Mixing
When does $(a,b)=(gcd(a,b))$ hold?
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 21 hours ago
amWhy
191k27223437
asked 22 hours ago
roi_saumon
1717
add a comment |
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 21 hours ago
amWhy
191k27223437
asked 22 hours ago
roi_saumon
1717
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
22 hours ago
For an application of this property, see math.stackexchange.com/questions/597543
– Watson
20 hours ago
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 21 hours ago
amWhy
191k27223437
asked 22 hours ago
roi_saumon
1717
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 21 hours ago
amWhy
191k27223437
asked 22 hours ago
roi_saumon
1717
edited 21 hours ago
amWhy
191k27223437
asked 22 hours ago
roi_saumon
1717
edited 21 hours ago
amWhy
191k27223437
edited 21 hours ago
amWhy
191k27223437
edited 21 hours ago
amWhy
191k27223437
191k27223437
asked 22 hours ago
roi_saumon
1717
asked 22 hours ago
roi_saumon
1717
asked 22 hours ago
roi_saumon
1717
1717
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
22 hours ago
For an application of this property, see math.stackexchange.com/questions/597543
– Watson
20 hours ago
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
22 hours ago
For an application of this property, see math.stackexchange.com/questions/597543
– Watson
20 hours ago
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
22 hours ago
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
22 hours ago
For an application of this property, see math.stackexchange.com/questions/597543
– Watson
20 hours ago
For an application of this property, see math.stackexchange.com/questions/597543
– Watson
20 hours ago
add a comment |
1 Answer
1
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 22 hours ago
Matthew Towers
7,20922244
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 22 hours ago
Matthew Towers
7,20922244
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 22 hours ago
Matthew Towers
7,20922244
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 22 hours ago
Matthew Towers
7,20922244
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 22 hours ago
Matthew Towers
7,20922244
edited 22 hours ago
answered 22 hours ago
Matthew Towers
7,20922244
answered 22 hours ago
Matthew Towers
7,20922244
answered 22 hours ago
Matthew Towers
7,20922244
7,20922244
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004825%2fwhen-does-a-b-gcda-b-hold%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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
22 hours ago
For an application of this property, see math.stackexchange.com/questions/597543
– Watson
20 hours ago