Mixing
Localization Preserves Euclidean Domains
$begingroup$
I'm wanting to prove that given a ring $A$ (by "ring" I mean a commutative ring with identity) and a multiplicative subset $S subset A$:
if $A$ is an Euclidean Domain, and $0 notin S$ then $S^{-1}A$ (localization of A at S) is also an Euclidean Domain.
I'm trying to produce an Euclidean Function in $S^{-1}A$ using the Euclidean Function $N:A rightarrow mathbb{N}$, that I already have from $A$ but I'm having trouble trying to define it in a way that works and verifies the properties an Euclidean Function must verify.
Does any one mind giving me hints? I don't really want a solution.. I would like to work it myself.
Thanks in advance. :)
abstract-algebra ring-theory euclidean-algorithm localization
asked Apr 18 '16 at 14:12
rierie
365110
$endgroup$
add a comment |
$begingroup$
I'm wanting to prove that given a ring $A$ (by "ring" I mean a commutative ring with identity) and a multiplicative subset $S subset A$:
if $A$ is an Euclidean Domain, and $0 notin S$ then $S^{-1}A$ (localization of A at S) is also an Euclidean Domain.
I'm trying to produce an Euclidean Function in $S^{-1}A$ using the Euclidean Function $N:A rightarrow mathbb{N}$, that I already have from $A$ but I'm having trouble trying to define it in a way that works and verifies the properties an Euclidean Function must verify.
Does any one mind giving me hints? I don't really want a solution.. I would like to work it myself.
Thanks in advance. :)
abstract-algebra ring-theory euclidean-algorithm localization
asked Apr 18 '16 at 14:12
rierie
365110
$endgroup$
add a comment |
$begingroup$
I'm wanting to prove that given a ring $A$ (by "ring" I mean a commutative ring with identity) and a multiplicative subset $S subset A$:
if $A$ is an Euclidean Domain, and $0 notin S$ then $S^{-1}A$ (localization of A at S) is also an Euclidean Domain.
I'm trying to produce an Euclidean Function in $S^{-1}A$ using the Euclidean Function $N:A rightarrow mathbb{N}$, that I already have from $A$ but I'm having trouble trying to define it in a way that works and verifies the properties an Euclidean Function must verify.
Does any one mind giving me hints? I don't really want a solution.. I would like to work it myself.
Thanks in advance. :)
abstract-algebra ring-theory euclidean-algorithm localization
asked Apr 18 '16 at 14:12
rierie
365110
$endgroup$
I'm wanting to prove that given a ring $A$ (by "ring" I mean a commutative ring with identity) and a multiplicative subset $S subset A$:
if $A$ is an Euclidean Domain, and $0 notin S$ then $S^{-1}A$ (localization of A at S) is also an Euclidean Domain.
I'm trying to produce an Euclidean Function in $S^{-1}A$ using the Euclidean Function $N:A rightarrow mathbb{N}$, that I already have from $A$ but I'm having trouble trying to define it in a way that works and verifies the properties an Euclidean Function must verify.
Does any one mind giving me hints? I don't really want a solution.. I would like to work it myself.
Thanks in advance. :)
abstract-algebra ring-theory euclidean-algorithm localization
abstract-algebra ring-theory euclidean-algorithm localization
asked Apr 18 '16 at 14:12
rierie
365110
asked Apr 18 '16 at 14:12
rierie
365110
asked Apr 18 '16 at 14:12
rierie
365110
asked Apr 18 '16 at 14:12
rierie
365110
asked Apr 18 '16 at 14:12
rierie
365110
365110
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In wikipedia's language, we may assume that $N$ satisfies $N(a)le N(ab)$ for $a,bin A$. Let us denote the candidate function for the localization by $N_Scolon (S^{-1}A)setminus{0}tomathbb N$.
We will also replace $S$ by its saturation, i.e. by $S_{mathrm{sat}}:={ ain A mid exists bin A: abin S}$. Notice that $S_{mathrm{sat}}^{-1}A=S^{-1}A$ because for any $ain S_{mathrm{sat}}$, we have $a^{-1}=frac{b}{s}in S^{-1}A$ where $bin A$ and $sin S$ are such that $s= ab$. Hence, assume henceforth that $S$ is saturated in the sense that for any $ain A$, if there exists some $bin A$ with $abin S$, then we have $ain S$.
