Extension of 2-adic valuation to the real numbersElementary results with p-adic numbersValuations on tensor...
Extension of 2-adic valuation to the real numbers
Elementary results with p-adic numbersValuations on tensor productsValuations given by flags on a variety and valuations of maximal rational rankp-adic valuation of a sumExtension of the product formula for valuations to a simultaneous completionCompletion of a finite field extension is also finite?Structure of valuations on $mathbb{F}_q(X,Y)$?2-adic valuation of odd harmonic sumsRelation between valuation of p-adic regulator of totally real field and its finite p-unramified abelian extensionsThe localization of the integral closure of a valuation ring is a valuation ring
$begingroup$
I just want to know what properties of valuations extend to $mathbb R$...
Denote an extension of the 2-adic valuation from $mathbb Q$ to $mathbb R$ by $nu$.
Suppose $nu(x)=nu(y)=0$.
Is it true that $nu(x+y)ne 0$?
What about $nu(x^2+y^2)le 1$?
I'm interested in knowing both whether these are true for every extension, as well as knowing whether there is some extension for which they are true (for every $x$ and $y$).
nt.number-theory ra.rings-and-algebras valuation-theory valuation-rings
edited 7 hours ago
Glorfindel
1,31241221
asked 18 hours ago
domotorpdomotorp
10k3390
$endgroup$
add a comment |
$begingroup$
I just want to know what properties of valuations extend to $mathbb R$...
Denote an extension of the 2-adic valuation from $mathbb Q$ to $mathbb R$ by $nu$.
Suppose $nu(x)=nu(y)=0$.
Is it true that $nu(x+y)ne 0$?
What about $nu(x^2+y^2)le 1$?
I'm interested in knowing both whether these are true for every extension, as well as knowing whether there is some extension for which they are true (for every $x$ and $y$).
nt.number-theory ra.rings-and-algebras valuation-theory valuation-rings
edited 7 hours ago
Glorfindel
1,31241221
asked 18 hours ago
domotorpdomotorp
10k3390
$endgroup$
7
$begingroup$
IMHO the title is too much of a clickbait.
$endgroup$
– schematic_boi
13 hours ago
1
$begingroup$
I changed the title to something more appropriate, so no more clickbait.
$endgroup$
– KConrad
12 hours ago
5
$begingroup$
I saw nothing wrong with the title actually. Nothing wrong with a bit of humour.
$endgroup$
– RP_
11 hours ago
add a comment |
$begingroup$
I just want to know what properties of valuations extend to $mathbb R$...
Denote an extension of the 2-adic valuation from $mathbb Q$ to $mathbb R$ by $nu$.
Suppose $nu(x)=nu(y)=0$.
Is it true that $nu(x+y)ne 0$?
What about $nu(x^2+y^2)le 1$?
I'm interested in knowing both whether these are true for every extension, as well as knowing whether there is some extension for which they are true (for every $x$ and $y$).
nt.number-theory ra.rings-and-algebras valuation-theory valuation-rings
edited 7 hours ago
Glorfindel
1,31241221
asked 18 hours ago
domotorpdomotorp
10k3390
$endgroup$
I just want to know what properties of valuations extend to $mathbb R$...
Denote an extension of the 2-adic valuation from $mathbb Q$ to $mathbb R$ by $nu$.
Suppose $nu(x)=nu(y)=0$.
Is it true that $nu(x+y)ne 0$?
What about $nu(x^2+y^2)le 1$?
I'm interested in knowing both whether these are true for every extension, as well as knowing whether there is some extension for which they are true (for every $x$ and $y$).
nt.number-theory ra.rings-and-algebras valuation-theory valuation-rings
nt.number-theory ra.rings-and-algebras valuation-theory valuation-rings
edited 7 hours ago
Glorfindel
1,31241221
asked 18 hours ago
domotorpdomotorp
10k3390
edited 7 hours ago
Glorfindel
1,31241221
asked 18 hours ago
domotorpdomotorp
10k3390
edited 7 hours ago
Glorfindel
1,31241221
edited 7 hours ago
Glorfindel
1,31241221
edited 7 hours ago
Glorfindel
1,31241221
1,31241221
asked 18 hours ago
domotorpdomotorp
10k3390
asked 18 hours ago
domotorpdomotorp
10k3390
asked 18 hours ago
domotorpdomotorp
10k3390
10k3390
7
$begingroup$
IMHO the title is too much of a clickbait.
