Are the distributive and abelian properties inherited from a ring to a SUBSET of the ring?(The subset is...
I was reading the proof of the theorem for proving subring and it assumed this.
can someone verify.
the proof i was reading about(which i mentioned in the description) was about proving that a subset with some particular properties is a subring. it inferred from the fact that closure under both operations means it follows distributivity and also the commutative property under addition. I was confused how they inferred this directly, and that's what I asked in the question
abstract-algebra ring-theory
asked Nov 28 at 17:54
Cosmic
6110
closed as off-topic by Brahadeesh, amWhy, Lord_Farin, Leucippus, Shailesh Nov 29 at 0:30
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, amWhy, Lord_Farin, Leucippus, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
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I was reading the proof of the theorem for proving subring and it assumed this.
can someone verify.
the proof i was reading about(which i mentioned in the description) was about proving that a subset with some particular properties is a subring. it inferred from the fact that closure under both operations means it follows distributivity and also the commutative property under addition. I was confused how they inferred this directly, and that's what I asked in the question
abstract-algebra ring-theory
asked Nov 28 at 17:54
Cosmic
6110
closed as off-topic by Brahadeesh, amWhy, Lord_Farin, Leucippus, Shailesh Nov 29 at 0:30
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, amWhy, Lord_Farin, Leucippus, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
If the subset is not closed under multiplication and addition, what does it mean to be "distributive"?
– Morgan Rodgers
Nov 28 at 18:03
as a matter of fact it is, sorry forgot to mention that.
– Cosmic
Nov 28 at 18:05
If it's closed under multiplication and addition, then isn't it a subring?
– Morgan Rodgers
Nov 29 at 6:46
the proof i was reading about(which i mentioned in the description) was about proving that a subset with some particular properties is a subring. it inferred from the fact that closure under both operations means it follows distributivity and also the commutative property under addition. I was confused how they inferred this directly, and that's what I asked in the question.
– Cosmic
Nov 29 at 15:02
Notice that the distributive and abelian statements are only phrased in terms of specific elements, and having the distributive/abelian "property" means the statement holds for ALL elements of the ring (and so for any collection of elements in a subset).
– Morgan Rodgers
Nov 29 at 23:14
|
show 1 more comment
I was reading the proof of the theorem for proving subring and it assumed this.
can someone verify.
the proof i was reading about(which i mentioned in the description) was about proving that a subset with some particular properties is a subring. it inferred from the fact that closure under both operations means it follows distributivity and also the commutative property under addition. I was confused how they inferred this directly, and that's what I asked in the question
abstract-algebra ring-theory
asked Nov 28 at 17:54
Cosmic
6110
I was reading the proof of the theorem for proving subring and it assumed this.
can someone verify.
the proof i was reading about(which i mentioned in the description) was about proving that a subset with some particular properties is a subring. it inferred from the fact that closure under both operations means it follows distributivity and also the commutative property under addition. I was confused how they inferred this directly, and that's what I asked in the question
abstract-algebra ring-theory
abstract-algebra ring-theory
asked Nov 28 at 17:54
Cosmic
6110
asked Nov 28 at 17:54
Cosmic
6110
edited Nov 29 at 15:18
asked Nov 28 at 17:54
Cosmic
6110
asked Nov 28 at 17:54
Cosmic
6110
asked Nov 28 at 17:54
Cosmic
6110
6110
closed as off-topic by Brahadeesh, amWhy, Lord_Farin, Leucippus, Shailesh Nov 29 at 0:30
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, amWhy, Lord_Farin, Leucippus, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Brahadeesh, amWhy, Lord_Farin, Leucippus, Shailesh Nov 29 at 0:30
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, amWhy, Lord_Farin, Leucippus, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
If the subset is not closed under multiplication and addition, what does it mean to be "distributive"?
– Morgan Rodgers
Nov 28 at 18:03
as a matter of fact it is, sorry forgot to mention that.
– Cosmic
Nov 28 at 18:05
If it's closed under multiplication and addition, then isn't it a subring?
