localization of a polynomial ring at a prime ideal
$begingroup$
My question is about the hint of exercise 5.1 of Matsumura:
Let $k$ be a field,and $R=k[X_{1},dots,X_{n}]$ and let $mathfrak{p}in operatorname{Spec} R$.
Set $k[X_{1},dots,X_{n}]/mathfrak{p}=k[x_{1},dots,x_{n}]$. Suppose that $x_{1},dots,x_{r}$ is a transcendence basis of $k(x)$ over $k$. Set $K=k[X_{1},dots,X_{r}]$. Then the localization of $k[X_{1},dots,X_{n}]$ at $mathfrak{p}$ is the localization of $K[X_{r+1},dots,X_{n}]$ at a prime ideal $P$.
My question is: Why does $k[X_{1},dots,X_{n}]_{mathfrak{p}}=K[X_{r+1},dots,X_{n}]_{P}$ for some prime ideal $P$? And what does $P$ look like?
commutative-algebra localization
edited Dec 10 '18 at 9:21
user26857
39.3k124183
asked Dec 9 '18 at 16:40
SchröchinSchröchin
955
$endgroup$
add a comment |
$begingroup$
My question is about the hint of exercise 5.1 of Matsumura:
Let $k$ be a field,and $R=k[X_{1},dots,X_{n}]$ and let $mathfrak{p}in operatorname{Spec} R$.
Set $k[X_{1},dots,X_{n}]/mathfrak{p}=k[x_{1},dots,x_{n}]$. Suppose that $x_{1},dots,x_{r}$ is a transcendence basis of $k(x)$ over $k$. Set $K=k[X_{1},dots,X_{r}]$. Then the localization of $k[X_{1},dots,X_{n}]$ at $mathfrak{p}$ is the localization of $K[X_{r+1},dots,X_{n}]$ at a prime ideal $P$.
My question is: Why does $k[X_{1},dots,X_{n}]_{mathfrak{p}}=K[X_{r+1},dots,X_{n}]_{P}$ for some prime ideal $P$? And what does $P$ look like?
commutative-algebra localization
edited Dec 10 '18 at 9:21
user26857
39.3k124183
asked Dec 9 '18 at 16:40
SchröchinSchröchin
955
$endgroup$
$begingroup$
Hint. $R_psimeq (S^{-1}R)_{S^{-1}p}$ where $Ssubset R$ is a multiplicative set such that $Scap p=emptyset$. Can you choose an appropriate $S$?
$endgroup$
– user26857
Dec 9 '18 at 23:06
$begingroup$
@user26857 Let $S=k[X_{1},dots,X_{r}]-{0}$.Thank you!
$endgroup$
– Schröchin
Dec 10 '18 at 4:44
add a comment |
$begingroup$
My question is about the hint of exercise 5.1 of Matsumura:
Let $k$ be a field,and $R=k[X_{1},dots,X_{n}]$ and let $mathfrak{p}in operatorname{Spec} R$.
Set $k[X_{1},dots,X_{n}]/mathfrak{p}=k[x_{1},dots,x_{n}]$. Suppose that $x_{1},dots,x_{r}$ is a transcendence basis of $k(x)$ over $k$. Set $K=k[X_{1},dots,X_{r}]$. Then the localization of $k[X_{1},dots,X_{n}]$ at $mathfrak{p}$ is the localization of $K[X_{r+1},dots,X_{n}]$ at a prime ideal $P$.
My question is: Why does $k[X_{1},dots,X_{n}]_{mathfrak{p}}=K[X_{r+1},dots,X_{n}]_{P}$ for some prime ideal $P$? And what does $P$ look like?
commutative-algebra localization
edited Dec 10 '18 at 9:21
user26857
39.3k124183
asked Dec 9 '18 at 16:40
SchröchinSchröchin
955
$endgroup$
My question is about the hint of exercise 5.1 of Matsumura:
Let $k$ be a field,and $R=k[X_{1},dots,X_{n}]$ and let $mathfrak{p}in operatorname{Spec} R$.
