G.C.D. of two elements in a U.F.D. The 2019 Stack Overflow Developer Survey Results Are InThe height of a principal prime idealPolynomial rings — Inherited properties from coefficient ringAn integral domain with the factorization property and gcd for every two elements is a UFDIdeals in $mathbbZ[X]$ with three generators (and not with two)Let $a$ and $b$ be nonzero elements of the Unique Factorization Domain R. Prove that $a$ and $b$ have a least common multipleShow that there are finitely many different principal idealsSub-modules of free modules: twisting question a littleProper Ideal is a Product of Maximal IdealsConstructing a non principal ideal in a ring which is not a UFDShowing an Artinian ring, all of whose maximal ideals are principal, is a principal ideal ring.
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G.C.D. of two elements in a U.F.D.
The 2019 Stack Overflow Developer Survey Results Are InThe height of a principal prime idealPolynomial rings — Inherited properties from coefficient ringAn integral domain with the factorization property and gcd for every two elements is a UFDIdeals in $mathbbZ[X]$ with three generators (and not with two)Let $a$ and $b$ be nonzero elements of the Unique Factorization Domain R. Prove that $a$ and $b$ have a least common multipleShow that there are finitely many different principal idealsSub-modules of free modules: twisting question a littleProper Ideal is a Product of Maximal IdealsConstructing a non principal ideal in a ring which is not a UFDShowing an Artinian ring, all of whose maximal ideals are principal, is a principal ideal ring.
$begingroup$
The following Theorems are already known:
1)If $R$ is a U.F.D., then g.c.d. of any two elements exists.
2)In a P.I.D., two ideals $(a)$ and $(b)$ are co-maximal iff g.c.d$(a,b)=1.$
My Question:
If $R$ is a U.F.D. and $a,bin Rsetminus$$0$.
Define: $N:$$Rsetminus$$0$$rightarrow mathbbNcup$$0$.
Now, Suppose that g.c.d.$(N(a),N(b))>1.$
My Question:
I want to prove that g.c.d.$(a,b)neq 1.$
My Intuition:
It holds because:
By Theorem 1 above, g.c.d.$(a,b)$ exists.
Suppose, g.c.d.$(a,b)= 1.$
$implies (a),(b)$ are two co-maximal ideals of $R$. [ by Theorem 2 above]
$implies$ for any $ain (a)$ and $b in (b)$
$(N(a),N(b))=1$
$implies$ Contradiction to the assumption that g.c.d$(N(a),N(b))>1.$
Is my Intuition correct? Please provide me some Hints and Insights.
Another Question:
Can the above results be generalized for Integral domains that may not be U.F.D if g.c.d of two elements exists?
Note:
Here, $R$ contains $1neq 0.$
U.F.D.=Unique Factorization Domain
P.I.D.=Principal Ideal Domain
abstract-algebra unique-factorization-domains
edited Mar 30 at 10:22
Bernard
124k741117
asked Mar 30 at 9:30
KumarKumar
519
$endgroup$
|
show 1 more comment
$begingroup$
The following Theorems are already known:
1)If $R$ is a U.F.D., then g.c.d. of any two elements exists.
2)In a P.I.D., two ideals $(a)$ and $(b)$ are co-maximal iff g.c.d$(a,b)=1.$
My Question:
If $R$ is a U.F.D. and $a,bin Rsetminus$$0$.
Define: $N:$$Rsetminus$$0$$rightarrow mathbbNcup$$0$.
Now, Suppose that g.c.d.$(N(a),N(b))>1.$
My Question:
I want to prove that g.c.d.$(a,b)neq 1.$
My Intuition:
It holds because:
By Theorem 1 above, g.c.d.$(a,b)$ exists.
Suppose, g.c.d.$(a,b)= 1.$
$implies (a),(b)$ are two co-maximal ideals of $R$. [ by Theorem 2 above]
$implies$ for any $ain (a)$ and $b in (b)$
$(N(a),N(b))=1$
$implies$ Contradiction to the assumption that g.c.d$(N(a),N(b))>1.$
Is my Intuition correct? Please provide me some Hints and Insights.
