When do we have equality in the third axiom of valuations? [duplicate]
This question already has an answer here:
How do I compute the discrete valuation of the sum of two elements
1 answer
Let $R$ be a ring and $v: R to mathbb Z cup {+infty}$ a map that meets following axioms:
- $v(a) = +infty iff a=0$
- $v(ab) = v(a)+v(b)$
- $v(a+b) geq min{v(a),v(b)}$
I have to show that $v(a) neq v(b)$ implies $v(a+b) = min{v(a),v(b)}$, but I have no idea where to begin, can anyone give me some hints?
abstract-algebra valuation-theory
asked Oct 27 '15 at 9:39
flawr
11.2k32445
marked as duplicate by André 3000, Leucippus, Cesareo, Brahadeesh, max_zorn Nov 27 at 6:01
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
How do I compute the discrete valuation of the sum of two elements
1 answer
Let $R$ be a ring and $v: R to mathbb Z cup {+infty}$ a map that meets following axioms:
- $v(a) = +infty iff a=0$
- $v(ab) = v(a)+v(b)$
- $v(a+b) geq min{v(a),v(b)}$
I have to show that $v(a) neq v(b)$ implies $v(a+b) = min{v(a),v(b)}$, but I have no idea where to begin, can anyone give me some hints?
abstract-algebra valuation-theory
asked Oct 27 '15 at 9:39
flawr
11.2k32445
marked as duplicate by André 3000, Leucippus, Cesareo, Brahadeesh, max_zorn Nov 27 at 6:01
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
How do I compute the discrete valuation of the sum of two elements
1 answer
Let $R$ be a ring and $v: R to mathbb Z cup {+infty}$ a map that meets following axioms:
- $v(a) = +infty iff a=0$
- $v(ab) = v(a)+v(b)$
- $v(a+b) geq min{v(a),v(b)}$
I have to show that $v(a) neq v(b)$ implies $v(a+b) = min{v(a),v(b)}$, but I have no idea where to begin, can anyone give me some hints?
abstract-algebra valuation-theory
asked Oct 27 '15 at 9:39
flawr
11.2k32445
This question already has an answer here:
How do I compute the discrete valuation of the sum of two elements
1 answer
Let $R$ be a ring and $v: R to mathbb Z cup {+infty}$ a map that meets following axioms:
- $v(a) = +infty iff a=0$
- $v(ab) = v(a)+v(b)$
- $v(a+b) geq min{v(a),v(b)}$
I have to show that $v(a) neq v(b)$ implies $v(a+b) = min{v(a),v(b)}$, but I have no idea where to begin, can anyone give me some hints?
This question already has an answer here:
How do I compute the discrete valuation of the sum of two elements
1 answer
abstract-algebra valuation-theory
abstract-algebra valuation-theory
asked Oct 27 '15 at 9:39
flawr
11.2k32445
asked Oct 27 '15 at 9:39
flawr
11.2k32445
edited Nov 26 at 22:28
asked Oct 27 '15 at 9:39
flawr
11.2k32445
asked Oct 27 '15 at 9:39
flawr
11.2k32445
asked Oct 27 '15 at 9:39
flawr
11.2k32445
11.2k32445
marked as duplicate by André 3000, Leucippus, Cesareo, Brahadeesh, max_zorn Nov 27 at 6:01
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by André 3000, Leucippus, Cesareo, Brahadeesh, max_zorn Nov 27 at 6:01
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
add a comment |
1 Answer
1
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oldest
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Assume $v(a)>v(b)$ and $v(a+b)>v(b)$.
Then
$$ v(b)=v((a+b)+(-a))gemin{v(a+b),v(-a)}=min{v(a+b),v(a)}>v(b).$$
answered Oct 27 '15 at 9:49
Hagen von Eitzen
276k21268495
1
You should also prove that $v(-a)=v(a)$, but it's easy.
– egreg
Oct 27 '15 at 9:58
@egreg This would just have been my follow up question, I tried to prove this but I just do not get anywhere, can you give me another hint how to prove $v(-a) = v(a)$?
– flawr
Oct 27 '15 at 10:04
@hagenvoneitzen Thank you very much for your elegant answer so far!
– flawr
Oct 27 '15 at 10:05
1
@flawr Prove $v(1)=0$, then $v(-1)=0$ follows; therefore $v(-a)=v((-1)a)=v(-1)+v(a)$.
– egreg
Oct 27 '15 at 10:21
Thank you very much! This seems so easy when you know the solution, but it seems you just have to see it or you'll never see it =)
– flawr
Oct 27 '15 at 10:31
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Assume $v(a)>v(b)$ and $v(a+b)>v(b)$.
Then
$$ v(b)=v((a+b)+(-a))gemin{v(a+b),v(-a)}=min{v(a+b),v(a)}>v(b).$$
answered Oct 27 '15 at 9:49
Hagen von Eitzen
276k21268495
1
You should also prove that $v(-a)=v(a)$, but it's easy.
– egreg
Oct 27 '15 at 9:58
@egreg This would just have been my follow up question, I tried to prove this but I just do not get anywhere, can you give me another hint how to prove $v(-a) = v(a)$?
– flawr
Oct 27 '15 at 10:04
@hagenvoneitzen Thank you very much for your elegant answer so far!
– flawr
Oct 27 '15 at 10:05
1
@flawr Prove $v(1)=0$, then $v(-1)=0$ follows; therefore $v(-a)=v((-1)a)=v(-1)+v(a)$.
