Extension of 2-adic valuation to the real numbersElementary 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 $mathbbF_q(X,Y)$?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
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 $mathbbF_q(X,Y)$?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 Apr 26 at 19:56
Glorfindel
1,32441221
asked Apr 26 at 8:42
domotorpdomotorp
10.1k3390
$endgroup$
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 Apr 26 at 19:56
Glorfindel
1,32441221
asked Apr 26 at 8:42
domotorpdomotorp
10.1k3390
$endgroup$
2
$begingroup$
I changed the title to something more appropriate, so no more clickbait.
$endgroup$
– KConrad
Apr 26 at 15:23
8
$begingroup$
I saw nothing wrong with the title actually. Nothing wrong with a bit of humour.
$endgroup$
– RP_
Apr 26 at 16:03
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 Apr 26 at 19:56
Glorfindel
1,32441221
asked Apr 26 at 8:42
domotorpdomotorp
10.1k3390
$endgroup$
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 Apr 26 at 19:56
Glorfindel
1,32441221
asked Apr 26 at 8:42
domotorpdomotorp
10.1k3390
edited Apr 26 at 19:56
Glorfindel
1,32441221
asked Apr 26 at 8:42
domotorpdomotorp
10.1k3390
edited Apr 26 at 19:56
Glorfindel
1,32441221
edited Apr 26 at 19:56
Glorfindel
1,32441221
edited Apr 26 at 19:56
Glorfindel
1,32441221
1,32441221
asked Apr 26 at 8:42
domotorpdomotorp
10.1k3390
asked Apr 26 at 8:42
domotorpdomotorp
10.1k3390
asked Apr 26 at 8:42
domotorpdomotorp
10.1k3390
10.1k3390
2
$begingroup$
I changed the title to something more appropriate, so no more clickbait.
$endgroup$
– KConrad
Apr 26 at 15:23
8
$begingroup$
I saw nothing wrong with the title actually. Nothing wrong with a bit of humour.
$endgroup$
– RP_
Apr 26 at 16:03
add a comment |
2
$begingroup$
I changed the title to something more appropriate, so no more clickbait.
$endgroup$
– KConrad
Apr 26 at 15:23
8
$begingroup$
I saw nothing wrong with the title actually. Nothing wrong with a bit of humour.
$endgroup$
– RP_
Apr 26 at 16:03
2
2
$begingroup$
I changed the title to something more appropriate, so no more clickbait.
$endgroup$
– KConrad
Apr 26 at 15:23
$begingroup$
I changed the title to something more appropriate, so no more clickbait.
$endgroup$
– KConrad
Apr 26 at 15:23
8
8
$begingroup$
I saw nothing wrong with the title actually. Nothing wrong with a bit of humour.
$endgroup$
– RP_
Apr 26 at 16:03
$begingroup$
I saw nothing wrong with the title actually. Nothing wrong with a bit of humour.
$endgroup$
– RP_
Apr 26 at 16:03
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 $v: K to mathbbR$ 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 $mathbbR$. I'll work in $K = mathbbQ(sqrt5)$ for the first question and in $K = mathbbQ(sqrt3)$ for the second.
The ring of integers in $mathbbQ[sqrt5]$ is $mathbbZ[tau]$ where $tau = tfrac1+sqrt52$, with minimal polynomial $tau^2=tau+1$. Note that $mathcalO_K/(2 mathcalO_K)$ is the field $mathbbF_4$ with four elements. Your first statement is true in $mathbbQ$ only because $mathbbZ/(2 mathbbZ)$ has two elements.
Specifically, both $1$ and $tau$ are in $mathcalO_K$ but not $2 mathcalO_K$, so $v(1) = v(tau) = 0$, but $1+tau$ is also not in $2 mathcalO_K$ so $v(1+tau)=0$ as well.
Similarly, the ring of integers in $mathbbQ(sqrt3)$ is $mathbbZ[sqrt3]$ and the prime $2$ is ramified, with $2 = (1+sqrt3)^2 (2-sqrt3)$ (note that $2-sqrt3$ is a unit). We have $v(1) = v(sqrt3) = 0$, but $v(1+sqrt3^2) = 2$. In this case, the result is true in $mathbbQ$ because $2$ is unramified.
answered Apr 26 at 10:10
David E SpeyerDavid E Speyer
108k9287546
$endgroup$
2
$begingroup$
"the field $Bbb F_4$ with four elements"?
