Why is the residue field of $mathbbQ_p$ isomorphic to $mathbbF_p$? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Quotients of a valuation ring in the completion of a number fieldDiscrete valuations of a functional field have discrete valuation rings.Is the residue field of an algebraically closed field with respect to a non-trivial valuation infinite?Localization of a valuation ring at a prime is abstractly isomorphic to the original ringLocal subring of a DVR and finite residue field extension$p$-adic field with infinite residue field.Frobenius automorphism only defined on valuation ring?Residue field of finite extension of p adic numbers is a finite field.Absolute value of a generator of the differentRing of integers of $mathbbC_p$
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Why is the residue field of $mathbbQ_p$ isomorphic to $mathbbF_p$?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Quotients of a valuation ring in the completion of a number fieldDiscrete valuations of a functional field have discrete valuation rings.Is the residue field of an algebraically closed field with respect to a non-trivial valuation infinite?Localization of a valuation ring at a prime is abstractly isomorphic to the original ringLocal subring of a DVR and finite residue field extension$p$-adic field with infinite residue field.Frobenius automorphism only defined on valuation ring?Residue field of finite extension of p adic numbers is a finite field.Absolute value of a generator of the differentRing of integers of $mathbbC_p$
$begingroup$
$mathcalO=xin mathbbQ_p:v(x)geq0$ is a valuation ring.
$mathfrakm=xin mathbbQ_p: v(x)>0$ is the maximal ideal of $mathcalO$.
Why is $K=mathcalO/mathfrakm$ isomorphic to $mathbbF_p$, the finite field with p elements?
number-theory field-theory finite-fields p-adic-number-theory valuation-theory
edited Apr 2 at 8:58
Sam Streeter
1,504418
asked Feb 20 at 23:42
LawrdyLawrdLawrdyLawrd
212
$endgroup$
add a comment |
$begingroup$
$mathcalO=xin mathbbQ_p:v(x)geq0$ is a valuation ring.
$mathfrakm=xin mathbbQ_p: v(x)>0$ is the maximal ideal of $mathcalO$.
Why is $K=mathcalO/mathfrakm$ isomorphic to $mathbbF_p$, the finite field with p elements?
number-theory field-theory finite-fields p-adic-number-theory valuation-theory
edited Apr 2 at 8:58
Sam Streeter
1,504418
asked Feb 20 at 23:42
LawrdyLawrdLawrdyLawrd
212
$endgroup$
$begingroup$
Because $O = mathbbZ_p, mathfrakM = bigcup_a=0^p-1 a+pmathbbZ_p$. If you define $mathbbZ_p$ as the completion of $mathbbZ$ for $|x|_p = p^-v(x)$ then that $O/mathfrakM = (O cap mathbbZ)/(mathfrakMcap mathbbZ) = mathbbZ/pmathbbZ$ is a consequence of that $v$ is a discrete valuation
$endgroup$
– reuns
Feb 21 at 0:23
2
$begingroup$
How you come to see the truth of this claim may depend on which definition of $Bbb Q_p$ you’re using.
$endgroup$
– Lubin
Feb 21 at 5:12
add a comment |
$begingroup$
$mathcalO=xin mathbbQ_p:v(x)geq0$ is a valuation ring.
$mathfrakm=xin mathbbQ_p: v(x)>0$ is the maximal ideal of $mathcalO$.
Why is $K=mathcalO/mathfrakm$ isomorphic to $mathbbF_p$, the finite field with p elements?
number-theory field-theory finite-fields p-adic-number-theory valuation-theory
edited Apr 2 at 8:58
Sam Streeter
1,504418
asked Feb 20 at 23:42
LawrdyLawrdLawrdyLawrd
212
$endgroup$
$mathcalO=xin mathbbQ_p:v(x)geq0$ is a valuation ring.
$mathfrakm=xin mathbbQ_p: v(x)>0$ is the maximal ideal of $mathcalO$.
Why is $K=mathcalO/mathfrakm$ isomorphic to $mathbbF_p$, the finite field with p elements?
number-theory field-theory finite-fields p-adic-number-theory valuation-theory
number-theory field-theory finite-fields p-adic-number-theory valuation-theory
edited Apr 2 at 8:58
Sam Streeter
1,504418
asked Feb 20 at 23:42
LawrdyLawrdLawrdyLawrd
212
edited Apr 2 at 8:58
Sam Streeter
1,504418
asked Feb 20 at 23:42
LawrdyLawrdLawrdyLawrd
212
edited Apr 2 at 8:58
Sam Streeter
1,504418
edited Apr 2 at 8:58
Sam Streeter
1,504418
edited Apr 2 at 8:58
Sam Streeter
1,504418
1,504418
asked Feb 20 at 23:42
LawrdyLawrdLawrdyLawrd
212
asked Feb 20 at 23:42
LawrdyLawrdLawrdyLawrd
212
asked Feb 20 at 23:42
LawrdyLawrdLawrdyLawrd
212
212
$begingroup$
Because $O = mathbbZ_p, mathfrakM = bigcup_a=0^p-1 a+pmathbbZ_p$. If you define $mathbbZ_p$ as the completion of $mathbbZ$ for $|x|_p = p^-v(x)$ then that $O/mathfrakM = (O cap mathbbZ)/(mathfrakMcap mathbbZ) = mathbbZ/pmathbbZ$ is a consequence of that $v$ is a discrete valuation
$endgroup$
– reuns
Feb 21 at 0:23
2
$begingroup$
How you come to see the truth of this claim may depend on which definition of $Bbb Q_p$ you’re using.