Hint:
First, note that you may assume $N_S(s)=1$ for all $sin S$. Indeed, for any $ain S^{-1}A$, you have $N_S(s)le N_S(frac ascdot s)=N_S(a)$. Hence, $N_S(s)$ must be minimal. Argue similarly that $N_S(frac 1s)=1$ for $sin S$. Now use the fact that $A$ is a unique factorization domain.
Full spoiler, hover for reveal:
We first note that an Element $sin S$ can not have any prime factor in $Asetminus S$. Indeed, let $s=s_1cdots s_n$ be the prime factors of $s$. Then, $s_1in S$ and $s_2cdots s_nin S$ because $S$ is saturated. Proceed by induction.
For $frac{ta}{s}in S^{-1}A$, with $t,sin S$ and $a$ not divisible by any element of $S$, let $N_Sleft(frac{ta}{s}right):=N(a)$. This is well-defined because if $frac{t_1a_1}{s_1}=frac{t_2a_2}{s_2}$, then $s_1t_2a_2=s_2t_1a_1$. Since $a_1$ is not divisible by any element of $S$, it contains no prime factor in $S$. Since $s_2t_1in S$, it contains no prime factor in $Asetminus S$. This argument symmetrically works for $s_1t_2cdot a_1$ and it follows that $s_1t_2=s_2t_1$ and (more importantly) $a_1=a_2$.
Now we prove that $N_S$ yields a degree function turning $S^{-1}A$ into a Euclidean ring. Given $frac{t_1a_1}{s_1},frac{t_2a_2}{s_2}in S^{-1}A$, we perform division with remainder $s_2t_1a_1 = qa_2 + r$ such that either $r=0$ or $N(r)<N(a_2)$. Hence, $$frac{t_1a_1}{s_1} = frac{q}{s_1t_2}cdot frac{t_2a_2}{s_2} + frac{r}{s_1s_2}$$ and we have $N_{S}left(frac r{s_1s_2}right)=N(r)<N(a_2)=Nleft(frac{t_2a_2}{s_2}right)$.
edited Jan 31 at 19:29
dantopa
6,67442245
answered Apr 18 '16 at 14:40
Jesko HüttenhainJesko Hüttenhain
10.6k12358
$endgroup$
$begingroup$
The problem with making $N_S$ with that property is that first you must show that an arbitrary Euclidean function exists in $S^{-1}A$ and then you can create another one having that property (namely, $N(s) leq N(as)$, $(*)$), using the first one. As I see it, here you are assuming you can define an euclidean function $N_S$ that, a priori, satisfies property $(*)$ and then "construct" it using this property. The only problem I have is that, as I understand, this property requires first the existence of an euclidean function on $S^{-1}A$ and that's what we are wanting to prove, isn't it?
$endgroup$
– rie
Apr 18 '16 at 15:01
1
$begingroup$
No, my first statement refers to $N$, not $N_S$. Then I make some hyopthetical statements that are supposed to help you define $N_S$. After all, if any $N_S$ exists, then a multiplicative one exists as well. In fact, assuming $N$ multiplicative you can define $N_S$ which is multiplicative right away. You wanted a hint only. If you want the full solution, read the spoiler.
$endgroup$
– Jesko Hüttenhain
Apr 18 '16 at 15:03
$begingroup$
@rie: I changed the answer and corrected my mistake thanks to your insightful questions, indeed you have to make sure that $S$ is saturated for the argument to work.
$endgroup$
– Jesko Hüttenhain
Apr 19 '16 at 10:49
$begingroup$
Thank you very much Jesko, this helped me learn a lot.
$endgroup$
– rie
Apr 19 '16 at 14:58
$begingroup$
Me too actually. I knew the statement but had never thought about the proof before =).
$endgroup$
– Jesko Hüttenhain
Apr 19 '16 at 15:14
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In wikipedia's language, we may assume that $N$ satisfies $N(a)le N(ab)$ for $a,bin A$. Let us denote the candidate function for the localization by $N_Scolon (S^{-1}A)setminus{0}tomathbb N$.
We will also replace $S$ by its saturation, i.e. by $S_{mathrm{sat}}:={ ain A mid exists bin A: abin S}$. Notice that $S_{mathrm{sat}}^{-1}A=S^{-1}A$ because for any $ain S_{mathrm{sat}}$, we have $a^{-1}=frac{b}{s}in S^{-1}A$ where $bin A$ and $sin S$ are such that $s= ab$. Hence, assume henceforth that $S$ is saturated in the sense that for any $ain A$, if there exists some $bin A$ with $abin S$, then we have $ain S$.
