- #1
snoble
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I am wondering about the general definition of relatively prime in terms of commutative rings.
Specifically if I have my first definition being that given a commutative ring R if r_1 and r_2 are relatively prime then if [tex]r_1 k\in r_2R[/tex] then [tex]k \in r_2 R[/tex]. And vice versa. In other words if [tex]r_2 | kr_1[/tex] then [tex]r_2 | k[/tex].
My second definition is if r_1 and r_2 are relatively prime then [tex]\exists x,y\in R[/tex] such that [tex]xr_1 +yr_2 = 1[/tex] (yes I'm assuming all rings have a unit)
So I'm wondering for what types of commutative rings are these two definitions equivalent (I'm guessing always or almost always) and where can I find a proof of that. Notice the Euclidean algorithm depends on an ordering which you may not have here (at least as I know the algorithm).
Another way to think of the problem is how do I show given the first definition that <r_1> and <r_2> are comaximal: ie [tex]<r_1>+<r_2>=R[/tex]. This is the actual problem I've been thinking about.
Thanks,
Steven
Specifically if I have my first definition being that given a commutative ring R if r_1 and r_2 are relatively prime then if [tex]r_1 k\in r_2R[/tex] then [tex]k \in r_2 R[/tex]. And vice versa. In other words if [tex]r_2 | kr_1[/tex] then [tex]r_2 | k[/tex].
My second definition is if r_1 and r_2 are relatively prime then [tex]\exists x,y\in R[/tex] such that [tex]xr_1 +yr_2 = 1[/tex] (yes I'm assuming all rings have a unit)
So I'm wondering for what types of commutative rings are these two definitions equivalent (I'm guessing always or almost always) and where can I find a proof of that. Notice the Euclidean algorithm depends on an ordering which you may not have here (at least as I know the algorithm).
Another way to think of the problem is how do I show given the first definition that <r_1> and <r_2> are comaximal: ie [tex]<r_1>+<r_2>=R[/tex]. This is the actual problem I've been thinking about.
Thanks,
Steven