$v_{p}(n):=e$ if $p^{e}|n$ and $p^{e+1} nmid n$. Show that $v_{p}(n*m)= v_{p}(n)+ v_{p}(m)$ and...
Happy new year. This seems like an easy to solve task but I'm stuck right now, so any help is appreciated.
Let $pin mathbb{Z}$ be a prime number. The function $v_{p}(n)geq 0$ is defined as $v_{p}(n):=e$ if $p^{e}|n$ and $p^{e+1} nmid n$. $v_{p}(0):=infty$ and $infty + e = e + infty = infty$ for all $ein mathbb{Z}$.
Show that i) $v_{p}(n*m)= v_{p}(n)+ v_{p}(m)$ for all $n,m in mathbb{Z}$ and
ii)$v_{p}(n+m)=v_{p}(n)$ if $v_{p}(n)<v_{p}(m)$ and $v_{p}(n+m)geq v_{p}(n)$ if $v_{p}(n)=v_{p}(m)$.
I would also like to know if this function has a specific name so that I can find more about it and read it. Thanks in advance for any help.
abstract-algebra
asked yesterday
Anzu
146
add a comment |
Happy new year. This seems like an easy to solve task but I'm stuck right now, so any help is appreciated.
Let $pin mathbb{Z}$ be a prime number. The function $v_{p}(n)geq 0$ is defined as $v_{p}(n):=e$ if $p^{e}|n$ and $p^{e+1} nmid n$. $v_{p}(0):=infty$ and $infty + e = e + infty = infty$ for all $ein mathbb{Z}$.
Show that i) $v_{p}(n*m)= v_{p}(n)+ v_{p}(m)$ for all $n,m in mathbb{Z}$ and
ii)$v_{p}(n+m)=v_{p}(n)$ if $v_{p}(n)<v_{p}(m)$ and $v_{p}(n+m)geq v_{p}(n)$ if $v_{p}(n)=v_{p}(m)$.
I would also like to know if this function has a specific name so that I can find more about it and read it. Thanks in advance for any help.
abstract-algebra
asked yesterday
Anzu
146
One of its names is "$p$-adic valuation". Regarding the proofs - what's the difficulty you see?
– metamorphy
yesterday
Indeed, which parts of the instantly accepted answer below was representing the least difficulty to you?
– Did
yesterday
I'm usually having troubles to find the right way to start. I could find the proper way to define $n$ and $m$ with the name of the function and was editing my post with a solution when the answer showed up. Thanks again.
– Anzu
yesterday
add a comment |
Happy new year. This seems like an easy to solve task but I'm stuck right now, so any help is appreciated.
Let $pin mathbb{Z}$ be a prime number. The function $v_{p}(n)geq 0$ is defined as $v_{p}(n):=e$ if $p^{e}|n$ and $p^{e+1} nmid n$. $v_{p}(0):=infty$ and $infty + e = e + infty = infty$ for all $ein mathbb{Z}$.
Show that i) $v_{p}(n*m)= v_{p}(n)+ v_{p}(m)$ for all $n,m in mathbb{Z}$ and
ii)$v_{p}(n+m)=v_{p}(n)$ if $v_{p}(n)<v_{p}(m)$ and $v_{p}(n+m)geq v_{p}(n)$ if $v_{p}(n)=v_{p}(m)$.
I would also like to know if this function has a specific name so that I can find more about it and read it. Thanks in advance for any help.
abstract-algebra
asked yesterday
Anzu
146
Happy new year. This seems like an easy to solve task but I'm stuck right now, so any help is appreciated.
Let $pin mathbb{Z}$ be a prime number. The function $v_{p}(n)geq 0$ is defined as $v_{p}(n):=e$ if $p^{e}|n$ and $p^{e+1} nmid n$. $v_{p}(0):=infty$ and $infty + e = e + infty = infty$ for all $ein mathbb{Z}$.
Show that i) $v_{p}(n*m)= v_{p}(n)+ v_{p}(m)$ for all $n,m in mathbb{Z}$ and
ii)$v_{p}(n+m)=v_{p}(n)$ if $v_{p}(n)<v_{p}(m)$ and $v_{p}(n+m)geq v_{p}(n)$ if $v_{p}(n)=v_{p}(m)$.
I would also like to know if this function has a specific name so that I can find more about it and read it. Thanks in advance for any help.
abstract-algebra
abstract-algebra
asked yesterday
Anzu
146
asked yesterday
Anzu
146
asked yesterday
Anzu
146
asked yesterday
Anzu
146
asked yesterday
Anzu
146
146
One of its names is "$p$-adic valuation". Regarding the proofs - what's the difficulty you see?
