When does $(a,b)=(gcd(a,b))$ hold?
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I had a look here to understand why $K[X,Y]$ is not a PID. So one of the conclusions was that $(x,y) neq (1) = gcd(x,y)$, but I thought that $(a,b)=gcd(a,b)$ was always true so obviously I was wrong. But when exactly does this relation hold then, if $ a,b in R$ and $R$ is a ring?
ring-theory ideals greatest-common-divisor principal-ideal-domains
edited Nov 19 at 13:22
amWhy
191k27223437
asked Nov 19 at 11:45
roi_saumon
31117
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up vote
0
down vote
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I had a look here to understand why $K[X,Y]$ is not a PID. So one of the conclusions was that $(x,y) neq (1) = gcd(x,y)$, but I thought that $(a,b)=gcd(a,b)$ was always true so obviously I was wrong. But when exactly does this relation hold then, if $ a,b in R$ and $R$ is a ring?
ring-theory ideals greatest-common-divisor principal-ideal-domains
edited Nov 19 at 13:22
amWhy
191k27223437
asked Nov 19 at 11:45
roi_saumon
31117
2
It holds iff $gcdleft(a,bright)$ is expressible as an $R$-linear combination of $a$ and $b$. It certainly holds when $R$ is a Euclidean domain, then, as the Euclidean algorithm furnishes you with such an expression. I know this is not an answer to your question, but I hope it helps towards finding one.
– Sam Streeter
Nov 19 at 11:53
For an application of this property, see math.stackexchange.com/questions/597543
– Watson
Nov 19 at 13:30
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I had a look here to understand why $K[X,Y]$ is not a PID. So one of the conclusions was that $(x,y) neq (1) = gcd(x,y)$, but I thought that $(a,b)=gcd(a,b)$ was always true so obviously I was wrong. But when exactly does this relation hold then, if $ a,b in R$ and $R$ is a ring?
ring-theory ideals greatest-common-divisor principal-ideal-domains
edited Nov 19 at 13:22
amWhy
191k27223437
asked Nov 19 at 11:45
roi_saumon
31117
I had a look here to understand why $K[X,Y]$ is not a PID. So one of the conclusions was that $(x,y) neq (1) = gcd(x,y)$, but I thought that $(a,b)=gcd(a,b)$ was always true so obviously I was wrong. But when exactly does this relation hold then, if $ a,b in R$ and $R$ is a ring?
ring-theory ideals greatest-common-divisor principal-ideal-domains
ring-theory ideals greatest-common-divisor principal-ideal-domains
edited Nov 19 at 13:22
amWhy
191k27223437
asked Nov 19 at 11:45
roi_saumon
31117
edited Nov 19 at 13:22
amWhy
191k27223437
asked Nov 19 at 11:45
roi_saumon
31117
edited Nov 19 at 13:22
amWhy
191k27223437
edited Nov 19 at 13:22
amWhy
191k27223437
edited Nov 19 at 13:22
amWhy
191k27223437
191k27223437
asked Nov 19 at 11:45
roi_saumon
31117
asked Nov 19 at 11:45
roi_saumon
31117
asked Nov 19 at 11:45
roi_saumon
31117
31117
2
It holds iff $gcdleft(a,bright)$ is expressible as an $R$-linear combination of $a$ and $b$. It certainly holds when $R$ is a Euclidean domain, then, as the Euclidean algorithm furnishes you with such an expression. I know this is not an answer to your question, but I hope it helps towards finding one.
– Sam Streeter
Nov 19 at 11:53
For an application of this property, see math.stackexchange.com/questions/597543
– Watson
Nov 19 at 13:30
add a comment |
2
It holds iff $gcdleft(a,bright)$ is expressible as an $R$-linear combination of $a$ and $b$. It certainly holds when $R$ is a Euclidean domain, then, as the Euclidean algorithm furnishes you with such an expression. I know this is not an answer to your question, but I hope it helps towards finding one.
– Sam Streeter
Nov 19 at 11:53
For an application of this property, see math.stackexchange.com/questions/597543
– Watson
Nov 19 at 13:30
2
2
It holds iff $gcdleft(a,bright)$ is expressible as an $R$-linear combination of $a$ and $b$. It certainly holds when $R$ is a Euclidean domain, then, as the Euclidean algorithm furnishes you with such an expression. I know this is not an answer to your question, but I hope it helps towards finding one.
– Sam Streeter
Nov 19 at 11:53
It holds iff $gcdleft(a,bright)$ is expressible as an $R$-linear combination of $a$ and $b$. It certainly holds when $R$ is a Euclidean domain, then, as the Euclidean algorithm furnishes you with such an expression. I know this is not an answer to your question, but I hope it helps towards finding one.
