Existence of Principal Ideal
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My first question is DOES a Principal Ideal always exist in a ring?
(My thoughts: By it's very definition any element can generate a PI and hence it exists)
Follow-up question: (If answer to 1st question is YES)
Please have a look at the alternate proof I have in mind for the following question:
Question: Prove that a commutative ring R with unity whose only ideals are <0> and R itself is a field.
PROOF: Let "a" be any element belonging to R. (Assume a to be non-zero)
Consider < a >.
Now since only ideals are <0> and R itself, and since a is non-trivial, < a > must be equal to R.
< a > contains elements of the form {r1a, r2a, ...} for all r belonging to R.
But since < a > = R, one of these elements MUST be UNITY!
Hence there exists ri such that r1a = 1.
Hence showing the existence of inverse of a.
Furthermore since a is ANY element belonging to R we have shown that for every element an inverse exists. Hence it's a field (other properties already satisfied)
(Left inverse proof is trivial from above)
What is the problem in this PROOF?
abstract-algebra ring-theory ideals principal-ideal-domains
asked Nov 16 at 5:31
PLAP_
266
|
show 1 more comment
up vote
1
down vote
favorite
My first question is DOES a Principal Ideal always exist in a ring?
(My thoughts: By it's very definition any element can generate a PI and hence it exists)
Follow-up question: (If answer to 1st question is YES)
Please have a look at the alternate proof I have in mind for the following question:
Question: Prove that a commutative ring R with unity whose only ideals are <0> and R itself is a field.
PROOF: Let "a" be any element belonging to R. (Assume a to be non-zero)
Consider < a >.
Now since only ideals are <0> and R itself, and since a is non-trivial, < a > must be equal to R.
< a > contains elements of the form {r1a, r2a, ...} for all r belonging to R.
But since < a > = R, one of these elements MUST be UNITY!
Hence there exists ri such that r1a = 1.
Hence showing the existence of inverse of a.
Furthermore since a is ANY element belonging to R we have shown that for every element an inverse exists. Hence it's a field (other properties already satisfied)
(Left inverse proof is trivial from above)
What is the problem in this PROOF?
abstract-algebra ring-theory ideals principal-ideal-domains
asked Nov 16 at 5:31
PLAP_
266
What makes you think there is a problem?
– Eric Wofsey
Nov 16 at 5:35
Because the solution given in the book and the professor is "supposedly" the shortest solution, but is longer than this. Hence the doubt.
– PLAP_
Nov 16 at 5:36
1
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Nov 16 at 5:44
1
Your proof looks fine to me. To answer the question in your first paragraph, yes, every element generates a principal ideal.
– Bungo
Nov 16 at 5:46
1
Better to think of $langle arangle$ as $aR$. This is the right principal ideal generated by $a$, but with a commutative ring it's just the principal ideal. Then your proof becomes $1in R=aR$, so $a$ has a right inverse.
– vadim123
Nov 16 at 5:54
|
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
My first question is DOES a Principal Ideal always exist in a ring?
(My thoughts: By it's very definition any element can generate a PI and hence it exists)
Follow-up question: (If answer to 1st question is YES)
Please have a look at the alternate proof I have in mind for the following question:
Question: Prove that a commutative ring R with unity whose only ideals are <0> and R itself is a field.
PROOF: Let "a" be any element belonging to R. (Assume a to be non-zero)
Consider < a >.
Now since only ideals are <0> and R itself, and since a is non-trivial, < a > must be equal to R.
< a > contains elements of the form {r1a, r2a, ...} for all r belonging to R.
But since < a > = R, one of these elements MUST be UNITY!
Hence there exists ri such that r1a = 1.
Hence showing the existence of inverse of a.
Furthermore since a is ANY element belonging to R we have shown that for every element an inverse exists. Hence it's a field (other properties already satisfied)
(Left inverse proof is trivial from above)
What is the problem in this PROOF?
abstract-algebra ring-theory ideals principal-ideal-domains
asked Nov 16 at 5:31
PLAP_
266
My first question is DOES a Principal Ideal always exist in a ring?
(My thoughts: By it's very definition any element can generate a PI and hence it exists)
Follow-up question: (If answer to 1st question is YES)
Please have a look at the alternate proof I have in mind for the following question:
Question: Prove that a commutative ring R with unity whose only ideals are <0> and R itself is a field.
PROOF: Let "a" be any element belonging to R. (Assume a to be non-zero)
Consider < a >.