$endgroup$
– Lubin
Feb 21 at 5:12
add a comment |
$begingroup$
Because $O = mathbbZ_p, mathfrakM = bigcup_a=0^p-1 a+pmathbbZ_p$. If you define $mathbbZ_p$ as the completion of $mathbbZ$ for $|x|_p = p^-v(x)$ then that $O/mathfrakM = (O cap mathbbZ)/(mathfrakMcap mathbbZ) = mathbbZ/pmathbbZ$ is a consequence of that $v$ is a discrete valuation
$endgroup$
– reuns
Feb 21 at 0:23
2
$begingroup$
How you come to see the truth of this claim may depend on which definition of $Bbb Q_p$ you’re using.
$endgroup$
– Lubin
Feb 21 at 5:12
$begingroup$
Because $O = mathbbZ_p, mathfrakM = bigcup_a=0^p-1 a+pmathbbZ_p$. If you define $mathbbZ_p$ as the completion of $mathbbZ$ for $|x|_p = p^-v(x)$ then that $O/mathfrakM = (O cap mathbbZ)/(mathfrakMcap mathbbZ) = mathbbZ/pmathbbZ$ is a consequence of that $v$ is a discrete valuation
$endgroup$
– reuns
Feb 21 at 0:23
$begingroup$
Because $O = mathbbZ_p, mathfrakM = bigcup_a=0^p-1 a+pmathbbZ_p$. If you define $mathbbZ_p$ as the completion of $mathbbZ$ for $|x|_p = p^-v(x)$ then that $O/mathfrakM = (O cap mathbbZ)/(mathfrakMcap mathbbZ) = mathbbZ/pmathbbZ$ is a consequence of that $v$ is a discrete valuation
$endgroup$
– reuns
Feb 21 at 0:23
2
2
$begingroup$
How you come to see the truth of this claim may depend on which definition of $Bbb Q_p$ you’re using.
$endgroup$
– Lubin
Feb 21 at 5:12
$begingroup$
How you come to see the truth of this claim may depend on which definition of $Bbb Q_p$ you’re using.
$endgroup$
– Lubin
Feb 21 at 5:12
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We have the following exact sequence
$$
0rightarrow mathbbZ_prightarrow mathbbZ_prightarrow mathbbZ/p^nmathbbZrightarrow 0,
$$
where the first map is multiplication by $p^n$ and the second sends $x=(x_i)in mathbbZ_p=lim_leftarrowmathbbZ/p^nmathbbZ$ to its $n$th term. Thus $ mathbbZ_p/p^n mathbbZ_pcong mathbbZ/p^nmathbbZ$, so take $n=1$.
answered Feb 21 at 0:04
ArbutusArbutus
760715
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We have the following exact sequence
$$
0rightarrow mathbbZ_prightarrow mathbbZ_prightarrow mathbbZ/p^nmathbbZrightarrow 0,
$$
where the first map is multiplication by $p^n$ and the second sends $x=(x_i)in mathbbZ_p=lim_leftarrowmathbbZ/p^nmathbbZ$ to its $n$th term. Thus $ mathbbZ_p/p^n mathbbZ_pcong mathbbZ/p^nmathbbZ$, so take $n=1$.
answered Feb 21 at 0:04
ArbutusArbutus
760715
$endgroup$
add a comment |
$begingroup$
We have the following exact sequence
$$
0rightarrow mathbbZ_prightarrow mathbbZ_prightarrow mathbbZ/p^nmathbbZrightarrow 0,
$$
where the first map is multiplication by $p^n$ and the second sends $x=(x_i)in mathbbZ_p=lim_leftarrowmathbbZ/p^nmathbbZ$ to its $n$th term. Thus $ mathbbZ_p/p^n mathbbZ_pcong mathbbZ/p^nmathbbZ$, so take $n=1$.
answered Feb 21 at 0:04
ArbutusArbutus
760715
$endgroup$
add a comment |
$begingroup$
We have the following exact sequence
$$
0rightarrow mathbbZ_prightarrow mathbbZ_prightarrow mathbbZ/p^nmathbbZrightarrow 0,
$$
where the first map is multiplication by $p^n$ and the second sends $x=(x_i)in mathbbZ_p=lim_leftarrowmathbbZ/p^nmathbbZ$ to its $n$th term. Thus $ mathbbZ_p/p^n mathbbZ_pcong mathbbZ/p^nmathbbZ$, so take $n=1$.
answered Feb 21 at 0:04
ArbutusArbutus
760715
$endgroup$
We have the following exact sequence
$$
0rightarrow mathbbZ_prightarrow mathbbZ_prightarrow mathbbZ/p^nmathbbZrightarrow 0,
$$
where the first map is multiplication by $p^n$ and the second sends $x=(x_i)in mathbbZ_p=lim_leftarrowmathbbZ/p^nmathbbZ$ to its $n$th term. Thus $ mathbbZ_p/p^n mathbbZ_pcong mathbbZ/p^nmathbbZ$, so take $n=1$.
answered Feb 21 at 0:04
ArbutusArbutus
760715
answered Feb 21 at 0:04
ArbutusArbutus
760715
answered Feb 21 at 0:04
ArbutusArbutus
760715
answered Feb 21 at 0:04
ArbutusArbutus
760715
760715
add a comment |
add a comment |
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$begingroup$
Because $O = mathbbZ_p, mathfrakM = bigcup_a=0^p-1 a+pmathbbZ_p$. If you define $mathbbZ_p$ as the completion of $mathbbZ$ for $|x|_p = p^-v(x)$ then that $O/mathfrakM = (O cap mathbbZ)/(mathfrakMcap mathbbZ) = mathbbZ/pmathbbZ$ is a consequence of that $v$ is a discrete valuation
$endgroup$
– reuns
Feb 21 at 0:23
2
$begingroup$
How you come to see the truth of this claim may depend on which definition of $Bbb Q_p$ you’re using.
$endgroup$
– Lubin
Feb 21 at 5:12