Another Question:
Can the above results be generalized for Integral domains that may not be U.F.D if g.c.d of two elements exists?
Note:
Here, $R$ contains $1neq 0.$
U.F.D.=Unique Factorization Domain
P.I.D.=Principal Ideal Domain
abstract-algebra unique-factorization-domains
edited Mar 30 at 10:22
Bernard
124k741117
asked Mar 30 at 9:30
KumarKumar
519
$endgroup$
2
$begingroup$
$N$ is any old map from nonzero elements of $R$ to nonnegative integers?
$endgroup$
– Gerry Myerson
Mar 30 at 10:43
$begingroup$
Yes, But As per my thought, constant maps won't be useful ones for the above question. So, Please avoid those.
$endgroup$
– Kumar
Mar 30 at 12:51
$begingroup$
Please double check the statement of the exercise since it is likely you misread it (or missed some context that further defines $N$)
$endgroup$
– Bill Dubuque
Mar 30 at 23:34
$begingroup$
@BillDubuque it's my original question. I haven't taken it from some exercise. But the definition of $N$ is correct.
$endgroup$
– Kumar
Mar 31 at 1:25
$begingroup$
What motivated the question? Experience with some specific maps $N$?
$endgroup$
– Bill Dubuque
Apr 1 at 0:32
|
show 1 more comment
$begingroup$
The following Theorems are already known:
1)If $R$ is a U.F.D., then g.c.d. of any two elements exists.
2)In a P.I.D., two ideals $(a)$ and $(b)$ are co-maximal iff g.c.d$(a,b)=1.$
My Question:
If $R$ is a U.F.D. and $a,bin Rsetminus$$0$.
Define: $N:$$Rsetminus$$0$$rightarrow mathbbNcup$$0$.
Now, Suppose that g.c.d.$(N(a),N(b))>1.$
My Question:
I want to prove that g.c.d.$(a,b)neq 1.$
My Intuition:
It holds because:
By Theorem 1 above, g.c.d.$(a,b)$ exists.
Suppose, g.c.d.$(a,b)= 1.$
$implies (a),(b)$ are two co-maximal ideals of $R$. [ by Theorem 2 above]
$implies$ for any $ain (a)$ and $b in (b)$
$(N(a),N(b))=1$
$implies$ Contradiction to the assumption that g.c.d$(N(a),N(b))>1.$
Is my Intuition correct? Please provide me some Hints and Insights.
Another Question:
Can the above results be generalized for Integral domains that may not be U.F.D if g.c.d of two elements exists?
Note:
Here, $R$ contains $1neq 0.$
U.F.D.=Unique Factorization Domain
P.I.D.=Principal Ideal Domain
abstract-algebra unique-factorization-domains
edited Mar 30 at 10:22
Bernard
124k741117
asked Mar 30 at 9:30
KumarKumar
519
$endgroup$
The following Theorems are already known:
1)If $R$ is a U.F.D., then g.c.d. of any two elements exists.
2)In a P.I.D., two ideals $(a)$ and $(b)$ are co-maximal iff g.c.d$(a,b)=1.$
My Question:
If $R$ is a U.F.D. and $a,bin Rsetminus$$0$.
Define: $N:$$Rsetminus$$0$$rightarrow mathbbNcup$$0$.
Now, Suppose that g.c.d.$(N(a),N(b))>1.$
My Question:
I want to prove that g.c.d.$(a,b)neq 1.$
My Intuition:
It holds because:
By Theorem 1 above, g.c.d.$(a,b)$ exists.