Now since only ideals are <0> and R itself, and since a is non-trivial, < a > must be equal to R.
< a > contains elements of the form {r1a, r2a, ...} for all r belonging to R.
But since < a > = R, one of these elements MUST be UNITY!
Hence there exists ri such that r1a = 1.
Hence showing the existence of inverse of a.
Furthermore since a is ANY element belonging to R we have shown that for every element an inverse exists. Hence it's a field (other properties already satisfied)
(Left inverse proof is trivial from above)
What is the problem in this PROOF?
abstract-algebra ring-theory ideals principal-ideal-domains
abstract-algebra ring-theory ideals principal-ideal-domains
asked Nov 16 at 5:31
PLAP_
266
asked Nov 16 at 5:31
PLAP_
266
asked Nov 16 at 5:31
PLAP_
266
asked Nov 16 at 5:31
PLAP_
266
asked Nov 16 at 5:31
PLAP_
266
266
What makes you think there is a problem?
– Eric Wofsey
Nov 16 at 5:35
Because the solution given in the book and the professor is "supposedly" the shortest solution, but is longer than this. Hence the doubt.
– PLAP_
Nov 16 at 5:36
1
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Nov 16 at 5:44
1
Your proof looks fine to me. To answer the question in your first paragraph, yes, every element generates a principal ideal.
– Bungo
Nov 16 at 5:46
1
Better to think of $langle arangle$ as $aR$. This is the right principal ideal generated by $a$, but with a commutative ring it's just the principal ideal. Then your proof becomes $1in R=aR$, so $a$ has a right inverse.
– vadim123
Nov 16 at 5:54
|
show 1 more comment
What makes you think there is a problem?
– Eric Wofsey
Nov 16 at 5:35
Because the solution given in the book and the professor is "supposedly" the shortest solution, but is longer than this. Hence the doubt.
– PLAP_
Nov 16 at 5:36
1
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Nov 16 at 5:44
1
Your proof looks fine to me. To answer the question in your first paragraph, yes, every element generates a principal ideal.
– Bungo
Nov 16 at 5:46
1
Better to think of $langle arangle$ as $aR$. This is the right principal ideal generated by $a$, but with a commutative ring it's just the principal ideal. Then your proof becomes $1in R=aR$, so $a$ has a right inverse.
– vadim123
Nov 16 at 5:54
What makes you think there is a problem?
– Eric Wofsey
Nov 16 at 5:35
What makes you think there is a problem?
– Eric Wofsey
Nov 16 at 5:35
Because the solution given in the book and the professor is "supposedly" the shortest solution, but is longer than this. Hence the doubt.
– PLAP_
Nov 16 at 5:36
Because the solution given in the book and the professor is "supposedly" the shortest solution, but is longer than this. Hence the doubt.
– PLAP_
Nov 16 at 5:36
1
1
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Nov 16 at 5:44
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Nov 16 at 5:44
1
1
Your proof looks fine to me. To answer the question in your first paragraph, yes, every element generates a principal ideal.
– Bungo
Nov 16 at 5:46
Your proof looks fine to me. To answer the question in your first paragraph, yes, every element generates a principal ideal.
– Bungo
Nov 16 at 5:46
1
1
Better to think of $langle arangle$ as $aR$. This is the right principal ideal generated by $a$, but with a commutative ring it's just the principal ideal. Then your proof becomes $1in R=aR$, so $a$ has a right inverse.
– vadim123
Nov 16 at 5:54
Better to think of $langle arangle$ as $aR$. This is the right principal ideal generated by $a$, but with a commutative ring it's just the principal ideal. Then your proof becomes $1in R=aR$, so $a$ has a right inverse.
– vadim123
Nov 16 at 5:54
|
show 1 more comment
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What makes you think there is a problem?
– Eric Wofsey
Nov 16 at 5:35
Because the solution given in the book and the professor is "supposedly" the shortest solution, but is longer than this. Hence the doubt.
– PLAP_
Nov 16 at 5:36
1
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Nov 16 at 5:44
1
Your proof looks fine to me. To answer the question in your first paragraph, yes, every element generates a principal ideal.
– Bungo
Nov 16 at 5:46
1
Better to think of $langle arangle$ as $aR$. This is the right principal ideal generated by $a$, but with a commutative ring it's just the principal ideal. Then your proof becomes $1in R=aR$, so $a$ has a right inverse.
– vadim123
Nov 16 at 5:54