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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?

ring-theory commutative-algebra

share|cite|improve this question

asked Mar 16 at 9:55

rabotarabota

14.3k32782

$endgroup$

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  $begingroup$
  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
  

  
  
  
  

add a comment | 

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?

ring-theory commutative-algebra

share|cite|improve this question

asked Mar 16 at 9:55

rabotarabota

14.3k32782

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

$begingroup$

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?

ring-theory commutative-algebra

share|cite|improve this question

asked Mar 16 at 9:55

rabotarabota

14.3k32782

$endgroup$

share|cite|improve this question

asked Mar 16 at 9:55

rabotarabota

14.3k32782

share|cite|improve this question

asked Mar 16 at 9:55

rabotarabota

14.3k32782

asked Mar 16 at 9:55

rabotarabota

14.3k32782

asked Mar 16 at 9:55

rabotarabota

14.3k32782