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In which rings does this multiplicative analogue of Bézout's theorem hold?


International Zhautykov Olympiad 2019 problem 6Dedekind domain with a finite number of prime ideals is principalClassifying Unital Commutative Rings of Order $p^2$$mathbbZ[X]$ is noetherianProve that if R and S are nonzero rings then $Rtimes S$ is never a field.Consequence of Chinese Remainder TheoremConditions for $(I+K)cap (J+K) = Icap J +K $ to hold for ideals of ring $R$Does the Dorroh Extension Theorem simplify ring theory to the study of rings with identity?Chinese remainder theorem for rings applied to Z/nZDo both versions (invariant and primary) of the Fundamental Theorem for Finitely Generated Abelian Groups hold at the same time?Analogue for Group Rings for multiplicative sets













2












$begingroup$


When I was thinking about this question: International Zhautykov Olympiad 2019 problem 6 I learned that, when $0 < a,b$ are integers that divide $n >0$ and $d$ is their $gcd$, then if for a divisor $m mid n$, $K_m$ denotes the kernel of the reduction map $(mathbb Z/n)^times to (mathbb Z/m)^times$, we have $K_aK_b = K_d$.



Note that Bézout's theorem (mod $n$) says precisely that the same is true for the reduction maps $(mathbb Z/n, +)to (mathbb Z/m, +)$.



The proof I have for the multiplicative version uses the Chinese remainder theorem: do the case where $n$ is a prime power first, then glue the solutions for a given element of $K_d$ using the CRT. Slightly more generally, it shows that any PID $R$ satisfies the following:




If $0 neq N subset A, B$ are ideals of $R$, then $$1+(A+B) = (1+A)(1+B) + N$$




I'm a bit dissatisfied with the proof for PID's, and would like to understand the equality more conceptually. So my question is:
Which other commutative rings $R$ satisfy this property? Is there a characterization?










share|cite|improve this question









$endgroup$







  • 1




    $begingroup$
    Is the left set $1+a+bmid ain A, bin B$ and the right set is $(1+a)(1+b)+nmid a,in A, bin B,nin N$? I ask because i'm not sure how you meant to define the product $(1+A)(1+B)$.
    $endgroup$
    – rschwieb
    Mar 17 at 0:42










  • $begingroup$
    @rschwieb Yes. I denote $a * B = a*b : b in B $ and $A*B = a*b : a in A, b in B $ for an operation $*$ - which indeed kills distributivity.
    $endgroup$
    – rabota
    Mar 17 at 10:20















2












$begingroup$


When I was thinking about this question: International Zhautykov Olympiad 2019 problem 6 I learned that, when $0 < a,b$ are integers that divide $n >0$ and $d$ is their $gcd$, then if for a divisor $m mid n$, $K_m$ denotes the kernel of the reduction map $(mathbb Z/n)^times to (mathbb Z/m)^times$, we have $K_aK_b = K_d$.



Note that Bézout's theorem (mod $n$) says precisely that the same is true for the reduction maps $(mathbb Z/n, +)to (mathbb Z/m, +)$.



The proof I have for the multiplicative version uses the Chinese remainder theorem: do the case where $n$ is a prime power first, then glue the solutions for a given element of $K_d$ using the CRT. Slightly more generally, it shows that any PID $R$ satisfies the following:




If $0 neq N subset A, B$ are ideals of $R$, then $$1+(A+B) = (1+A)(1+B) + N$$




I'm a bit dissatisfied with the proof for PID's, and would like to understand the equality more conceptually. So my question is:
Which other commutative rings $R$ satisfy this property? Is there a characterization?










share|cite|improve this question









$endgroup$







  • 1




    $begingroup$
    Is the left set $1+a+bmid ain A, bin B$ and the right set is $(1+a)(1+b)+nmid a,in A, bin B,nin N$? I ask because i'm not sure how you meant to define the product $(1+A)(1+B)$.
    $endgroup$
    – rschwieb
    Mar 17 at 0:42










  • $begingroup$
    @rschwieb Yes. I denote $a * B = a*b : b in B $ and $A*B = a*b : a in A, b in B $ for an operation $*$ - which indeed kills distributivity.
    $endgroup$
    – rabota
    Mar 17 at 10:20













2












2








2





$begingroup$


When I was thinking about this question: International Zhautykov Olympiad 2019 problem 6 I learned that, when $0 < a,b$ are integers that divide $n >0$ and $d$ is their $gcd$, then if for a divisor $m mid n$, $K_m$ denotes the kernel of the reduction map $(mathbb Z/n)^times to (mathbb Z/m)^times$, we have $K_aK_b = K_d$.



