$v_{p}(n):=e$ if $p^{e}|n$ and $p^{e+1} nmid n$. Show that $v_{p}(n*m)= v_{p}(n)+ v_{p}(m)$ and...












0














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







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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
















0














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







share|cite|improve this question






















  • 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














0












0








0







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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • 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










1 Answer
1






active

oldest

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1














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.






share|cite|improve this answer





















  • 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











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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.






share|cite|improve this answer





















  • 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
















1














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.






share|cite|improve this answer





















  • 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














1












1








1






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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • 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


















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