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







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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
    22 hours ago










  • For an application of this property, see math.stackexchange.com/questions/597543
    – Watson
    20 hours ago















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







share|cite|improve this question




















  • 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













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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













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edited 21 hours ago









amWhy

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191k27223437










asked 22 hours ago









roi_saumon

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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














  • 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










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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






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    1 Answer
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    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






    share|cite|improve this answer



























      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






      share|cite|improve this answer

























        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






        share|cite|improve this answer














        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







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 22 hours ago

























        answered 22 hours ago









        Matthew Towers

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