Note that Bézout's theorem (mod $n$) says precisely that the same is true for the reduction maps $(mathbb Z/n, +)to (mathbb Z/m, +)$.



The proof I have for the multiplicative version uses the Chinese remainder theorem: do the case where $n$ is a prime power first, then glue the solutions for a given element of $K_d$ using the CRT. Slightly more generally, it shows that any PID $R$ satisfies the following:




If $0 neq N subset A, B$ are ideals of $R$, then $$1+(A+B) = (1+A)(1+B) + N$$




I'm a bit dissatisfied with the proof for PID's, and would like to understand the equality more conceptually. So my question is:
Which other commutative rings $R$ satisfy this property? Is there a characterization?










share|cite|improve this question









$endgroup$




When I was thinking about this question: International Zhautykov Olympiad 2019 problem 6 I learned that, when $0 < a,b$ are integers that divide $n >0$ and $d$ is their $gcd$, then if for a divisor $m mid n$, $K_m$ denotes the kernel of the reduction map $(mathbb Z/n)^times to (mathbb Z/m)^times$, we have $K_aK_b = K_d$.



Note that Bézout's theorem (mod $n$) says precisely that the same is true for the reduction maps $(mathbb Z/n, +)to (mathbb Z/m, +)$.



The proof I have for the multiplicative version uses the Chinese remainder theorem: do the case where $n$ is a prime power first, then glue the solutions for a given element of $K_d$ using the CRT. Slightly more generally, it shows that any PID $R$ satisfies the following:




If $0 neq N subset A, B$ are ideals of $R$, then $$1+(A+B) = (1+A)(1+B) + N$$




I'm a bit dissatisfied with the proof for PID's, and would like to understand the equality more conceptually. So my question is:
Which other commutative rings $R$ satisfy this property? Is there a characterization?







ring-theory commutative-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 16 at 9:55









rabotarabota

14.3k32782




14.3k32782







  • 1




    $begingroup$
    Is the left set $1+a+bmid ain A, bin B$ and the right set is $(1+a)(1+b)+nmid a,in A, bin B,nin N$? I ask because i'm not sure how you meant to define the product $(1+A)(1+B)$.
    $endgroup$
    – rschwieb
    Mar 17 at 0:42










  • $begingroup$
    @rschwieb Yes. I denote $a * B = a*b : b in B $ and $A*B = a*b : a in A, b in B $ for an operation $*$ - which indeed kills distributivity.
    $endgroup$
    – rabota
    Mar 17 at 10:20












  • 1




    $begingroup$
    Is the left set $1+a+bmid ain A, bin B$ and the right set is $(1+a)(1+b)+nmid a,in A, bin B,nin N$? I ask because i'm not sure how you meant to define the product $(1+A)(1+B)$.
    $endgroup$
    – rschwieb
    Mar 17 at 0:42










  • $begingroup$
    @rschwieb Yes. I denote $a * B = a*b : b in B $ and $A*B = a*b : a in A, b in B $ for an operation $*$ - which indeed kills distributivity.
    $endgroup$
    – rabota
    Mar 17 at 10:20







1




1




$begingroup$
Is the left set $1+a+bmid ain A, bin B$ and the right set is $(1+a)(1+b)+nmid a,in A, bin B,nin N$? I ask because i'm not sure how you meant to define the product $(1+A)(1+B)$.
$endgroup$
– rschwieb
Mar 17 at 0:42




$begingroup$
Is the left set $1+a+bmid ain A, bin B$ and the right set is $(1+a)(1+b)+nmid a,in A, bin B,nin N$? I ask because i'm not sure how you meant to define the product $(1+A)(1+B)$.
$endgroup$
– rschwieb
Mar 17 at 0:42












$begingroup$
@rschwieb Yes. I denote $a * B = a*b : b in B $ and $A*B = a*b : a in A, b in B $ for an operation $*$ - which indeed kills distributivity.
$endgroup$
– rabota
Mar 17 at 10:20




$begingroup$
@rschwieb Yes. I denote $a * B = a*b : b in B $ and $A*B = a*b : a in A, b in B $ for an operation $*$ - which indeed kills distributivity.
$endgroup$
– rabota
Mar 17 at 10:20










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