– egreg
Oct 27 '15 at 10:21
Thank you very much! This seems so easy when you know the solution, but it seems you just have to see it or you'll never see it =)
– flawr
Oct 27 '15 at 10:31
add a comment |
Assume $v(a)>v(b)$ and $v(a+b)>v(b)$.
Then
$$ v(b)=v((a+b)+(-a))gemin{v(a+b),v(-a)}=min{v(a+b),v(a)}>v(b).$$
answered Oct 27 '15 at 9:49
Hagen von Eitzen
276k21268495
1
You should also prove that $v(-a)=v(a)$, but it's easy.
– egreg
Oct 27 '15 at 9:58
@egreg This would just have been my follow up question, I tried to prove this but I just do not get anywhere, can you give me another hint how to prove $v(-a) = v(a)$?
– flawr
Oct 27 '15 at 10:04
@hagenvoneitzen Thank you very much for your elegant answer so far!
– flawr
Oct 27 '15 at 10:05
1
@flawr Prove $v(1)=0$, then $v(-1)=0$ follows; therefore $v(-a)=v((-1)a)=v(-1)+v(a)$.
– egreg
Oct 27 '15 at 10:21
Thank you very much! This seems so easy when you know the solution, but it seems you just have to see it or you'll never see it =)
– flawr
Oct 27 '15 at 10:31
add a comment |
Assume $v(a)>v(b)$ and $v(a+b)>v(b)$.
Then
$$ v(b)=v((a+b)+(-a))gemin{v(a+b),v(-a)}=min{v(a+b),v(a)}>v(b).$$
answered Oct 27 '15 at 9:49
Hagen von Eitzen
276k21268495
Assume $v(a)>v(b)$ and $v(a+b)>v(b)$.
Then
$$ v(b)=v((a+b)+(-a))gemin{v(a+b),v(-a)}=min{v(a+b),v(a)}>v(b).$$
answered Oct 27 '15 at 9:49
Hagen von Eitzen
276k21268495
answered Oct 27 '15 at 9:49
Hagen von Eitzen
276k21268495
answered Oct 27 '15 at 9:49
Hagen von Eitzen
276k21268495
answered Oct 27 '15 at 9:49
Hagen von Eitzen
276k21268495
276k21268495
1
You should also prove that $v(-a)=v(a)$, but it's easy.
– egreg
Oct 27 '15 at 9:58
@egreg This would just have been my follow up question, I tried to prove this but I just do not get anywhere, can you give me another hint how to prove $v(-a) = v(a)$?
– flawr
Oct 27 '15 at 10:04
@hagenvoneitzen Thank you very much for your elegant answer so far!
– flawr
Oct 27 '15 at 10:05
1
@flawr Prove $v(1)=0$, then $v(-1)=0$ follows; therefore $v(-a)=v((-1)a)=v(-1)+v(a)$.
– egreg
Oct 27 '15 at 10:21
Thank you very much! This seems so easy when you know the solution, but it seems you just have to see it or you'll never see it =)
– flawr
Oct 27 '15 at 10:31
add a comment |
1
You should also prove that $v(-a)=v(a)$, but it's easy.
– egreg
Oct 27 '15 at 9:58
@egreg This would just have been my follow up question, I tried to prove this but I just do not get anywhere, can you give me another hint how to prove $v(-a) = v(a)$?
– flawr
Oct 27 '15 at 10:04
@hagenvoneitzen Thank you very much for your elegant answer so far!
– flawr
Oct 27 '15 at 10:05
1
@flawr Prove $v(1)=0$, then $v(-1)=0$ follows; therefore $v(-a)=v((-1)a)=v(-1)+v(a)$.
– egreg
Oct 27 '15 at 10:21
Thank you very much! This seems so easy when you know the solution, but it seems you just have to see it or you'll never see it =)
– flawr
Oct 27 '15 at 10:31
1
1
You should also prove that $v(-a)=v(a)$, but it's easy.
– egreg
Oct 27 '15 at 9:58
You should also prove that $v(-a)=v(a)$, but it's easy.
– egreg
Oct 27 '15 at 9:58
@egreg This would just have been my follow up question, I tried to prove this but I just do not get anywhere, can you give me another hint how to prove $v(-a) = v(a)$?
– flawr
Oct 27 '15 at 10:04
@egreg This would just have been my follow up question, I tried to prove this but I just do not get anywhere, can you give me another hint how to prove $v(-a) = v(a)$?
– flawr
Oct 27 '15 at 10:04
@hagenvoneitzen Thank you very much for your elegant answer so far!
– flawr
Oct 27 '15 at 10:05
@hagenvoneitzen Thank you very much for your elegant answer so far!
– flawr
Oct 27 '15 at 10:05
1
1
@flawr Prove $v(1)=0$, then $v(-1)=0$ follows; therefore $v(-a)=v((-1)a)=v(-1)+v(a)$.
– egreg
Oct 27 '15 at 10:21
@flawr Prove $v(1)=0$, then $v(-1)=0$ follows; therefore $v(-a)=v((-1)a)=v(-1)+v(a)$.
– egreg
Oct 27 '15 at 10:21
Thank you very much! This seems so easy when you know the solution, but it seems you just have to see it or you'll never see it =)
– flawr
Oct 27 '15 at 10:31
Thank you very much! This seems so easy when you know the solution, but it seems you just have to see it or you'll never see it =)
– flawr
Oct 27 '15 at 10:31
add a comment |