– metamorphy
yesterday
Indeed, which parts of the instantly accepted answer below was representing the least difficulty to you?
– Did
yesterday
I'm usually having troubles to find the right way to start. I could find the proper way to define $n$ and $m$ with the name of the function and was editing my post with a solution when the answer showed up. Thanks again.
– Anzu
yesterday
add a comment |
One of its names is "$p$-adic valuation". Regarding the proofs - what's the difficulty you see?
– metamorphy
yesterday
Indeed, which parts of the instantly accepted answer below was representing the least difficulty to you?
– Did
yesterday
I'm usually having troubles to find the right way to start. I could find the proper way to define $n$ and $m$ with the name of the function and was editing my post with a solution when the answer showed up. Thanks again.
– Anzu
yesterday
One of its names is "$p$-adic valuation". Regarding the proofs - what's the difficulty you see?
– metamorphy
yesterday
One of its names is "$p$-adic valuation". Regarding the proofs - what's the difficulty you see?
– metamorphy
yesterday
Indeed, which parts of the instantly accepted answer below was representing the least difficulty to you?
– Did
yesterday
Indeed, which parts of the instantly accepted answer below was representing the least difficulty to you?
– Did
yesterday
I'm usually having troubles to find the right way to start. I could find the proper way to define $n$ and $m$ with the name of the function and was editing my post with a solution when the answer showed up. Thanks again.
– Anzu
yesterday
I'm usually having troubles to find the right way to start. I could find the proper way to define $n$ and $m$ with the name of the function and was editing my post with a solution when the answer showed up. Thanks again.
– Anzu
yesterday
add a comment |
1 Answer
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This is called the $p$-adic valuation. I will write $v(n) = v_p(n)$ for the remainder of the answer.
Let $n,m in Bbb{Z}$ and write $n = p^{v(n)}a$, $m = p^{v(m)}b$ with $a,b$ coprime to $p$. Then $$nm = p^{v(m)+v(n)}ab$$ and $ab$ is coprime to $p$ so $v(nm) = v(n) + v(m)$. This proves $(i)$. For $(ii)$, without loss of generality suppose that $v(n) leq v(m)$. Then $$n + m = p^{v(n)}(a + p^{v(n)-v(m)}b)$$ If $v(n) > v(m)$ then $p nmid a + p^{v(n)-v(m)}b$ and hence $v(n+m) = v(n)$. If $v(n) = v(m)$ then $v(n+m) = v(n) + v(a + b)$ - in particular, since $v(a + b) geq 0$, $v(n+m) geq v(n)$.
Note that your $(ii)$ is often more succinctly written as $v(n + m) geq min(v(n),v(m))$ with equality if $v(n) neq v(m)$. Another thing to note is that $v_p$ can be uniquely extended to $Bbb{Q}$ be $v_p(frac{a}{b}) = v_p(a) - v_p(b)$.
As a final thing, the notion of a valuation on a field (or ring) generalises the notion of the $p$-adic valuation. If $K$ is a field, then a valuation $v$ on $K$ is a function $v : K rightarrow Bbb{R} cup infty$ satisfying:
$v(x) = infty$ if and only if $x = 0$
$v(xy) = v(x) + v(y)$ for all $x,y in K$
$v(x + y) geq min(v(x),v(y))$ for all $x,y in K$ with equality if and only if $v(x) neq v(y)$
So in fact all we have done is defined a function using the two properties that you've been asked to prove for the $p$-adic valuation. These valuations are incredibly powerful tools which are studied and used extensively in modern number theory.
answered yesterday
ODF
1,361510
Thank you for the detailed answer. I found the right way to define $n$ and $m$ with metamorphy's answer and was editing my post with a semi solution when your answer showed up.
– Anzu
yesterday
add a comment |
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1 Answer
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This is called the $p$-adic valuation. I will write $v(n) = v_p(n)$ for the remainder of the answer.