Suppose, g.c.d.$(a,b)= 1.$
$implies (a),(b)$ are two co-maximal ideals of $R$. [ by Theorem 2 above]
$implies$ for any $ain (a)$ and $b in (b)$
$(N(a),N(b))=1$
$implies$ Contradiction to the assumption that g.c.d$(N(a),N(b))>1.$
Is my Intuition correct? Please provide me some Hints and Insights.
Another Question:
Can the above results be generalized for Integral domains that may not be U.F.D if g.c.d of two elements exists?
Note:
Here, $R$ contains $1neq 0.$
U.F.D.=Unique Factorization Domain
P.I.D.=Principal Ideal Domain
abstract-algebra unique-factorization-domains
abstract-algebra unique-factorization-domains
edited Mar 30 at 10:22
Bernard
124k741117
asked Mar 30 at 9:30
KumarKumar
519
edited Mar 30 at 10:22
Bernard
124k741117
asked Mar 30 at 9:30
KumarKumar
519
edited Mar 30 at 10:22
Bernard
124k741117
edited Mar 30 at 10:22
Bernard
124k741117
edited Mar 30 at 10:22
Bernard
124k741117
124k741117
asked Mar 30 at 9:30
KumarKumar
519
asked Mar 30 at 9:30
KumarKumar
519
asked Mar 30 at 9:30
KumarKumar
519
519
2
$begingroup$
$N$ is any old map from nonzero elements of $R$ to nonnegative integers?
$endgroup$
– Gerry Myerson
Mar 30 at 10:43
$begingroup$
Yes, But As per my thought, constant maps won't be useful ones for the above question. So, Please avoid those.
$endgroup$
– Kumar
Mar 30 at 12:51
$begingroup$
Please double check the statement of the exercise since it is likely you misread it (or missed some context that further defines $N$)
$endgroup$
– Bill Dubuque
Mar 30 at 23:34
$begingroup$
@BillDubuque it's my original question. I haven't taken it from some exercise. But the definition of $N$ is correct.
$endgroup$
– Kumar
Mar 31 at 1:25
$begingroup$
What motivated the question? Experience with some specific maps $N$?
$endgroup$
– Bill Dubuque
Apr 1 at 0:32
|
show 1 more comment
2
$begingroup$
$N$ is any old map from nonzero elements of $R$ to nonnegative integers?
$endgroup$
– Gerry Myerson
Mar 30 at 10:43
$begingroup$
Yes, But As per my thought, constant maps won't be useful ones for the above question. So, Please avoid those.
$endgroup$
– Kumar
Mar 30 at 12:51
$begingroup$
Please double check the statement of the exercise since it is likely you misread it (or missed some context that further defines $N$)
$endgroup$
– Bill Dubuque
Mar 30 at 23:34
$begingroup$
@BillDubuque it's my original question. I haven't taken it from some exercise. But the definition of $N$ is correct.
$endgroup$
– Kumar
Mar 31 at 1:25
$begingroup$
What motivated the question? Experience with some specific maps $N$?
$endgroup$
– Bill Dubuque
Apr 1 at 0:32
2
2
$begingroup$
$N$ is any old map from nonzero elements of $R$ to nonnegative integers?
$endgroup$
– Gerry Myerson
Mar 30 at 10:43
$begingroup$
$N$ is any old map from nonzero elements of $R$ to nonnegative integers?
$endgroup$
– Gerry Myerson
Mar 30 at 10:43
$begingroup$
Yes, But As per my thought, constant maps won't be useful ones for the above question. So, Please avoid those.
$endgroup$
– Kumar
Mar 30 at 12:51
$begingroup$
Yes, But As per my thought, constant maps won't be useful ones for the above question. So, Please avoid those.
$endgroup$
– Kumar
Mar 30 at 12:51
$begingroup$
Please double check the statement of the exercise since it is likely you misread it (or missed some context that further defines $N$)
$endgroup$
– Bill Dubuque
Mar 30 at 23:34
$begingroup$
Please double check the statement of the exercise since it is likely you misread it (or missed some context that further defines $N$)
$endgroup$
– Bill Dubuque
Mar 30 at 23:34
$begingroup$
@BillDubuque it's my original question. I haven't taken it from some exercise. But the definition of $N$ is correct.