$endgroup$
– schematic_boi
13 hours ago
1
$begingroup$
I changed the title to something more appropriate, so no more clickbait.
$endgroup$
– KConrad
12 hours ago
5
$begingroup$
I saw nothing wrong with the title actually. Nothing wrong with a bit of humour.
$endgroup$
– RP_
11 hours ago
add a comment |
7
$begingroup$
IMHO the title is too much of a clickbait.
$endgroup$
– schematic_boi
13 hours ago
1
$begingroup$
I changed the title to something more appropriate, so no more clickbait.
$endgroup$
– KConrad
12 hours ago
5
$begingroup$
I saw nothing wrong with the title actually. Nothing wrong with a bit of humour.
$endgroup$
– RP_
11 hours ago
7
7
$begingroup$
IMHO the title is too much of a clickbait.
$endgroup$
– schematic_boi
13 hours ago
$begingroup$
IMHO the title is too much of a clickbait.
$endgroup$
– schematic_boi
13 hours ago
1
1
$begingroup$
I changed the title to something more appropriate, so no more clickbait.
$endgroup$
– KConrad
12 hours ago
$begingroup$
I changed the title to something more appropriate, so no more clickbait.
$endgroup$
– KConrad
12 hours ago
5
5
$begingroup$
I saw nothing wrong with the title actually. Nothing wrong with a bit of humour.
$endgroup$
– RP_
11 hours ago
$begingroup$
I saw nothing wrong with the title actually. Nothing wrong with a bit of humour.
$endgroup$
– RP_
11 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No. The important thing to know is that, if $K subseteq L$ is a field extension and $v: K to mathbb{R}$ is a valuation, then $v$ can be extended to $L$. So I can answer all of your questions by working in some easy to handle subfield of $mathbb{R}$. I'll work in $K = mathbb{Q}(sqrt{5})$ for the first question and in $K = mathbb{Q}(sqrt{3})$ for the second.
The ring of integers in $mathbb{Q}[sqrt{5}]$ is $mathbb{Z}[tau]$ where $tau = tfrac{1+sqrt{5}}{2}$, with minimal polynomial $tau^2=tau+1$. Note that $mathcal{O}_K/(2 mathcal{O}_K)$ is the field $mathbb{F}_4$ with four elements. Your first statement is true in $mathbb{Q}$ only because $mathbb{Z}/(2 mathbb{Z})$ has two elements.
Specifically, both $1$ and $tau$ are in $mathcal{O}_K$ but not $2 mathcal{O}_K$, so $v(1) = v(tau) = 0$, but $1+tau$ is also not in $2 mathcal{O}_K$ so $v(1+tau)=0$ as well.
Similarly, the ring of integers in $mathbb{Q}(sqrt{3})$ is $mathbb{Z}[sqrt{3}]$ and the prime $2$ is ramified, with $2 = (1+sqrt{3})^2 (2-sqrt{3})$ (note that $2-sqrt{3}$ is a unit). We have $v(1) = v(sqrt{3}) = 0$, but $v(1+sqrt{3}^2) = 2$. In this case, the result is true in $mathbb{Q}$ because $2$ is unramified.
answered 17 hours ago
David E SpeyerDavid E Speyer
108k9285543
$endgroup$
2
$begingroup$
"the field $Bbb F_4$ with four elements"?
$endgroup$
– Greg Martin
10 hours ago
$begingroup$
Thanks for the correction! @GregMartin
$endgroup$
– David E Speyer
7 hours ago
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No. The important thing to know is that, if $K subseteq L$ is a field extension and $v: K to mathbb{R}$ is a valuation, then $v$ can be extended to $L$. So I can answer all of your questions by working in some easy to handle subfield of $mathbb{R}$. I'll work in $K = mathbb{Q}(sqrt{5})$ for the first question and in $K = mathbb{Q}(sqrt{3})$ for the second.