Let $n,m in Bbb{Z}$ and write $n = p^{v(n)}a$, $m = p^{v(m)}b$ with $a,b$ coprime to $p$. Then $$nm = p^{v(m)+v(n)}ab$$ and $ab$ is coprime to $p$ so $v(nm) = v(n) + v(m)$. This proves $(i)$. For $(ii)$, without loss of generality suppose that $v(n) leq v(m)$. Then $$n + m = p^{v(n)}(a + p^{v(n)-v(m)}b)$$ If $v(n) > v(m)$ then $p nmid a + p^{v(n)-v(m)}b$ and hence $v(n+m) = v(n)$. If $v(n) = v(m)$ then $v(n+m) = v(n) + v(a + b)$ - in particular, since $v(a + b) geq 0$, $v(n+m) geq v(n)$.
Note that your $(ii)$ is often more succinctly written as $v(n + m) geq min(v(n),v(m))$ with equality if $v(n) neq v(m)$. Another thing to note is that $v_p$ can be uniquely extended to $Bbb{Q}$ be $v_p(frac{a}{b}) = v_p(a) - v_p(b)$.
As a final thing, the notion of a valuation on a field (or ring) generalises the notion of the $p$-adic valuation. If $K$ is a field, then a valuation $v$ on $K$ is a function $v : K rightarrow Bbb{R} cup infty$ satisfying:
$v(x) = infty$ if and only if $x = 0$
$v(xy) = v(x) + v(y)$ for all $x,y in K$
$v(x + y) geq min(v(x),v(y))$ for all $x,y in K$ with equality if and only if $v(x) neq v(y)$
So in fact all we have done is defined a function using the two properties that you've been asked to prove for the $p$-adic valuation. These valuations are incredibly powerful tools which are studied and used extensively in modern number theory.
answered yesterday
ODF
1,361510
Thank you for the detailed answer. I found the right way to define $n$ and $m$ with metamorphy's answer and was editing my post with a semi solution when your answer showed up.
– Anzu
yesterday
add a comment |
This is called the $p$-adic valuation. I will write $v(n) = v_p(n)$ for the remainder of the answer.
Let $n,m in Bbb{Z}$ and write $n = p^{v(n)}a$, $m = p^{v(m)}b$ with $a,b$ coprime to $p$. Then $$nm = p^{v(m)+v(n)}ab$$ and $ab$ is coprime to $p$ so $v(nm) = v(n) + v(m)$. This proves $(i)$. For $(ii)$, without loss of generality suppose that $v(n) leq v(m)$. Then $$n + m = p^{v(n)}(a + p^{v(n)-v(m)}b)$$ If $v(n) > v(m)$ then $p nmid a + p^{v(n)-v(m)}b$ and hence $v(n+m) = v(n)$. If $v(n) = v(m)$ then $v(n+m) = v(n) + v(a + b)$ - in particular, since $v(a + b) geq 0$, $v(n+m) geq v(n)$.
Note that your $(ii)$ is often more succinctly written as $v(n + m) geq min(v(n),v(m))$ with equality if $v(n) neq v(m)$. Another thing to note is that $v_p$ can be uniquely extended to $Bbb{Q}$ be $v_p(frac{a}{b}) = v_p(a) - v_p(b)$.
As a final thing, the notion of a valuation on a field (or ring) generalises the notion of the $p$-adic valuation. If $K$ is a field, then a valuation $v$ on $K$ is a function $v : K rightarrow Bbb{R} cup infty$ satisfying:
$v(x) = infty$ if and only if $x = 0$
$v(xy) = v(x) + v(y)$ for all $x,y in K$
$v(x + y) geq min(v(x),v(y))$ for all $x,y in K$ with equality if and only if $v(x) neq v(y)$
So in fact all we have done is defined a function using the two properties that you've been asked to prove for the $p$-adic valuation. These valuations are incredibly powerful tools which are studied and used extensively in modern number theory.
answered yesterday
ODF
1,361510
Thank you for the detailed answer. I found the right way to define $n$ and $m$ with metamorphy's answer and was editing my post with a semi solution when your answer showed up.
– Anzu
yesterday
add a comment |
This is called the $p$-adic valuation. I will write $v(n) = v_p(n)$ for the remainder of the answer.
Let $n,m in Bbb{Z}$ and write $n = p^{v(n)}a$, $m = p^{v(m)}b$ with $a,b$ coprime to $p$. Then $$nm = p^{v(m)+v(n)}ab$$ and $ab$ is coprime to $p$ so $v(nm) = v(n) + v(m)$. This proves $(i)$. For $(ii)$, without loss of generality suppose that $v(n) leq v(m)$. Then $$n + m = p^{v(n)}(a + p^{v(n)-v(m)}b)$$ If $v(n) > v(m)$ then $p nmid a + p^{v(n)-v(m)}b$ and hence $v(n+m) = v(n)$. If $v(n) = v(m)$ then $v(n+m) = v(n) + v(a + b)$ - in particular, since $v(a + b) geq 0$, $v(n+m) geq v(n)$.
