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0

$begingroup$

Are the following two statements equivalent to each other ?

1) $(R,+,*)$ is a field

2) $(R,+)$ is an abelian group and $(Rsetminus0_R,*)$ is an abelian group

If not give an example.

group-theory ring-theory field-theory

share|cite|improve this question

edited Mar 12 at 19:46

Scott

asked Mar 12 at 19:35

ScottScott

347

$endgroup$

closed as off-topic by Servaes, Randall, Shaun, verret, Lee David Chung Lin Mar 13 at 0:33

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Servaes, Randall, Shaun, verret, Lee David Chung Lin

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Are you trying to typeset $mathbbR$?
  $endgroup$
  – Randall
  Mar 12 at 19:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No, they are not equivalent. $(1) implies (2)$, but one also needs that multiplication distributes across addition to go the other way.
  $endgroup$
  – jawheele
  Mar 12 at 19:40
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Are the following two statements equivalent to each other ?

1) $(R,+,*)$ is a field

2) $(R,+)$ is an abelian group and $(Rsetminus0_R,*)$ is an abelian group

If not give an example.

group-theory ring-theory field-theory

share|cite|improve this question

edited Mar 12 at 19:46

Scott

asked Mar 12 at 19:35

ScottScott

347

$endgroup$

closed as off-topic by Servaes, Randall, Shaun, verret, Lee David Chung Lin Mar 13 at 0:33

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Servaes, Randall, Shaun, verret, Lee David Chung Lin

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Are you trying to typeset $mathbbR$?
  $endgroup$
  – Randall
  Mar 12 at 19:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No, they are not equivalent. $(1) implies (2)$, but one also needs that multiplication distributes across addition to go the other way.
  $endgroup$
  – jawheele
  Mar 12 at 19:40
  
  

  
  
  
  

add a comment | 

$begingroup$

Are the following two statements equivalent to each other ?

1) $(R,+,*)$ is a field

2) $(R,+)$ is an abelian group and $(Rsetminus0_R,*)$ is an abelian group

If not give an example.

group-theory ring-theory field-theory

share|cite|improve this question

edited Mar 12 at 19:46

Scott

asked Mar 12 at 19:35

ScottScott

347

$endgroup$

share|cite|improve this question

edited Mar 12 at 19:46

Scott

asked Mar 12 at 19:35

ScottScott

347

share|cite|improve this question

edited Mar 12 at 19:46

Scott

asked Mar 12 at 19:35

ScottScott

347

asked Mar 12 at 19:35

ScottScott

347

asked Mar 12 at 19:35

ScottScott

347

1 Answer
1

active

oldest

votes

4

$begingroup$

No. There is one more property satisfied in a field: the distributivity between $+$ and $times$.

If the distributivity is satisfied, then $A$ is a ring in which non zero elements are $times$-invertible. It is, indeed, a field.

share|cite|improve this answer

answered Mar 12 at 19:39

LeaningLeaning

1,331718

$endgroup$

add a comment | 

4

$begingroup$

No. There is one more property satisfied in a field: the distributivity between $+$ and $times$.

If the distributivity is satisfied, then $A$ is a ring in which non zero elements are $times$-invertible. It is, indeed, a field.

share|cite|improve this answer

answered Mar 12 at 19:39

LeaningLeaning

1,331718

$endgroup$

add a comment | 

4

$begingroup$

No. There is one more property satisfied in a field: the distributivity between $+$ and $times$.

If the distributivity is satisfied, then $A$ is a ring in which non zero elements are $times$-invertible. It is, indeed, a field.

share|cite|improve this answer

answered Mar 12 at 19:39

LeaningLeaning

1,331718

$endgroup$

add a comment | 

$begingroup$

No. There is one more property satisfied in a field: the distributivity between $+$ and $times$.

If the distributivity is satisfied, then $A$ is a ring in which non zero elements are $times$-invertible. It is, indeed, a field.

share|cite|improve this answer

answered Mar 12 at 19:39

LeaningLeaning

1,331718

$endgroup$

share|cite|improve this answer

answered Mar 12 at 19:39

LeaningLeaning

1,331718

answered Mar 12 at 19:39

LeaningLeaning

1,331718

answered Mar 12 at 19:39

LeaningLeaning

1,331718