Existence of Principal Ideal
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 2 more comments
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 2 more comments
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 2 more comments
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 2 more comments
1 Answer
1
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votes
This was already said in the comments, but I second it: the argument looks fine to me.
To address another point in the comments:
Yes, a prinicpal ideal is essentially the "rings" version of a cyclic subgroup. However, one subtlety worth noting is that for groups you have "cyclic groups" (not thought of as subgroups of some larger group). Cyclic groups basically look like cyclic subgroups, but in ring theory it's different: the analogous notion of a "cyclic ring" is quite different-looking from an ideal.
[ The group-theoretic analogy is stronger for related objects called "modules" than for rings per se. In fact, this is no accident: abelian groups are precisely the same as modules over $Bbb Z$. ]
This answer exists primarily to remove this question from the Unanswered queue. Please upvote or give it Best Answer to complete this process.
– aleph_two
2 days ago
add a comment |
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This was already said in the comments, but I second it: the argument looks fine to me.
To address another point in the comments:
Yes, a prinicpal ideal is essentially the "rings" version of a cyclic subgroup. However, one subtlety worth noting is that for groups you have "cyclic groups" (not thought of as subgroups of some larger group). Cyclic groups basically look like cyclic subgroups, but in ring theory it's different: the analogous notion of a "cyclic ring" is quite different-looking from an ideal.
[ The group-theoretic analogy is stronger for related objects called "modules" than for rings per se. In fact, this is no accident: abelian groups are precisely the same as modules over $Bbb Z$. ]
This answer exists primarily to remove this question from the Unanswered queue. Please upvote or give it Best Answer to complete this process.
– aleph_two
2 days ago
add a comment |
This was already said in the comments, but I second it: the argument looks fine to me.
To address another point in the comments:
Yes, a prinicpal ideal is essentially the "rings" version of a cyclic subgroup. However, one subtlety worth noting is that for groups you have "cyclic groups" (not thought of as subgroups of some larger group). Cyclic groups basically look like cyclic subgroups, but in ring theory it's different: the analogous notion of a "cyclic ring" is quite different-looking from an ideal.
[ The group-theoretic analogy is stronger for related objects called "modules" than for rings per se. In fact, this is no accident: abelian groups are precisely the same as modules over $Bbb Z$. ]
This answer exists primarily to remove this question from the Unanswered queue. Please upvote or give it Best Answer to complete this process.
– aleph_two
2 days ago
add a comment |
This was already said in the comments, but I second it: the argument looks fine to me.
To address another point in the comments:
Yes, a prinicpal ideal is essentially the "rings" version of a cyclic subgroup. However, one subtlety worth noting is that for groups you have "cyclic groups" (not thought of as subgroups of some larger group). Cyclic groups basically look like cyclic subgroups, but in ring theory it's different: the analogous notion of a "cyclic ring" is quite different-looking from an ideal.
[ The group-theoretic analogy is stronger for related objects called "modules" than for rings per se. In fact, this is no accident: abelian groups are precisely the same as modules over $Bbb Z$. ]
This was already said in the comments, but I second it: the argument looks fine to me.
To address another point in the comments:
Yes, a prinicpal ideal is essentially the "rings" version of a cyclic subgroup. However, one subtlety worth noting is that for groups you have "cyclic groups" (not thought of as subgroups of some larger group). Cyclic groups basically look like cyclic subgroups, but in ring theory it's different: the analogous notion of a "cyclic ring" is quite different-looking from an ideal.
[ The group-theoretic analogy is stronger for related objects called "modules" than for rings per se. In fact, this is no accident: abelian groups are precisely the same as modules over $Bbb Z$. ]
answered 2 days ago
community wiki
aleph_two
This answer exists primarily to remove this question from the Unanswered queue. Please upvote or give it Best Answer to complete this process.
– aleph_two
2 days ago
add a comment |
This answer exists primarily to remove this question from the Unanswered queue. Please upvote or give it Best Answer to complete this process.
– aleph_two
2 days ago
This answer exists primarily to remove this question from the Unanswered queue. Please upvote or give it Best Answer to complete this process.
– aleph_two
2 days ago
This answer exists primarily to remove this question from the Unanswered queue. Please upvote or give it Best Answer to complete this process.
– aleph_two
2 days ago
add a 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