Can two odd numbers sum up to an odd number?Generating random smooth numbers (or, “What do random smooth numbers look like?”)Complete extensions of valuations from Q to R.Valuations 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 $mathbbF_q(X,Y)$?Relation between valuation of p-adic regulator of totally real field and its finite p-unramified abelian extensionsOstrowski's Theorem for topological rings?
Can two odd numbers sum up to an odd number?
Generating random smooth numbers (or, “What do random smooth numbers look like?”)Complete extensions of valuations from Q to R.Valuations 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 $mathbbF_q(X,Y)$?Relation between valuation of p-adic regulator of totally real field and its finite p-unramified abelian extensionsOstrowski's Theorem for topological rings?
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Okay, the title is a bit of a clickbait, 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
$endgroup$
add a comment |
$begingroup$
Okay, the title is a bit of a clickbait, 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
$endgroup$
add a comment |
$begingroup$
Okay, the title is a bit of a clickbait, 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
$endgroup$
Okay, the title is a bit of a clickbait, 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
asked 3 hours ago
domotorpdomotorp
10k3388
10k3388
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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 two 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.
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$begingroup$
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 two 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.
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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 two 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.
$endgroup$
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 two 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.
$endgroup$
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 two 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.
edited 2 hours ago
answered 2 hours ago
David E SpeyerDavid E Speyer
108k9284542
108k9284542
add a comment |
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