The ring of integers in $mathbb{Q}[sqrt{5}]$ is $mathbb{Z}[tau]$ where $tau = tfrac{1+sqrt{5}}{2}$, with minimal polynomial $tau^2=tau+1$. Note that $mathcal{O}_K/(2 mathcal{O}_K)$ is the field $mathbb{F}_4$ with four elements. Your first statement is true in $mathbb{Q}$ only because $mathbb{Z}/(2 mathbb{Z})$ has two elements.
Specifically, both $1$ and $tau$ are in $mathcal{O}_K$ but not $2 mathcal{O}_K$, so $v(1) = v(tau) = 0$, but $1+tau$ is also not in $2 mathcal{O}_K$ so $v(1+tau)=0$ as well.
Similarly, the ring of integers in $mathbb{Q}(sqrt{3})$ is $mathbb{Z}[sqrt{3}]$ and the prime $2$ is ramified, with $2 = (1+sqrt{3})^2 (2-sqrt{3})$ (note that $2-sqrt{3}$ is a unit). We have $v(1) = v(sqrt{3}) = 0$, but $v(1+sqrt{3}^2) = 2$. In this case, the result is true in $mathbb{Q}$ because $2$ is unramified.
answered 17 hours ago
David E SpeyerDavid E Speyer
108k9285543
$endgroup$
2
$begingroup$
"the field $Bbb F_4$ with four elements"?
$endgroup$
– Greg Martin
10 hours ago
$begingroup$
Thanks for the correction! @GregMartin
$endgroup$
– David E Speyer
7 hours ago
add a comment |
$begingroup$
No. The important thing to know is that, if $K subseteq L$ is a field extension and $v: K to mathbb{R}$ is a valuation, then $v$ can be extended to $L$. So I can answer all of your questions by working in some easy to handle subfield of $mathbb{R}$. I'll work in $K = mathbb{Q}(sqrt{5})$ for the first question and in $K = mathbb{Q}(sqrt{3})$ for the second.
The ring of integers in $mathbb{Q}[sqrt{5}]$ is $mathbb{Z}[tau]$ where $tau = tfrac{1+sqrt{5}}{2}$, with minimal polynomial $tau^2=tau+1$. Note that $mathcal{O}_K/(2 mathcal{O}_K)$ is the field $mathbb{F}_4$ with four elements. Your first statement is true in $mathbb{Q}$ only because $mathbb{Z}/(2 mathbb{Z})$ has two elements.
Specifically, both $1$ and $tau$ are in $mathcal{O}_K$ but not $2 mathcal{O}_K$, so $v(1) = v(tau) = 0$, but $1+tau$ is also not in $2 mathcal{O}_K$ so $v(1+tau)=0$ as well.
Similarly, the ring of integers in $mathbb{Q}(sqrt{3})$ is $mathbb{Z}[sqrt{3}]$ and the prime $2$ is ramified, with $2 = (1+sqrt{3})^2 (2-sqrt{3})$ (note that $2-sqrt{3}$ is a unit). We have $v(1) = v(sqrt{3}) = 0$, but $v(1+sqrt{3}^2) = 2$. In this case, the result is true in $mathbb{Q}$ because $2$ is unramified.
answered 17 hours ago
David E SpeyerDavid E Speyer
108k9285543
$endgroup$
2
$begingroup$
"the field $Bbb F_4$ with four elements"?
$endgroup$
– Greg Martin
10 hours ago
$begingroup$
Thanks for the correction! @GregMartin
$endgroup$
– David E Speyer
7 hours ago
add a comment |
$begingroup$
No. The important thing to know is that, if $K subseteq L$ is a field extension and $v: K to mathbb{R}$ is a valuation, then $v$ can be extended to $L$. So I can answer all of your questions by working in some easy to handle subfield of $mathbb{R}$. I'll work in $K = mathbb{Q}(sqrt{5})$ for the first question and in $K = mathbb{Q}(sqrt{3})$ for the second.
The ring of integers in $mathbb{Q}[sqrt{5}]$ is $mathbb{Z}[tau]$ where $tau = tfrac{1+sqrt{5}}{2}$, with minimal polynomial $tau^2=tau+1$. Note that $mathcal{O}_K/(2 mathcal{O}_K)$ is the field $mathbb{F}_4$ with four elements. Your first statement is true in $mathbb{Q}$ only because $mathbb{Z}/(2 mathbb{Z})$ has two elements.