$endgroup$
– Kumar
Mar 31 at 1:25
$begingroup$
@BillDubuque it's my original question. I haven't taken it from some exercise. But the definition of $N$ is correct.
$endgroup$
– Kumar
Mar 31 at 1:25
$begingroup$
What motivated the question? Experience with some specific maps $N$?
$endgroup$
– Bill Dubuque
Apr 1 at 0:32
$begingroup$
What motivated the question? Experience with some specific maps $N$?
$endgroup$
– Bill Dubuque
Apr 1 at 0:32
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
As OP confirms that any old nonconstant map $N$ will do, let $R$ be any UFD, let $a,b$ be any two nonzero elements of $R$ with $gcd(a,b)=1$, define $N$ by $N(a)=6$, $N(b)=10$, and $N(c)$ is whatever you like for $cne a$, $cne b$. Then $gcd(N(a),N(b))=gcd(6,10)=2ne1$, but $gcd(a,b)=1$.
answered Mar 30 at 21:28
Gerry MyersonGerry Myerson
148k8152306
$endgroup$
3
$begingroup$
Please excuse my rudeness, @Kumar, but if an example that accords with the definition you have established is not a counterexample to your intuition, that means that your definition does not adequately represent your intuition.
$endgroup$
– Lubin
Mar 31 at 2:11
1
$begingroup$
What I have written, Kumar, doesn't depend on the countability or otherwise of $R$. It works as long as $R$ has more than two elements.
$endgroup$
– Gerry Myerson
Mar 31 at 2:22
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
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active
oldest
votes
$begingroup$
As OP confirms that any old nonconstant map $N$ will do, let $R$ be any UFD, let $a,b$ be any two nonzero elements of $R$ with $gcd(a,b)=1$, define $N$ by $N(a)=6$, $N(b)=10$, and $N(c)$ is whatever you like for $cne a$, $cne b$. Then $gcd(N(a),N(b))=gcd(6,10)=2ne1$, but $gcd(a,b)=1$.
answered Mar 30 at 21:28
Gerry MyersonGerry Myerson
148k8152306
$endgroup$
3
$begingroup$
Please excuse my rudeness, @Kumar, but if an example that accords with the definition you have established is not a counterexample to your intuition, that means that your definition does not adequately represent your intuition.
$endgroup$
– Lubin
Mar 31 at 2:11
1
$begingroup$
What I have written, Kumar, doesn't depend on the countability or otherwise of $R$. It works as long as $R$ has more than two elements.
$endgroup$
– Gerry Myerson
Mar 31 at 2:22
add a comment |
$begingroup$
As OP confirms that any old nonconstant map $N$ will do, let $R$ be any UFD, let $a,b$ be any two nonzero elements of $R$ with $gcd(a,b)=1$, define $N$ by $N(a)=6$, $N(b)=10$, and $N(c)$ is whatever you like for $cne a$, $cne b$. Then $gcd(N(a),N(b))=gcd(6,10)=2ne1$, but $gcd(a,b)=1$.
answered Mar 30 at 21:28
Gerry MyersonGerry Myerson
148k8152306
$endgroup$
3
$begingroup$
Please excuse my rudeness, @Kumar, but if an example that accords with the definition you have established is not a counterexample to your intuition, that means that your definition does not adequately represent your intuition.
$endgroup$
– Lubin
Mar 31 at 2:11
1
$begingroup$
What I have written, Kumar, doesn't depend on the countability or otherwise of $R$. It works as long as $R$ has more than two elements.