– Morgan Rodgers
Nov 29 at 6:46
the proof i was reading about(which i mentioned in the description) was about proving that a subset with some particular properties is a subring. it inferred from the fact that closure under both operations means it follows distributivity and also the commutative property under addition. I was confused how they inferred this directly, and that's what I asked in the question.
– Cosmic
Nov 29 at 15:02
Notice that the distributive and abelian statements are only phrased in terms of specific elements, and having the distributive/abelian "property" means the statement holds for ALL elements of the ring (and so for any collection of elements in a subset).
– Morgan Rodgers
Nov 29 at 23:14
|
show 1 more comment
If the subset is not closed under multiplication and addition, what does it mean to be "distributive"?
– Morgan Rodgers
Nov 28 at 18:03
as a matter of fact it is, sorry forgot to mention that.
– Cosmic
Nov 28 at 18:05
If it's closed under multiplication and addition, then isn't it a subring?
– Morgan Rodgers
Nov 29 at 6:46
the proof i was reading about(which i mentioned in the description) was about proving that a subset with some particular properties is a subring. it inferred from the fact that closure under both operations means it follows distributivity and also the commutative property under addition. I was confused how they inferred this directly, and that's what I asked in the question.
– Cosmic
Nov 29 at 15:02
Notice that the distributive and abelian statements are only phrased in terms of specific elements, and having the distributive/abelian "property" means the statement holds for ALL elements of the ring (and so for any collection of elements in a subset).
– Morgan Rodgers
Nov 29 at 23:14
If the subset is not closed under multiplication and addition, what does it mean to be "distributive"?
– Morgan Rodgers
Nov 28 at 18:03
If the subset is not closed under multiplication and addition, what does it mean to be "distributive"?
– Morgan Rodgers
Nov 28 at 18:03
as a matter of fact it is, sorry forgot to mention that.
– Cosmic
Nov 28 at 18:05
as a matter of fact it is, sorry forgot to mention that.
– Cosmic
Nov 28 at 18:05
If it's closed under multiplication and addition, then isn't it a subring?
– Morgan Rodgers
Nov 29 at 6:46
If it's closed under multiplication and addition, then isn't it a subring?
– Morgan Rodgers
Nov 29 at 6:46
the proof i was reading about(which i mentioned in the description) was about proving that a subset with some particular properties is a subring. it inferred from the fact that closure under both operations means it follows distributivity and also the commutative property under addition. I was confused how they inferred this directly, and that's what I asked in the question.
– Cosmic
Nov 29 at 15:02
the proof i was reading about(which i mentioned in the description) was about proving that a subset with some particular properties is a subring. it inferred from the fact that closure under both operations means it follows distributivity and also the commutative property under addition. I was confused how they inferred this directly, and that's what I asked in the question.
– Cosmic
Nov 29 at 15:02
Notice that the distributive and abelian statements are only phrased in terms of specific elements, and having the distributive/abelian "property" means the statement holds for ALL elements of the ring (and so for any collection of elements in a subset).
– Morgan Rodgers
Nov 29 at 23:14
Notice that the distributive and abelian statements are only phrased in terms of specific elements, and having the distributive/abelian "property" means the statement holds for ALL elements of the ring (and so for any collection of elements in a subset).
– Morgan Rodgers
Nov 29 at 23:14
|
show 1 more comment
1 Answer
1
active
oldest
votes
"All of our cows are brown."
"What about the cows in that particular field, are they brown?"
"Yes, they are ours and they are brown."
Note that brown-ness and abielian-ness are properties that relate to individual elements (for brown-ness) or pairs of elements (for abielean-ness), and so they pass on almost automatically to subsets. It's good that you're worrying about this, as there are some properties that don't pass automatically to subsets. For example, "closure under inverses" does not.
answered Nov 28 at 17:59
JonathanZ
2,099613
Thanks for the help.
– Cosmic
Nov 28 at 18:07
Aw, I got a downvote. Who doesn't like brown cows?
– JonathanZ
Nov 30 at 2:32
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
"All of our cows are brown."
"What about the cows in that particular field, are they brown?"
