How to prove that in an integral domain $(x^k)=(x)^k$? [closed]
$begingroup$
Let $D$ be an integral domain. For $x$ in $D$ and for a k positive integer then $(x^k) = (x)^k$.
How do i prove this?
abstract-algebra ring-theory commutative-algebra ideals
edited Dec 18 '18 at 14:16
rschwieb
107k12103251
asked Dec 18 '18 at 12:34
GentianaGentiana
244
$endgroup$
closed as off-topic by user26857, quid♦ Dec 18 '18 at 21:14
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." – user26857, quid
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $D$ be an integral domain. For $x$ in $D$ and for a k positive integer then $(x^k) = (x)^k$.
How do i prove this?
abstract-algebra ring-theory commutative-algebra ideals
edited Dec 18 '18 at 14:16
rschwieb
107k12103251
asked Dec 18 '18 at 12:34
GentianaGentiana
244
$endgroup$
closed as off-topic by user26857, quid♦ Dec 18 '18 at 21:14
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." – user26857, quid
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $D$ be an integral domain. For $x$ in $D$ and for a k positive integer then $(x^k) = (x)^k$.
How do i prove this?
abstract-algebra ring-theory commutative-algebra ideals
edited Dec 18 '18 at 14:16
rschwieb
107k12103251
asked Dec 18 '18 at 12:34
GentianaGentiana
244
$endgroup$
Let $D$ be an integral domain. For $x$ in $D$ and for a k positive integer then $(x^k) = (x)^k$.
How do i prove this?
abstract-algebra ring-theory commutative-algebra ideals
abstract-algebra ring-theory commutative-algebra ideals
edited Dec 18 '18 at 14:16
rschwieb
107k12103251
asked Dec 18 '18 at 12:34
GentianaGentiana
244
edited Dec 18 '18 at 14:16
rschwieb
107k12103251
asked Dec 18 '18 at 12:34
GentianaGentiana
244
edited Dec 18 '18 at 14:16
rschwieb
107k12103251
edited Dec 18 '18 at 14:16
rschwieb
107k12103251
edited Dec 18 '18 at 14:16
rschwieb
107k12103251
107k12103251
asked Dec 18 '18 at 12:34
GentianaGentiana
244
asked Dec 18 '18 at 12:34
GentianaGentiana
244
asked Dec 18 '18 at 12:34
GentianaGentiana
244
244
closed as off-topic by user26857, quid♦ Dec 18 '18 at 21:14
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." – user26857, quid
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by user26857, quid♦ Dec 18 '18 at 21:14
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." – user26857, quid
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $I$ and $J$ are ideals generated by $a_i$ and $b_j$ respectively, then their product $IJ$ is generated by $a_i b_j$, its worth proving this, its not tricky. Believing that, given that $(x)^k$ is the $k$-fold product of the ideal $(x)$ with itself, so the generators of $(x)^k$ are the $k$-fold products of generators of $(x)$, but this is principle, so this is just $(x^k)$.
answered Dec 18 '18 at 12:43
user277182user277182
488212
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $I$ and $J$ are ideals generated by $a_i$ and $b_j$ respectively, then their product $IJ$ is generated by $a_i b_j$, its worth proving this, its not tricky. Believing that, given that $(x)^k$ is the $k$-fold product of the ideal $(x)$ with itself, so the generators of $(x)^k$ are the $k$-fold products of generators of $(x)$, but this is principle, so this is just $(x^k)$.
answered Dec 18 '18 at 12:43
user277182user277182
488212
$endgroup$
add a comment |
$begingroup$
If $I$ and $J$ are ideals generated by $a_i$ and $b_j$ respectively, then their product $IJ$ is generated by $a_i b_j$, its worth proving this, its not tricky. Believing that, given that $(x)^k$ is the $k$-fold product of the ideal $(x)$ with itself, so the generators of $(x)^k$ are the $k$-fold products of generators of $(x)$, but this is principle, so this is just $(x^k)$.
answered Dec 18 '18 at 12:43
user277182user277182
488212
$endgroup$
add a comment |
$begingroup$
If $I$ and $J$ are ideals generated by $a_i$ and $b_j$ respectively, then their product $IJ$ is generated by $a_i b_j$, its worth proving this, its not tricky. Believing that, given that $(x)^k$ is the $k$-fold product of the ideal $(x)$ with itself, so the generators of $(x)^k$ are the $k$-fold products of generators of $(x)$, but this is principle, so this is just $(x^k)$.
answered Dec 18 '18 at 12:43
user277182user277182
488212
$endgroup$
If $I$ and $J$ are ideals generated by $a_i$ and $b_j$ respectively, then their product $IJ$ is generated by $a_i b_j$, its worth proving this, its not tricky. Believing that, given that $(x)^k$ is the $k$-fold product of the ideal $(x)$ with itself, so the generators of $(x)^k$ are the $k$-fold products of generators of $(x)$, but this is principle, so this is just $(x^k)$.
answered Dec 18 '18 at 12:43
user277182user277182
488212
answered Dec 18 '18 at 12:43
user277182user277182
488212
answered Dec 18 '18 at 12:43
user277182user277182
488212
answered Dec 18 '18 at 12:43
user277182user277182
488212
488212
add a comment |
add a comment |