Hint:
First, note that you may assume $N_S(s)=1$ for all $sin S$. Indeed, for any $ain S^{-1}A$, you have $N_S(s)le N_S(frac ascdot s)=N_S(a)$. Hence, $N_S(s)$ must be minimal. Argue similarly that $N_S(frac 1s)=1$ for $sin S$. Now use the fact that $A$ is a unique factorization domain.
Full spoiler, hover for reveal:
We first note that an Element $sin S$ can not have any prime factor in $Asetminus S$. Indeed, let $s=s_1cdots s_n$ be the prime factors of $s$. Then, $s_1in S$ and $s_2cdots s_nin S$ because $S$ is saturated. Proceed by induction.
For $frac{ta}{s}in S^{-1}A$, with $t,sin S$ and $a$ not divisible by any element of $S$, let $N_Sleft(frac{ta}{s}right):=N(a)$. This is well-defined because if $frac{t_1a_1}{s_1}=frac{t_2a_2}{s_2}$, then $s_1t_2a_2=s_2t_1a_1$. Since $a_1$ is not divisible by any element of $S$, it contains no prime factor in $S$. Since $s_2t_1in S$, it contains no prime factor in $Asetminus S$. This argument symmetrically works for $s_1t_2cdot a_1$ and it follows that $s_1t_2=s_2t_1$ and (more importantly) $a_1=a_2$.
Now we prove that $N_S$ yields a degree function turning $S^{-1}A$ into a Euclidean ring. Given $frac{t_1a_1}{s_1},frac{t_2a_2}{s_2}in S^{-1}A$, we perform division with remainder $s_2t_1a_1 = qa_2 + r$ such that either $r=0$ or $N(r)<N(a_2)$. Hence, $$frac{t_1a_1}{s_1} = frac{q}{s_1t_2}cdot frac{t_2a_2}{s_2} + frac{r}{s_1s_2}$$ and we have $N_{S}left(frac r{s_1s_2}right)=N(r)<N(a_2)=Nleft(frac{t_2a_2}{s_2}right)$.
edited Jan 31 at 19:29
dantopa
6,67442245
answered Apr 18 '16 at 14:40
Jesko HüttenhainJesko Hüttenhain
10.6k12358
$endgroup$
$begingroup$
The problem with making $N_S$ with that property is that first you must show that an arbitrary Euclidean function exists in $S^{-1}A$ and then you can create another one having that property (namely, $N(s) leq N(as)$, $(*)$), using the first one. As I see it, here you are assuming you can define an euclidean function $N_S$ that, a priori, satisfies property $(*)$ and then "construct" it using this property. The only problem I have is that, as I understand, this property requires first the existence of an euclidean function on $S^{-1}A$ and that's what we are wanting to prove, isn't it?
$endgroup$
– rie
Apr 18 '16 at 15:01
1
$begingroup$
No, my first statement refers to $N$, not $N_S$. Then I make some hyopthetical statements that are supposed to help you define $N_S$. After all, if any $N_S$ exists, then a multiplicative one exists as well. In fact, assuming $N$ multiplicative you can define $N_S$ which is multiplicative right away. You wanted a hint only. If you want the full solution, read the spoiler.
$endgroup$
– Jesko Hüttenhain
Apr 18 '16 at 15:03
$begingroup$
@rie: I changed the answer and corrected my mistake thanks to your insightful questions, indeed you have to make sure that $S$ is saturated for the argument to work.
$endgroup$
– Jesko Hüttenhain
Apr 19 '16 at 10:49
$begingroup$
Thank you very much Jesko, this helped me learn a lot.
$endgroup$
– rie
Apr 19 '16 at 14:58
$begingroup$
Me too actually. I knew the statement but had never thought about the proof before =).
$endgroup$
– Jesko Hüttenhain
Apr 19 '16 at 15:14
add a comment |
$begingroup$
In wikipedia's language, we may assume that $N$ satisfies $N(a)le N(ab)$ for $a,bin A$. Let us denote the candidate function for the localization by $N_Scolon (S^{-1}A)setminus{0}tomathbb N$.