"Yes, they are ours and they are brown."
Note that brown-ness and abielian-ness are properties that relate to individual elements (for brown-ness) or pairs of elements (for abielean-ness), and so they pass on almost automatically to subsets. It's good that you're worrying about this, as there are some properties that don't pass automatically to subsets. For example, "closure under inverses" does not.
answered Nov 28 at 17:59
JonathanZ
2,099613
Thanks for the help.
– Cosmic
Nov 28 at 18:07
Aw, I got a downvote. Who doesn't like brown cows?
– JonathanZ
Nov 30 at 2:32
add a comment |
"All of our cows are brown."
"What about the cows in that particular field, are they brown?"
"Yes, they are ours and they are brown."
Note that brown-ness and abielian-ness are properties that relate to individual elements (for brown-ness) or pairs of elements (for abielean-ness), and so they pass on almost automatically to subsets. It's good that you're worrying about this, as there are some properties that don't pass automatically to subsets. For example, "closure under inverses" does not.
answered Nov 28 at 17:59
JonathanZ
2,099613
Thanks for the help.
– Cosmic
Nov 28 at 18:07
Aw, I got a downvote. Who doesn't like brown cows?
– JonathanZ
Nov 30 at 2:32
add a comment |
"All of our cows are brown."
"What about the cows in that particular field, are they brown?"
"Yes, they are ours and they are brown."
Note that brown-ness and abielian-ness are properties that relate to individual elements (for brown-ness) or pairs of elements (for abielean-ness), and so they pass on almost automatically to subsets. It's good that you're worrying about this, as there are some properties that don't pass automatically to subsets. For example, "closure under inverses" does not.
answered Nov 28 at 17:59
JonathanZ
2,099613
"All of our cows are brown."
"What about the cows in that particular field, are they brown?"
"Yes, they are ours and they are brown."
Note that brown-ness and abielian-ness are properties that relate to individual elements (for brown-ness) or pairs of elements (for abielean-ness), and so they pass on almost automatically to subsets. It's good that you're worrying about this, as there are some properties that don't pass automatically to subsets. For example, "closure under inverses" does not.
answered Nov 28 at 17:59
JonathanZ
2,099613
edited Nov 28 at 18:06
answered Nov 28 at 17:59
JonathanZ
2,099613
answered Nov 28 at 17:59
JonathanZ
2,099613
answered Nov 28 at 17:59
JonathanZ
2,099613
2,099613
Thanks for the help.
– Cosmic
Nov 28 at 18:07
Aw, I got a downvote. Who doesn't like brown cows?
– JonathanZ
Nov 30 at 2:32
add a comment |
Thanks for the help.
– Cosmic
Nov 28 at 18:07
Aw, I got a downvote. Who doesn't like brown cows?
– JonathanZ
Nov 30 at 2:32
Thanks for the help.
– Cosmic
Nov 28 at 18:07
Thanks for the help.
– Cosmic
Nov 28 at 18:07
Aw, I got a downvote. Who doesn't like brown cows?
– JonathanZ
Nov 30 at 2:32
Aw, I got a downvote. Who doesn't like brown cows?
– JonathanZ
Nov 30 at 2:32
add a comment |
If the subset is not closed under multiplication and addition, what does it mean to be "distributive"?
– Morgan Rodgers
Nov 28 at 18:03
as a matter of fact it is, sorry forgot to mention that.
– Cosmic
Nov 28 at 18:05
If it's closed under multiplication and addition, then isn't it a subring?
– Morgan Rodgers
Nov 29 at 6:46
the proof i was reading about(which i mentioned in the description) was about proving that a subset with some particular properties is a subring. it inferred from the fact that closure under both operations means it follows distributivity and also the commutative property under addition. I was confused how they inferred this directly, and that's what I asked in the question.
– Cosmic
Nov 29 at 15:02
Notice that the distributive and abelian statements are only phrased in terms of specific elements, and having the distributive/abelian "property" means the statement holds for ALL elements of the ring (and so for any collection of elements in a subset).
– Morgan Rodgers
Nov 29 at 23:14