Set $k[X_{1},dots,X_{n}]/mathfrak{p}=k[x_{1},dots,x_{n}]$. Suppose that $x_{1},dots,x_{r}$ is a transcendence basis of $k(x)$ over $k$. Set $K=k[X_{1},dots,X_{r}]$. Then the localization of $k[X_{1},dots,X_{n}]$ at $mathfrak{p}$ is the localization of $K[X_{r+1},dots,X_{n}]$ at a prime ideal $P$.
My question is: Why does $k[X_{1},dots,X_{n}]_{mathfrak{p}}=K[X_{r+1},dots,X_{n}]_{P}$ for some prime ideal $P$? And what does $P$ look like?
commutative-algebra localization
commutative-algebra localization
edited Dec 10 '18 at 9:21
user26857
39.3k124183
asked Dec 9 '18 at 16:40
SchröchinSchröchin
955
edited Dec 10 '18 at 9:21
user26857
39.3k124183
asked Dec 9 '18 at 16:40
SchröchinSchröchin
955
edited Dec 10 '18 at 9:21
user26857
39.3k124183
edited Dec 10 '18 at 9:21
user26857
39.3k124183
edited Dec 10 '18 at 9:21
user26857
39.3k124183
39.3k124183
asked Dec 9 '18 at 16:40
SchröchinSchröchin
955
asked Dec 9 '18 at 16:40
SchröchinSchröchin
955
asked Dec 9 '18 at 16:40
SchröchinSchröchin
955
955
$begingroup$
Hint. $R_psimeq (S^{-1}R)_{S^{-1}p}$ where $Ssubset R$ is a multiplicative set such that $Scap p=emptyset$. Can you choose an appropriate $S$?
$endgroup$
– user26857
Dec 9 '18 at 23:06
$begingroup$
@user26857 Let $S=k[X_{1},dots,X_{r}]-{0}$.Thank you!
$endgroup$
– Schröchin
Dec 10 '18 at 4:44
add a comment |
$begingroup$
Hint. $R_psimeq (S^{-1}R)_{S^{-1}p}$ where $Ssubset R$ is a multiplicative set such that $Scap p=emptyset$. Can you choose an appropriate $S$?
$endgroup$
– user26857
Dec 9 '18 at 23:06
$begingroup$
@user26857 Let $S=k[X_{1},dots,X_{r}]-{0}$.Thank you!
$endgroup$
– Schröchin
Dec 10 '18 at 4:44
$begingroup$
Hint. $R_psimeq (S^{-1}R)_{S^{-1}p}$ where $Ssubset R$ is a multiplicative set such that $Scap p=emptyset$. Can you choose an appropriate $S$?
$endgroup$
– user26857
Dec 9 '18 at 23:06
$begingroup$
Hint. $R_psimeq (S^{-1}R)_{S^{-1}p}$ where $Ssubset R$ is a multiplicative set such that $Scap p=emptyset$. Can you choose an appropriate $S$?
$endgroup$
– user26857
Dec 9 '18 at 23:06
$begingroup$
@user26857 Let $S=k[X_{1},dots,X_{r}]-{0}$.Thank you!
$endgroup$
– Schröchin
Dec 10 '18 at 4:44
$begingroup$
@user26857 Let $S=k[X_{1},dots,X_{r}]-{0}$.Thank you!
$endgroup$
– Schröchin
Dec 10 '18 at 4:44
add a comment |
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$begingroup$
Hint. $R_psimeq (S^{-1}R)_{S^{-1}p}$ where $Ssubset R$ is a multiplicative set such that $Scap p=emptyset$. Can you choose an appropriate $S$?
$endgroup$
– user26857
Dec 9 '18 at 23:06
$begingroup$
@user26857 Let $S=k[X_{1},dots,X_{r}]-{0}$.Thank you!
$endgroup$
– Schröchin
Dec 10 '18 at 4:44