We will also replace $S$ by its saturation, i.e. by $S_{mathrm{sat}}:={ ain A mid exists bin A: abin S}$. Notice that $S_{mathrm{sat}}^{-1}A=S^{-1}A$ because for any $ain S_{mathrm{sat}}$, we have $a^{-1}=frac{b}{s}in S^{-1}A$ where $bin A$ and $sin S$ are such that $s= ab$. Hence, assume henceforth that $S$ is saturated in the sense that for any $ain A$, if there exists some $bin A$ with $abin S$, then we have $ain S$.
Hint:
First, note that you may assume $N_S(s)=1$ for all $sin S$. Indeed, for any $ain S^{-1}A$, you have $N_S(s)le N_S(frac ascdot s)=N_S(a)$. Hence, $N_S(s)$ must be minimal. Argue similarly that $N_S(frac 1s)=1$ for $sin S$. Now use the fact that $A$ is a unique factorization domain.
Full spoiler, hover for reveal:
We first note that an Element $sin S$ can not have any prime factor in $Asetminus S$. Indeed, let $s=s_1cdots s_n$ be the prime factors of $s$. Then, $s_1in S$ and $s_2cdots s_nin S$ because $S$ is saturated. Proceed by induction.
For $frac{ta}{s}in S^{-1}A$, with $t,sin S$ and $a$ not divisible by any element of $S$, let $N_Sleft(frac{ta}{s}right):=N(a)$. This is well-defined because if $frac{t_1a_1}{s_1}=frac{t_2a_2}{s_2}$, then $s_1t_2a_2=s_2t_1a_1$. Since $a_1$ is not divisible by any element of $S$, it contains no prime factor in $S$. Since $s_2t_1in S$, it contains no prime factor in $Asetminus S$. This argument symmetrically works for $s_1t_2cdot a_1$ and it follows that $s_1t_2=s_2t_1$ and (more importantly) $a_1=a_2$.
Now we prove that $N_S$ yields a degree function turning $S^{-1}A$ into a Euclidean ring. Given $frac{t_1a_1}{s_1},frac{t_2a_2}{s_2}in S^{-1}A$, we perform division with remainder $s_2t_1a_1 = qa_2 + r$ such that either $r=0$ or $N(r)<N(a_2)$. Hence, $$frac{t_1a_1}{s_1} = frac{q}{s_1t_2}cdot frac{t_2a_2}{s_2} + frac{r}{s_1s_2}$$ and we have $N_{S}left(frac r{s_1s_2}right)=N(r)<N(a_2)=Nleft(frac{t_2a_2}{s_2}right)$.
edited Jan 31 at 19:29
dantopa
6,67442245
answered Apr 18 '16 at 14:40
Jesko HüttenhainJesko Hüttenhain
10.6k12358
$endgroup$
$begingroup$
The problem with making $N_S$ with that property is that first you must show that an arbitrary Euclidean function exists in $S^{-1}A$ and then you can create another one having that property (namely, $N(s) leq N(as)$, $(*)$), using the first one. As I see it, here you are assuming you can define an euclidean function $N_S$ that, a priori, satisfies property $(*)$ and then "construct" it using this property. The only problem I have is that, as I understand, this property requires first the existence of an euclidean function on $S^{-1}A$ and that's what we are wanting to prove, isn't it?
$endgroup$
– rie
Apr 18 '16 at 15:01
1
$begingroup$
No, my first statement refers to $N$, not $N_S$. Then I make some hyopthetical statements that are supposed to help you define $N_S$. After all, if any $N_S$ exists, then a multiplicative one exists as well. In fact, assuming $N$ multiplicative you can define $N_S$ which is multiplicative right away. You wanted a hint only. If you want the full solution, read the spoiler.
$endgroup$
– Jesko Hüttenhain
Apr 18 '16 at 15:03
$begingroup$
@rie: I changed the answer and corrected my mistake thanks to your insightful questions, indeed you have to make sure that $S$ is saturated for the argument to work.
$endgroup$
– Jesko Hüttenhain
Apr 19 '16 at 10:49
$begingroup$
Thank you very much Jesko, this helped me learn a lot.
$endgroup$
– rie
Apr 19 '16 at 14:58
$begingroup$
Me too actually. I knew the statement but had never thought about the proof before =).
