is truly a wild phenomenon!
$endgroup$
– Lubin
Dec 30 '18 at 18:15