Let $R$ be an integral domain and $P$ be a prime ideal in $R[x]$ such that $Pcap R={0}$. Show that...
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Let $R$ be an integral domain and $P$ be a prime ideal in $R[x]$ such that $P cap R={0}$. Show that $:operatorname{height}Ple 1$. please help to find its height...
commutative-algebra
edited 2 days ago
user26857
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asked Nov 13 at 18:50
Ravi Agarwal
121
put on hold as off-topic by user26857, user302797, jgon, Chinnapparaj R, Cesareo yesterday
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Let $R$ be an integral domain and $P$ be a prime ideal in $R[x]$ such that $P cap R={0}$. Show that $:operatorname{height}Ple 1$. please help to find its height...
commutative-algebra
edited 2 days ago
user26857
39.1k123882
asked Nov 13 at 18:50
Ravi Agarwal
121
put on hold as off-topic by user26857, user302797, jgon, Chinnapparaj R, Cesareo yesterday
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." – user26857, user302797, jgon, Chinnapparaj R, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
Please use MathJax math.meta.stackexchange.com/questions/5020/…
– Anurag A
Nov 13 at 18:56
Welcome to Maths SX! Do you know rings of fractions?
– Bernard
Nov 13 at 19:03
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $R$ be an integral domain and $P$ be a prime ideal in $R[x]$ such that $P cap R={0}$. Show that $:operatorname{height}Ple 1$. please help to find its height...
commutative-algebra
edited 2 days ago
user26857
39.1k123882
asked Nov 13 at 18:50
Ravi Agarwal
121
Let $R$ be an integral domain and $P$ be a prime ideal in $R[x]$ such that $P cap R={0}$. Show that $:operatorname{height}Ple 1$. please help to find its height...
commutative-algebra
commutative-algebra
edited 2 days ago
user26857
39.1k123882
asked Nov 13 at 18:50
Ravi Agarwal
121
edited 2 days ago
user26857
39.1k123882
asked Nov 13 at 18:50
Ravi Agarwal
121
edited 2 days ago
user26857
39.1k123882
edited 2 days ago
user26857
39.1k123882
edited 2 days ago
user26857
39.1k123882
39.1k123882
asked Nov 13 at 18:50
Ravi Agarwal
121
asked Nov 13 at 18:50
Ravi Agarwal
121
asked Nov 13 at 18:50
Ravi Agarwal
121
121
put on hold as off-topic by user26857, user302797, jgon, Chinnapparaj R, Cesareo yesterday
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." – user26857, user302797, jgon, Chinnapparaj R, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by user26857, user302797, jgon, Chinnapparaj R, Cesareo yesterday
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." – user26857, user302797, jgon, Chinnapparaj R, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
Please use MathJax math.meta.stackexchange.com/questions/5020/…
– Anurag A
Nov 13 at 18:56
Welcome to Maths SX! Do you know rings of fractions?
– Bernard
Nov 13 at 19:03
add a comment |
Please use MathJax math.meta.stackexchange.com/questions/5020/…
– Anurag A
Nov 13 at 18:56
Welcome to Maths SX! Do you know rings of fractions?
– Bernard
Nov 13 at 19:03
Please use MathJax math.meta.stackexchange.com/questions/5020/…
– Anurag A
Nov 13 at 18:56
Please use MathJax math.meta.stackexchange.com/questions/5020/…
– Anurag A
Nov 13 at 18:56
Welcome to Maths SX! Do you know rings of fractions?
– Bernard
Nov 13 at 19:03
Welcome to Maths SX! Do you know rings of fractions?
– Bernard
Nov 13 at 19:03
add a comment |
1 Answer
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Note that the statement holds if $R$ is a field. Now consider the localization of $R[x]$ at $R$, which is a polynomial ring over the field $K(R)$, where $K(R)$ is the field of fractions of $R$. Since $Pcap R={0}$, $P$ is still a prime ideal. It is easy to show that any kind proper inclusion $Qsubset P$ of prime ideals will hold in $K(R)[x]$. Hence the height of $P$ in $R[x]$ and in $K(R)[x]$ are the same. The latter one is at most one and the result follows.
answered Nov 13 at 19:08
Levent
3,386825
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Note that the statement holds if $R$ is a field. Now consider the localization of $R[x]$ at $R$, which is a polynomial ring over the field $K(R)$, where $K(R)$ is the field of fractions of $R$. Since $Pcap R={0}$, $P$ is still a prime ideal. It is easy to show that any kind proper inclusion $Qsubset P$ of prime ideals will hold in $K(R)[x]$. Hence the height of $P$ in $R[x]$ and in $K(R)[x]$ are the same. The latter one is at most one and the result follows.
answered Nov 13 at 19:08
Levent
3,386825
add a comment |
up vote
3
down vote
Note that the statement holds if $R$ is a field. Now consider the localization of $R[x]$ at $R$, which is a polynomial ring over the field $K(R)$, where $K(R)$ is the field of fractions of $R$. Since $Pcap R={0}$, $P$ is still a prime ideal. It is easy to show that any kind proper inclusion $Qsubset P$ of prime ideals will hold in $K(R)[x]$. Hence the height of $P$ in $R[x]$ and in $K(R)[x]$ are the same. The latter one is at most one and the result follows.
answered Nov 13 at 19:08
Levent
3,386825
add a comment |
up vote
3
down vote
up vote
3
down vote
Note that the statement holds if $R$ is a field. Now consider the localization of $R[x]$ at $R$, which is a polynomial ring over the field $K(R)$, where $K(R)$ is the field of fractions of $R$. Since $Pcap R={0}$, $P$ is still a prime ideal. It is easy to show that any kind proper inclusion $Qsubset P$ of prime ideals will hold in $K(R)[x]$. Hence the height of $P$ in $R[x]$ and in $K(R)[x]$ are the same. The latter one is at most one and the result follows.
answered Nov 13 at 19:08
Levent
3,386825
Note that the statement holds if $R$ is a field. Now consider the localization of $R[x]$ at $R$, which is a polynomial ring over the field $K(R)$, where $K(R)$ is the field of fractions of $R$. Since $Pcap R={0}$, $P$ is still a prime ideal. It is easy to show that any kind proper inclusion $Qsubset P$ of prime ideals will hold in $K(R)[x]$. Hence the height of $P$ in $R[x]$ and in $K(R)[x]$ are the same. The latter one is at most one and the result follows.
answered Nov 13 at 19:08
Levent
3,386825
answered Nov 13 at 19:08
Levent
3,386825
answered Nov 13 at 19:08
Levent
3,386825
answered Nov 13 at 19:08
Levent
3,386825
3,386825
add a comment |
add a comment |
Please use MathJax math.meta.stackexchange.com/questions/5020/…
– Anurag A
Nov 13 at 18:56
Welcome to Maths SX! Do you know rings of fractions?
– Bernard
Nov 13 at 19:03