Existence of Principal Ideal
$begingroup$
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 '18 at 5:31
PLAP_PLAP_
266
$endgroup$
|
show 2 more comments
$begingroup$
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 '18 at 5:31
PLAP_PLAP_
266
$endgroup$
$begingroup$
What makes you think there is a problem?
$endgroup$
– Eric Wofsey
Nov 16 '18 at 5:35
$begingroup$
Because the solution given in the book and the professor is "supposedly" the shortest solution, but is longer than this. Hence the doubt.
$endgroup$
– PLAP_
Nov 16 '18 at 5:36
1
$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Nov 16 '18 at 5:44
1
$begingroup$
Your proof looks fine to me. To answer the question in your first paragraph, yes, every element generates a principal ideal.
$endgroup$
– Bungo
Nov 16 '18 at 5:46
1
$begingroup$
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.
$endgroup$
– vadim123
Nov 16 '18 at 5:54
|
show 2 more comments
$begingroup$
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 '18 at 5:31
PLAP_PLAP_
266
$endgroup$
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 '18 at 5:31
PLAP_PLAP_
266
asked Nov 16 '18 at 5:31
PLAP_PLAP_
266
asked Nov 16 '18 at 5:31
PLAP_PLAP_
266
asked Nov 16 '18 at 5:31
PLAP_PLAP_
266
asked Nov 16 '18 at 5:31
PLAP_PLAP_
266
266
$begingroup$
What makes you think there is a problem?
$endgroup$
– Eric Wofsey
Nov 16 '18 at 5:35
$begingroup$
Because the solution given in the book and the professor is "supposedly" the shortest solution, but is longer than this. Hence the doubt.
$endgroup$
– PLAP_
Nov 16 '18 at 5:36
1
$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Nov 16 '18 at 5:44
1
$begingroup$
Your proof looks fine to me. To answer the question in your first paragraph, yes, every element generates a principal ideal.
$endgroup$
– Bungo
Nov 16 '18 at 5:46
1
$begingroup$
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.
$endgroup$
– vadim123
Nov 16 '18 at 5:54
|
show 2 more comments
$begingroup$
What makes you think there is a problem?
$endgroup$
– Eric Wofsey
Nov 16 '18 at 5:35
$begingroup$
Because the solution given in the book and the professor is "supposedly" the shortest solution, but is longer than this. Hence the doubt.
$endgroup$
– PLAP_
Nov 16 '18 at 5:36
1
$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Nov 16 '18 at 5:44
1
$begingroup$
Your proof looks fine to me. To answer the question in your first paragraph, yes, every element generates a principal ideal.
$endgroup$
– Bungo
Nov 16 '18 at 5:46
1
$begingroup$
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.
$endgroup$
– vadim123
Nov 16 '18 at 5:54
$begingroup$
What makes you think there is a problem?
$endgroup$
– Eric Wofsey
Nov 16 '18 at 5:35
$begingroup$
What makes you think there is a problem?
$endgroup$
– Eric Wofsey
Nov 16 '18 at 5:35
$begingroup$
Because the solution given in the book and the professor is "supposedly" the shortest solution, but is longer than this. Hence the doubt.
$endgroup$
– PLAP_
Nov 16 '18 at 5:36
$begingroup$
Because the solution given in the book and the professor is "supposedly" the shortest solution, but is longer than this. Hence the doubt.
$endgroup$
– PLAP_
Nov 16 '18 at 5:36
1
1
$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Nov 16 '18 at 5:44
$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Nov 16 '18 at 5:44
1
1
$begingroup$
Your proof looks fine to me. To answer the question in your first paragraph, yes, every element generates a principal ideal.
$endgroup$
– Bungo
Nov 16 '18 at 5:46
$begingroup$
Your proof looks fine to me. To answer the question in your first paragraph, yes, every element generates a principal ideal.
$endgroup$
– Bungo
Nov 16 '18 at 5:46
1
1
$begingroup$
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.
$endgroup$
– vadim123
Nov 16 '18 at 5:54
$begingroup$
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.
$endgroup$
– vadim123
Nov 16 '18 at 5:54
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
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 $Bbb Z$. ]
$endgroup$
$begingroup$
This answer exists primarily to remove this question from the Unanswered queue. Please upvote or give it Best Answer to complete this process.
$endgroup$
– aleph_two
Dec 25 '18 at 4:46
add a comment |
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$begingroup$
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 $Bbb Z$. ]
$endgroup$
$begingroup$
This answer exists primarily to remove this question from the Unanswered queue. Please upvote or give it Best Answer to complete this process.
$endgroup$
– aleph_two
Dec 25 '18 at 4:46
add a comment |
$begingroup$
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 $Bbb Z$. ]
$endgroup$
$begingroup$
This answer exists primarily to remove this question from the Unanswered queue. Please upvote or give it Best Answer to complete this process.
$endgroup$
– aleph_two
Dec 25 '18 at 4:46
add a comment |
$begingroup$
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 $Bbb Z$. ]
$endgroup$
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 Dec 25 '18 at 4:42
community wiki
aleph_two
$begingroup$
This answer exists primarily to remove this question from the Unanswered queue. Please upvote or give it Best Answer to complete this process.
$endgroup$
– aleph_two
Dec 25 '18 at 4:46
add a comment |
$begingroup$
This answer exists primarily to remove this question from the Unanswered queue. Please upvote or give it Best Answer to complete this process.
$endgroup$
– aleph_two
Dec 25 '18 at 4:46
$begingroup$
This answer exists primarily to remove this question from the Unanswered queue. Please upvote or give it Best Answer to complete this process.
$endgroup$
– aleph_two
Dec 25 '18 at 4:46
$begingroup$
This answer exists primarily to remove this question from the Unanswered queue. Please upvote or give it Best Answer to complete this process.
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– aleph_two
Dec 25 '18 at 4:46
add a comment |
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$begingroup$
What makes you think there is a problem?
$endgroup$
– Eric Wofsey
Nov 16 '18 at 5:35
$begingroup$
Because the solution given in the book and the professor is "supposedly" the shortest solution, but is longer than this. Hence the doubt.
$endgroup$
– PLAP_
Nov 16 '18 at 5:36
1
$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Nov 16 '18 at 5:44
1
$begingroup$
Your proof looks fine to me. To answer the question in your first paragraph, yes, every element generates a principal ideal.
$endgroup$
– Bungo
Nov 16 '18 at 5:46
1
$begingroup$
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.
$endgroup$
– vadim123
Nov 16 '18 at 5:54