$endgroup$
– Greg Martin
Apr 26 at 16:47
$begingroup$
Thanks for the correction! @GregMartin
$endgroup$
– David E Speyer
Apr 26 at 20:12
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
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 $v: K to mathbbR$ 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 $mathbbR$. I'll work in $K = mathbbQ(sqrt5)$ for the first question and in $K = mathbbQ(sqrt3)$ for the second.
The ring of integers in $mathbbQ[sqrt5]$ is $mathbbZ[tau]$ where $tau = tfrac1+sqrt52$, with minimal polynomial $tau^2=tau+1$. Note that $mathcalO_K/(2 mathcalO_K)$ is the field $mathbbF_4$ with four elements. Your first statement is true in $mathbbQ$ only because $mathbbZ/(2 mathbbZ)$ has two elements.
Specifically, both $1$ and $tau$ are in $mathcalO_K$ but not $2 mathcalO_K$, so $v(1) = v(tau) = 0$, but $1+tau$ is also not in $2 mathcalO_K$ so $v(1+tau)=0$ as well.
Similarly, the ring of integers in $mathbbQ(sqrt3)$ is $mathbbZ[sqrt3]$ and the prime $2$ is ramified, with $2 = (1+sqrt3)^2 (2-sqrt3)$ (note that $2-sqrt3$ is a unit). We have $v(1) = v(sqrt3) = 0$, but $v(1+sqrt3^2) = 2$. In this case, the result is true in $mathbbQ$ because $2$ is unramified.
answered Apr 26 at 10:10
David E SpeyerDavid E Speyer
108k9287546
$endgroup$
2
$begingroup$
"the field $Bbb F_4$ with four elements"?
$endgroup$
– Greg Martin
Apr 26 at 16:47
$begingroup$
Thanks for the correction! @GregMartin
$endgroup$
– David E Speyer
Apr 26 at 20:12
add a comment |
$begingroup$
No. The important thing to know is that, if $K subseteq L$ is a field extension and $v: K to mathbbR$ 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 $mathbbR$. I'll work in $K = mathbbQ(sqrt5)$ for the first question and in $K = mathbbQ(sqrt3)$ for the second.
The ring of integers in $mathbbQ[sqrt5]$ is $mathbbZ[tau]$ where $tau = tfrac1+sqrt52$, with minimal polynomial $tau^2=tau+1$. Note that $mathcalO_K/(2 mathcalO_K)$ is the field $mathbbF_4$ with four elements. Your first statement is true in $mathbbQ$ only because $mathbbZ/(2 mathbbZ)$ has two elements.
Specifically, both $1$ and $tau$ are in $mathcalO_K$ but not $2 mathcalO_K$, so $v(1) = v(tau) = 0$, but $1+tau$ is also not in $2 mathcalO_K$ so $v(1+tau)=0$ as well.
Similarly, the ring of integers in $mathbbQ(sqrt3)$ is $mathbbZ[sqrt3]$ and the prime $2$ is ramified, with $2 = (1+sqrt3)^2 (2-sqrt3)$ (note that $2-sqrt3$ is a unit). We have $v(1) = v(sqrt3) = 0$, but $v(1+sqrt3^2) = 2$. In this case, the result is true in $mathbbQ$ because $2$ is unramified.
answered Apr 26 at 10:10
David E SpeyerDavid E Speyer
108k9287546
$endgroup$
2
$begingroup$
"the field $Bbb F_4$ with four elements"?
$endgroup$
– Greg Martin
Apr 26 at 16:47
$begingroup$
Thanks for the correction! @GregMartin
$endgroup$
– David E Speyer
Apr 26 at 20:12
add a comment |
$begingroup$
No. The important thing to know is that, if $K subseteq L$ is a field extension and $v: K to mathbbR$ 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 $mathbbR$. I'll work in $K = mathbbQ(sqrt5)$ for the first question and in $K = mathbbQ(sqrt3)$ for the second.