Note that your $(ii)$ is often more succinctly written as $v(n + m) geq min(v(n),v(m))$ with equality if $v(n) neq v(m)$. Another thing to note is that $v_p$ can be uniquely extended to $Bbb{Q}$ be $v_p(frac{a}{b}) = v_p(a) - v_p(b)$.
As a final thing, the notion of a valuation on a field (or ring) generalises the notion of the $p$-adic valuation. If $K$ is a field, then a valuation $v$ on $K$ is a function $v : K rightarrow Bbb{R} cup infty$ satisfying:
$v(x) = infty$ if and only if $x = 0$
$v(xy) = v(x) + v(y)$ for all $x,y in K$
$v(x + y) geq min(v(x),v(y))$ for all $x,y in K$ with equality if and only if $v(x) neq v(y)$
So in fact all we have done is defined a function using the two properties that you've been asked to prove for the $p$-adic valuation. These valuations are incredibly powerful tools which are studied and used extensively in modern number theory.
answered yesterday
ODF
1,361510
This is called the $p$-adic valuation. I will write $v(n) = v_p(n)$ for the remainder of the answer.
Let $n,m in Bbb{Z}$ and write $n = p^{v(n)}a$, $m = p^{v(m)}b$ with $a,b$ coprime to $p$. Then $$nm = p^{v(m)+v(n)}ab$$ and $ab$ is coprime to $p$ so $v(nm) = v(n) + v(m)$. This proves $(i)$. For $(ii)$, without loss of generality suppose that $v(n) leq v(m)$. Then $$n + m = p^{v(n)}(a + p^{v(n)-v(m)}b)$$ If $v(n) > v(m)$ then $p nmid a + p^{v(n)-v(m)}b$ and hence $v(n+m) = v(n)$. If $v(n) = v(m)$ then $v(n+m) = v(n) + v(a + b)$ - in particular, since $v(a + b) geq 0$, $v(n+m) geq v(n)$.
Note that your $(ii)$ is often more succinctly written as $v(n + m) geq min(v(n),v(m))$ with equality if $v(n) neq v(m)$. Another thing to note is that $v_p$ can be uniquely extended to $Bbb{Q}$ be $v_p(frac{a}{b}) = v_p(a) - v_p(b)$.
As a final thing, the notion of a valuation on a field (or ring) generalises the notion of the $p$-adic valuation. If $K$ is a field, then a valuation $v$ on $K$ is a function $v : K rightarrow Bbb{R} cup infty$ satisfying:
$v(x) = infty$ if and only if $x = 0$
$v(xy) = v(x) + v(y)$ for all $x,y in K$
$v(x + y) geq min(v(x),v(y))$ for all $x,y in K$ with equality if and only if $v(x) neq v(y)$
So in fact all we have done is defined a function using the two properties that you've been asked to prove for the $p$-adic valuation. These valuations are incredibly powerful tools which are studied and used extensively in modern number theory.
answered yesterday
ODF
1,361510
answered yesterday
ODF
1,361510
answered yesterday
ODF
1,361510
answered yesterday
ODF
1,361510
1,361510
Thank you for the detailed answer. I found the right way to define $n$ and $m$ with metamorphy's answer and was editing my post with a semi solution when your answer showed up.
– Anzu
yesterday
add a comment |
Thank you for the detailed answer. I found the right way to define $n$ and $m$ with metamorphy's answer and was editing my post with a semi solution when your answer showed up.
– Anzu
yesterday
Thank you for the detailed answer. I found the right way to define $n$ and $m$ with metamorphy's answer and was editing my post with a semi solution when your answer showed up.
– Anzu
yesterday
Thank you for the detailed answer. I found the right way to define $n$ and $m$ with metamorphy's answer and was editing my post with a semi solution when your answer showed up.
– Anzu
yesterday
add a comment |
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One of its names is "$p$-adic valuation". Regarding the proofs - what's the difficulty you see?
– metamorphy
yesterday
Indeed, which parts of the instantly accepted answer below was representing the least difficulty to you?
– Did
yesterday
I'm usually having troubles to find the right way to start. I could find the proper way to define $n$ and $m$ with the name of the function and was editing my post with a solution when the answer showed up. Thanks again.
– Anzu
yesterday