Specifically, both $1$ and $tau$ are in $mathcal{O}_K$ but not $2 mathcal{O}_K$, so $v(1) = v(tau) = 0$, but $1+tau$ is also not in $2 mathcal{O}_K$ so $v(1+tau)=0$ as well.
Similarly, the ring of integers in $mathbb{Q}(sqrt{3})$ is $mathbb{Z}[sqrt{3}]$ and the prime $2$ is ramified, with $2 = (1+sqrt{3})^2 (2-sqrt{3})$ (note that $2-sqrt{3}$ is a unit). We have $v(1) = v(sqrt{3}) = 0$, but $v(1+sqrt{3}^2) = 2$. In this case, the result is true in $mathbb{Q}$ because $2$ is unramified.
answered 17 hours ago
David E SpeyerDavid E Speyer
108k9285543
$endgroup$
No. The important thing to know is that, if $K subseteq L$ is a field extension and $v: K to mathbb{R}$ is a valuation, then $v$ can be extended to $L$. So I can answer all of your questions by working in some easy to handle subfield of $mathbb{R}$. I'll work in $K = mathbb{Q}(sqrt{5})$ for the first question and in $K = mathbb{Q}(sqrt{3})$ for the second.
The ring of integers in $mathbb{Q}[sqrt{5}]$ is $mathbb{Z}[tau]$ where $tau = tfrac{1+sqrt{5}}{2}$, with minimal polynomial $tau^2=tau+1$. Note that $mathcal{O}_K/(2 mathcal{O}_K)$ is the field $mathbb{F}_4$ with four elements. Your first statement is true in $mathbb{Q}$ only because $mathbb{Z}/(2 mathbb{Z})$ has two elements.
Specifically, both $1$ and $tau$ are in $mathcal{O}_K$ but not $2 mathcal{O}_K$, so $v(1) = v(tau) = 0$, but $1+tau$ is also not in $2 mathcal{O}_K$ so $v(1+tau)=0$ as well.
Similarly, the ring of integers in $mathbb{Q}(sqrt{3})$ is $mathbb{Z}[sqrt{3}]$ and the prime $2$ is ramified, with $2 = (1+sqrt{3})^2 (2-sqrt{3})$ (note that $2-sqrt{3}$ is a unit). We have $v(1) = v(sqrt{3}) = 0$, but $v(1+sqrt{3}^2) = 2$. In this case, the result is true in $mathbb{Q}$ because $2$ is unramified.
answered 17 hours ago
David E SpeyerDavid E Speyer
108k9285543
edited 7 hours ago
answered 17 hours ago
David E SpeyerDavid E Speyer
108k9285543
answered 17 hours ago
David E SpeyerDavid E Speyer
108k9285543
answered 17 hours ago
David E SpeyerDavid E Speyer
108k9285543
108k9285543
2
$begingroup$
"the field $Bbb F_4$ with four elements"?
$endgroup$
– Greg Martin
10 hours ago
$begingroup$
Thanks for the correction! @GregMartin
$endgroup$
– David E Speyer
7 hours ago
add a comment |
2
$begingroup$
"the field $Bbb F_4$ with four elements"?
$endgroup$
– Greg Martin
10 hours ago
$begingroup$
Thanks for the correction! @GregMartin
$endgroup$
– David E Speyer
7 hours ago
2
2
$begingroup$
"the field $Bbb F_4$ with four elements"?
$endgroup$
– Greg Martin
10 hours ago
$begingroup$
"the field $Bbb F_4$ with four elements"?
$endgroup$
– Greg Martin
10 hours ago
$begingroup$
Thanks for the correction! @GregMartin
$endgroup$
– David E Speyer
7 hours ago
$begingroup$
Thanks for the correction! @GregMartin
$endgroup$
– David E Speyer
7 hours ago
add a comment |
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7
$begingroup$
IMHO the title is too much of a clickbait.
$endgroup$
– schematic_boi
13 hours ago
1
$begingroup$
I changed the title to something more appropriate, so no more clickbait.
$endgroup$
– KConrad
12 hours ago
5
$begingroup$
I saw nothing wrong with the title actually. Nothing wrong with a bit of humour.
$endgroup$
– RP_
11 hours ago