$endgroup$
– Gerry Myerson
Mar 31 at 2:22
add a comment |
$begingroup$
As OP confirms that any old nonconstant map $N$ will do, let $R$ be any UFD, let $a,b$ be any two nonzero elements of $R$ with $gcd(a,b)=1$, define $N$ by $N(a)=6$, $N(b)=10$, and $N(c)$ is whatever you like for $cne a$, $cne b$. Then $gcd(N(a),N(b))=gcd(6,10)=2ne1$, but $gcd(a,b)=1$.
answered Mar 30 at 21:28
Gerry MyersonGerry Myerson
148k8152306
$endgroup$
As OP confirms that any old nonconstant map $N$ will do, let $R$ be any UFD, let $a,b$ be any two nonzero elements of $R$ with $gcd(a,b)=1$, define $N$ by $N(a)=6$, $N(b)=10$, and $N(c)$ is whatever you like for $cne a$, $cne b$. Then $gcd(N(a),N(b))=gcd(6,10)=2ne1$, but $gcd(a,b)=1$.
answered Mar 30 at 21:28
Gerry MyersonGerry Myerson
148k8152306
answered Mar 30 at 21:28
Gerry MyersonGerry Myerson
148k8152306
answered Mar 30 at 21:28
Gerry MyersonGerry Myerson
148k8152306
answered Mar 30 at 21:28
Gerry MyersonGerry Myerson
148k8152306
148k8152306
3
$begingroup$
Please excuse my rudeness, @Kumar, but if an example that accords with the definition you have established is not a counterexample to your intuition, that means that your definition does not adequately represent your intuition.
$endgroup$
– Lubin
Mar 31 at 2:11
1
$begingroup$
What I have written, Kumar, doesn't depend on the countability or otherwise of $R$. It works as long as $R$ has more than two elements.
$endgroup$
– Gerry Myerson
Mar 31 at 2:22
add a comment |
3
$begingroup$
Please excuse my rudeness, @Kumar, but if an example that accords with the definition you have established is not a counterexample to your intuition, that means that your definition does not adequately represent your intuition.
$endgroup$
– Lubin
Mar 31 at 2:11
1
$begingroup$
What I have written, Kumar, doesn't depend on the countability or otherwise of $R$. It works as long as $R$ has more than two elements.
$endgroup$
– Gerry Myerson
Mar 31 at 2:22
3
3
$begingroup$
Please excuse my rudeness, @Kumar, but if an example that accords with the definition you have established is not a counterexample to your intuition, that means that your definition does not adequately represent your intuition.
$endgroup$
– Lubin
Mar 31 at 2:11
$begingroup$
Please excuse my rudeness, @Kumar, but if an example that accords with the definition you have established is not a counterexample to your intuition, that means that your definition does not adequately represent your intuition.
$endgroup$
– Lubin
Mar 31 at 2:11
1
1
$begingroup$
What I have written, Kumar, doesn't depend on the countability or otherwise of $R$. It works as long as $R$ has more than two elements.
$endgroup$
– Gerry Myerson
Mar 31 at 2:22
$begingroup$
What I have written, Kumar, doesn't depend on the countability or otherwise of $R$. It works as long as $R$ has more than two elements.
$endgroup$
– Gerry Myerson
Mar 31 at 2:22
add a comment |
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2
$begingroup$
$N$ is any old map from nonzero elements of $R$ to nonnegative integers?
$endgroup$
– Gerry Myerson
Mar 30 at 10:43
$begingroup$
Yes, But As per my thought, constant maps won't be useful ones for the above question. So, Please avoid those.
$endgroup$
– Kumar
Mar 30 at 12:51
$begingroup$
Please double check the statement of the exercise since it is likely you misread it (or missed some context that further defines $N$)
$endgroup$
– Bill Dubuque
Mar 30 at 23:34
$begingroup$
@BillDubuque it's my original question. I haven't taken it from some exercise. But the definition of $N$ is correct.
$endgroup$
– Kumar
Mar 31 at 1:25
$begingroup$
What motivated the question? Experience with some specific maps $N$?
$endgroup$
– Bill Dubuque
Apr 1 at 0:32