– Sam Streeter
Nov 19 at 11:53
For an application of this property, see math.stackexchange.com/questions/597543
– Watson
Nov 19 at 13:30
For an application of this property, see math.stackexchange.com/questions/597543
– Watson
Nov 19 at 13:30
add a comment |
1 Answer
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accepted
For commutative rings $R$, it holds if and only if $gcd(a,b) = alpha a + beta b$ for some $alpha,beta in R$. For if $gcd(a,b) = alpha a + beta b$ then clearly $(gcd(a,b)) subseteq (a,b)$, and on the other hand anything in $(a,b)$ has the form $ra + sb$ which is a multiple of $gcd(a,b)$ so lies in $(gcd(a,b))$.
Conversely if $(gcd(a,b)) = (a,b)$ then $gcd(a,b) in (a,b)$, so $gcd(a,b) = alpha a + beta b$ for some $alpha,beta in R$.
Every two elementa $a,b$ having a GCD which can be expressed as a linear combination of $a,b$ is equivalent to $R$ being a Bézout domain
answered Nov 19 at 11:52
Matthew Towers
7,24422244
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
For commutative rings $R$, it holds if and only if $gcd(a,b) = alpha a + beta b$ for some $alpha,beta in R$. For if $gcd(a,b) = alpha a + beta b$ then clearly $(gcd(a,b)) subseteq (a,b)$, and on the other hand anything in $(a,b)$ has the form $ra + sb$ which is a multiple of $gcd(a,b)$ so lies in $(gcd(a,b))$.
Conversely if $(gcd(a,b)) = (a,b)$ then $gcd(a,b) in (a,b)$, so $gcd(a,b) = alpha a + beta b$ for some $alpha,beta in R$.
Every two elementa $a,b$ having a GCD which can be expressed as a linear combination of $a,b$ is equivalent to $R$ being a Bézout domain
answered Nov 19 at 11:52
Matthew Towers
7,24422244
add a comment |
up vote
7
down vote
accepted
For commutative rings $R$, it holds if and only if $gcd(a,b) = alpha a + beta b$ for some $alpha,beta in R$. For if $gcd(a,b) = alpha a + beta b$ then clearly $(gcd(a,b)) subseteq (a,b)$, and on the other hand anything in $(a,b)$ has the form $ra + sb$ which is a multiple of $gcd(a,b)$ so lies in $(gcd(a,b))$.
Conversely if $(gcd(a,b)) = (a,b)$ then $gcd(a,b) in (a,b)$, so $gcd(a,b) = alpha a + beta b$ for some $alpha,beta in R$.
Every two elementa $a,b$ having a GCD which can be expressed as a linear combination of $a,b$ is equivalent to $R$ being a Bézout domain
answered Nov 19 at 11:52
Matthew Towers
7,24422244
add a comment |
up vote
7
down vote
accepted
up vote
7
down vote
accepted
For commutative rings $R$, it holds if and only if $gcd(a,b) = alpha a + beta b$ for some $alpha,beta in R$. For if $gcd(a,b) = alpha a + beta b$ then clearly $(gcd(a,b)) subseteq (a,b)$, and on the other hand anything in $(a,b)$ has the form $ra + sb$ which is a multiple of $gcd(a,b)$ so lies in $(gcd(a,b))$.
Conversely if $(gcd(a,b)) = (a,b)$ then $gcd(a,b) in (a,b)$, so $gcd(a,b) = alpha a + beta b$ for some $alpha,beta in R$.
Every two elementa $a,b$ having a GCD which can be expressed as a linear combination of $a,b$ is equivalent to $R$ being a Bézout domain
answered Nov 19 at 11:52
Matthew Towers
7,24422244
For commutative rings $R$, it holds if and only if $gcd(a,b) = alpha a + beta b$ for some $alpha,beta in R$. For if $gcd(a,b) = alpha a + beta b$ then clearly $(gcd(a,b)) subseteq (a,b)$, and on the other hand anything in $(a,b)$ has the form $ra + sb$ which is a multiple of $gcd(a,b)$ so lies in $(gcd(a,b))$.
Conversely if $(gcd(a,b)) = (a,b)$ then $gcd(a,b) in (a,b)$, so $gcd(a,b) = alpha a + beta b$ for some $alpha,beta in R$.
Every two elementa $a,b$ having a GCD which can be expressed as a linear combination of $a,b$ is equivalent to $R$ being a Bézout domain
answered Nov 19 at 11:52
Matthew Towers
7,24422244
edited Nov 19 at 12:22
answered Nov 19 at 11:52
Matthew Towers
7,24422244
answered Nov 19 at 11:52
Matthew Towers
7,24422244
answered Nov 19 at 11:52
Matthew Towers
7,24422244
7,24422244
add a comment |
add a comment |
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It holds iff $gcdleft(a,bright)$ is expressible as an $R$-linear combination of $a$ and $b$. It certainly holds when $R$ is a Euclidean domain, then, as the Euclidean algorithm furnishes you with such an expression. I know this is not an answer to your question, but I hope it helps towards finding one.
– Sam Streeter
Nov 19 at 11:53
For an application of this property, see math.stackexchange.com/questions/597543
– Watson
Nov 19 at 13:30