The ring of integers in $mathbbQ[sqrt5]$ is $mathbbZ[tau]$ where $tau = tfrac1+sqrt52$, with minimal polynomial $tau^2=tau+1$. Note that $mathcalO_K/(2 mathcalO_K)$ is the field $mathbbF_4$ with four elements. Your first statement is true in $mathbbQ$ only because $mathbbZ/(2 mathbbZ)$ has two elements.
Specifically, both $1$ and $tau$ are in $mathcalO_K$ but not $2 mathcalO_K$, so $v(1) = v(tau) = 0$, but $1+tau$ is also not in $2 mathcalO_K$ so $v(1+tau)=0$ as well.
Similarly, the ring of integers in $mathbbQ(sqrt3)$ is $mathbbZ[sqrt3]$ and the prime $2$ is ramified, with $2 = (1+sqrt3)^2 (2-sqrt3)$ (note that $2-sqrt3$ is a unit). We have $v(1) = v(sqrt3) = 0$, but $v(1+sqrt3^2) = 2$. In this case, the result is true in $mathbbQ$ because $2$ is unramified.
answered Apr 26 at 10:10
David E SpeyerDavid E Speyer
108k9287546
$endgroup$
No. The important thing to know is that, if $K subseteq L$ is a field extension and $v: K to mathbbR$ 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 $mathbbR$. I'll work in $K = mathbbQ(sqrt5)$ for the first question and in $K = mathbbQ(sqrt3)$ for the second.
The ring of integers in $mathbbQ[sqrt5]$ is $mathbbZ[tau]$ where $tau = tfrac1+sqrt52$, with minimal polynomial $tau^2=tau+1$. Note that $mathcalO_K/(2 mathcalO_K)$ is the field $mathbbF_4$ with four elements. Your first statement is true in $mathbbQ$ only because $mathbbZ/(2 mathbbZ)$ has two elements.
Specifically, both $1$ and $tau$ are in $mathcalO_K$ but not $2 mathcalO_K$, so $v(1) = v(tau) = 0$, but $1+tau$ is also not in $2 mathcalO_K$ so $v(1+tau)=0$ as well.
Similarly, the ring of integers in $mathbbQ(sqrt3)$ is $mathbbZ[sqrt3]$ and the prime $2$ is ramified, with $2 = (1+sqrt3)^2 (2-sqrt3)$ (note that $2-sqrt3$ is a unit). We have $v(1) = v(sqrt3) = 0$, but $v(1+sqrt3^2) = 2$. In this case, the result is true in $mathbbQ$ because $2$ is unramified.
answered Apr 26 at 10:10
David E SpeyerDavid E Speyer
108k9287546
edited Apr 26 at 20:12
answered Apr 26 at 10:10
David E SpeyerDavid E Speyer
108k9287546
answered Apr 26 at 10:10
David E SpeyerDavid E Speyer
108k9287546
answered Apr 26 at 10:10
David E SpeyerDavid E Speyer
108k9287546
108k9287546
2
$begingroup$
"the field $Bbb F_4$ with four elements"?
$endgroup$
– Greg Martin
Apr 26 at 16:47
$begingroup$
Thanks for the correction! @GregMartin
$endgroup$
– David E Speyer
Apr 26 at 20:12
add a comment |
2
$begingroup$
"the field $Bbb F_4$ with four elements"?
$endgroup$
– Greg Martin
Apr 26 at 16:47
$begingroup$
Thanks for the correction! @GregMartin
$endgroup$
– David E Speyer
Apr 26 at 20:12
2
2
$begingroup$
"the field $Bbb F_4$ with four elements"?
$endgroup$
– Greg Martin
Apr 26 at 16:47
$begingroup$
"the field $Bbb F_4$ with four elements"?
$endgroup$
– Greg Martin
Apr 26 at 16:47
$begingroup$
Thanks for the correction! @GregMartin
$endgroup$
– David E Speyer
Apr 26 at 20:12
$begingroup$
Thanks for the correction! @GregMartin
$endgroup$
– David E Speyer
Apr 26 at 20:12
add a comment |
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$begingroup$
I changed the title to something more appropriate, so no more clickbait.
$endgroup$
– KConrad
Apr 26 at 15:23
8
$begingroup$
I saw nothing wrong with the title actually. Nothing wrong with a bit of humour.
$endgroup$
– RP_
Apr 26 at 16:03