$endgroup$
– Jesko Hüttenhain
Apr 19 '16 at 15:14
add a comment |
$begingroup$
In wikipedia's language, we may assume that $N$ satisfies $N(a)le N(ab)$ for $a,bin A$. Let us denote the candidate function for the localization by $N_Scolon (S^{-1}A)setminus{0}tomathbb N$.
We will also replace $S$ by its saturation, i.e. by $S_{mathrm{sat}}:={ ain A mid exists bin A: abin S}$. Notice that $S_{mathrm{sat}}^{-1}A=S^{-1}A$ because for any $ain S_{mathrm{sat}}$, we have $a^{-1}=frac{b}{s}in S^{-1}A$ where $bin A$ and $sin S$ are such that $s= ab$. Hence, assume henceforth that $S$ is saturated in the sense that for any $ain A$, if there exists some $bin A$ with $abin S$, then we have $ain S$.
Hint:
First, note that you may assume $N_S(s)=1$ for all $sin S$. Indeed, for any $ain S^{-1}A$, you have $N_S(s)le N_S(frac ascdot s)=N_S(a)$. Hence, $N_S(s)$ must be minimal. Argue similarly that $N_S(frac 1s)=1$ for $sin S$. Now use the fact that $A$ is a unique factorization domain.
Full spoiler, hover for reveal:
We first note that an Element $sin S$ can not have any prime factor in $Asetminus S$. Indeed, let $s=s_1cdots s_n$ be the prime factors of $s$. Then, $s_1in S$ and $s_2cdots s_nin S$ because $S$ is saturated. Proceed by induction.
For $frac{ta}{s}in S^{-1}A$, with $t,sin S$ and $a$ not divisible by any element of $S$, let $N_Sleft(frac{ta}{s}right):=N(a)$. This is well-defined because if $frac{t_1a_1}{s_1}=frac{t_2a_2}{s_2}$, then $s_1t_2a_2=s_2t_1a_1$. Since $a_1$ is not divisible by any element of $S$, it contains no prime factor in $S$. Since $s_2t_1in S$, it contains no prime factor in $Asetminus S$. This argument symmetrically works for $s_1t_2cdot a_1$ and it follows that $s_1t_2=s_2t_1$ and (more importantly) $a_1=a_2$.
Now we prove that $N_S$ yields a degree function turning $S^{-1}A$ into a Euclidean ring. Given $frac{t_1a_1}{s_1},frac{t_2a_2}{s_2}in S^{-1}A$, we perform division with remainder $s_2t_1a_1 = qa_2 + r$ such that either $r=0$ or $N(r)<N(a_2)$. Hence, $$frac{t_1a_1}{s_1} = frac{q}{s_1t_2}cdot frac{t_2a_2}{s_2} + frac{r}{s_1s_2}$$ and we have $N_{S}left(frac r{s_1s_2}right)=N(r)<N(a_2)=Nleft(frac{t_2a_2}{s_2}right)$.
edited Jan 31 at 19:29
dantopa
6,67442245
answered Apr 18 '16 at 14:40
Jesko HüttenhainJesko Hüttenhain
10.6k12358
$endgroup$
In wikipedia's language, we may assume that $N$ satisfies $N(a)le N(ab)$ for $a,bin A$. Let us denote the candidate function for the localization by $N_Scolon (S^{-1}A)setminus{0}tomathbb N$.
We will also replace $S$ by its saturation, i.e. by $S_{mathrm{sat}}:={ ain A mid exists bin A: abin S}$. Notice that $S_{mathrm{sat}}^{-1}A=S^{-1}A$ because for any $ain S_{mathrm{sat}}$, we have $a^{-1}=frac{b}{s}in S^{-1}A$ where $bin A$ and $sin S$ are such that $s= ab$. Hence, assume henceforth that $S$ is saturated in the sense that for any $ain A$, if there exists some $bin A$ with $abin S$, then we have $ain S$.
Hint:
First, note that you may assume $N_S(s)=1$ for all $sin S$. Indeed, for any $ain S^{-1}A$, you have $N_S(s)le N_S(frac ascdot s)=N_S(a)$. Hence, $N_S(s)$ must be minimal. Argue similarly that $N_S(frac 1s)=1$ for $sin S$. Now use the fact that $A$ is a unique factorization domain.
Full spoiler, hover for reveal:
We first note that an Element $sin S$ can not have any prime factor in $Asetminus S$. Indeed, let $s=s_1cdots s_n$ be the prime factors of $s$. Then, $s_1in S$ and $s_2cdots s_nin S$ because $S$ is saturated. Proceed by induction.
For $frac{ta}{s}in S^{-1}A$, with $t,sin S$ and $a$ not divisible by any element of $S$, let $N_Sleft(frac{ta}{s}right):=N(a)$. This is well-defined because if $frac{t_1a_1}{s_1}=frac{t_2a_2}{s_2}$, then $s_1t_2a_2=s_2t_1a_1$. Since $a_1$ is not divisible by any element of $S$, it contains no prime factor in $S$. Since $s_2t_1in S$, it contains no prime factor in $Asetminus S$. This argument symmetrically works for $s_1t_2cdot a_1$ and it follows that $s_1t_2=s_2t_1$ and (more importantly) $a_1=a_2$.
Now we prove that $N_S$ yields a degree function turning $S^{-1}A$ into a Euclidean ring. Given $frac{t_1a_1}{s_1},frac{t_2a_2}{s_2}in S^{-1}A$, we perform division with remainder $s_2t_1a_1 = qa_2 + r$ such that either $r=0$ or $N(r)<N(a_2)$. Hence, $$frac{t_1a_1}{s_1} = frac{q}{s_1t_2}cdot frac{t_2a_2}{s_2} + frac{r}{s_1s_2}$$ and we have $N_{S}left(frac r{s_1s_2}right)=N(r)<N(a_2)=Nleft(frac{t_2a_2}{s_2}right)$.
edited Jan 31 at 19:29
dantopa
6,67442245
answered Apr 18 '16 at 14:40
Jesko HüttenhainJesko Hüttenhain
10.6k12358
edited Jan 31 at 19:29
dantopa
6,67442245
edited Jan 31 at 19:29
dantopa
6,67442245
edited Jan 31 at 19:29
dantopa
6,67442245
6,67442245
answered Apr 18 '16 at 14:40
Jesko HüttenhainJesko Hüttenhain
10.6k12358
answered Apr 18 '16 at 14:40
Jesko HüttenhainJesko Hüttenhain
10.6k12358
answered Apr 18 '16 at 14:40
Jesko HüttenhainJesko Hüttenhain
10.6k12358
10.6k12358
$begingroup$
The problem with making $N_S$ with that property is that first you must show that an arbitrary Euclidean function exists in $S^{-1}A$ and then you can create another one having that property (namely, $N(s) leq N(as)$, $(*)$), using the first one. As I see it, here you are assuming you can define an euclidean function $N_S$ that, a priori, satisfies property $(*)$ and then "construct" it using this property. The only problem I have is that, as I understand, this property requires first the existence of an euclidean function on $S^{-1}A$ and that's what we are wanting to prove, isn't it?
$endgroup$
– rie
Apr 18 '16 at 15:01
1
$begingroup$
No, my first statement refers to $N$, not $N_S$. Then I make some hyopthetical statements that are supposed to help you define $N_S$. After all, if any $N_S$ exists, then a multiplicative one exists as well. In fact, assuming $N$ multiplicative you can define $N_S$ which is multiplicative right away. You wanted a hint only. If you want the full solution, read the spoiler.
$endgroup$
– Jesko Hüttenhain
Apr 18 '16 at 15:03
$begingroup$
@rie: I changed the answer and corrected my mistake thanks to your insightful questions, indeed you have to make sure that $S$ is saturated for the argument to work.
$endgroup$
– Jesko Hüttenhain
Apr 19 '16 at 10:49
$begingroup$
Thank you very much Jesko, this helped me learn a lot.
$endgroup$
– rie
Apr 19 '16 at 14:58
$begingroup$
Me too actually. I knew the statement but had never thought about the proof before =).
$endgroup$
– Jesko Hüttenhain
Apr 19 '16 at 15:14
add a comment |
$begingroup$
The problem with making $N_S$ with that property is that first you must show that an arbitrary Euclidean function exists in $S^{-1}A$ and then you can create another one having that property (namely, $N(s) leq N(as)$, $(*)$), using the first one. As I see it, here you are assuming you can define an euclidean function $N_S$ that, a priori, satisfies property $(*)$ and then "construct" it using this property. The only problem I have is that, as I understand, this property requires first the existence of an euclidean function on $S^{-1}A$ and that's what we are wanting to prove, isn't it?
$endgroup$
– rie
Apr 18 '16 at 15:01
1
$begingroup$
No, my first statement refers to $N$, not $N_S$. Then I make some hyopthetical statements that are supposed to help you define $N_S$. After all, if any $N_S$ exists, then a multiplicative one exists as well. In fact, assuming $N$ multiplicative you can define $N_S$ which is multiplicative right away. You wanted a hint only. If you want the full solution, read the spoiler.
$endgroup$
– Jesko Hüttenhain
Apr 18 '16 at 15:03
$begingroup$
@rie: I changed the answer and corrected my mistake thanks to your insightful questions, indeed you have to make sure that $S$ is saturated for the argument to work.
$endgroup$
– Jesko Hüttenhain
Apr 19 '16 at 10:49
$begingroup$
Thank you very much Jesko, this helped me learn a lot.
$endgroup$
– rie
Apr 19 '16 at 14:58
$begingroup$
Me too actually. I knew the statement but had never thought about the proof before =).
$endgroup$
– Jesko Hüttenhain
Apr 19 '16 at 15:14
$begingroup$
The problem with making $N_S$ with that property is that first you must show that an arbitrary Euclidean function exists in $S^{-1}A$ and then you can create another one having that property (namely, $N(s) leq N(as)$, $(*)$), using the first one. As I see it, here you are assuming you can define an euclidean function $N_S$ that, a priori, satisfies property $(*)$ and then "construct" it using this property. The only problem I have is that, as I understand, this property requires first the existence of an euclidean function on $S^{-1}A$ and that's what we are wanting to prove, isn't it?
$endgroup$
– rie
Apr 18 '16 at 15:01
$begingroup$
The problem with making $N_S$ with that property is that first you must show that an arbitrary Euclidean function exists in $S^{-1}A$ and then you can create another one having that property (namely, $N(s) leq N(as)$, $(*)$), using the first one. As I see it, here you are assuming you can define an euclidean function $N_S$ that, a priori, satisfies property $(*)$ and then "construct" it using this property. The only problem I have is that, as I understand, this property requires first the existence of an euclidean function on $S^{-1}A$ and that's what we are wanting to prove, isn't it?
$endgroup$
– rie
Apr 18 '16 at 15:01
1
1
$begingroup$
No, my first statement refers to $N$, not $N_S$. Then I make some hyopthetical statements that are supposed to help you define $N_S$. After all, if any $N_S$ exists, then a multiplicative one exists as well. In fact, assuming $N$ multiplicative you can define $N_S$ which is multiplicative right away. You wanted a hint only. If you want the full solution, read the spoiler.
$endgroup$
– Jesko Hüttenhain
Apr 18 '16 at 15:03
$begingroup$
No, my first statement refers to $N$, not $N_S$. Then I make some hyopthetical statements that are supposed to help you define $N_S$. After all, if any $N_S$ exists, then a multiplicative one exists as well. In fact, assuming $N$ multiplicative you can define $N_S$ which is multiplicative right away. You wanted a hint only. If you want the full solution, read the spoiler.
$endgroup$
– Jesko Hüttenhain
Apr 18 '16 at 15:03
$begingroup$
@rie: I changed the answer and corrected my mistake thanks to your insightful questions, indeed you have to make sure that $S$ is saturated for the argument to work.
$endgroup$
– Jesko Hüttenhain
Apr 19 '16 at 10:49
$begingroup$
@rie: I changed the answer and corrected my mistake thanks to your insightful questions, indeed you have to make sure that $S$ is saturated for the argument to work.
$endgroup$
– Jesko Hüttenhain
Apr 19 '16 at 10:49
$begingroup$
Thank you very much Jesko, this helped me learn a lot.
$endgroup$
– rie
Apr 19 '16 at 14:58
$begingroup$
Thank you very much Jesko, this helped me learn a lot.
$endgroup$
– rie
Apr 19 '16 at 14:58
$begingroup$
Me too actually. I knew the statement but had never thought about the proof before =).
$endgroup$
– Jesko Hüttenhain
Apr 19 '16 at 15:14
$begingroup$
Me too actually. I knew the statement but had never thought about the proof before =).
$endgroup$
– Jesko Hüttenhain
Apr 19 '16 at 15:14
add a comment |
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