Is there an isomorphism between $ text{Spec}(R_{mathfrak{p}}) $ and the prime ideals of $ R $ which are...
$begingroup$
Suppose $ R $ is a ring, and $ mathfrak{p} in text{Spec}(R). $
I have been told that $ text{Spec}(R_{mathfrak{p}}) cong lbrace mathfrak{q} in text{Spec}(R);| mathfrak{q} subset mathfrak{p} rbrace, $ and I am trying to figure out whether this is true or not, but I am stuck trying to show that there is an injective map $ alpha : text{Spec}(R_{mathfrak{p}}) longrightarrow lbrace mathfrak{q} in text{Spec}(R);| mathfrak{q} subset mathfrak{p} rbrace.$
I suspect that there is a way to show this by exploiting the fact that any (prime) ideal of $ text{Spec}(R_{p}) $ cannot contain units of $ R_{mathfrak{p}} $, but I can't find a way through.
Any assistance would be much appreciated.
commutative-algebra
asked Jan 11 at 2:53
Addled StudentAddled Student
749
$endgroup$
add a comment |
$begingroup$
Suppose $ R $ is a ring, and $ mathfrak{p} in text{Spec}(R). $
I have been told that $ text{Spec}(R_{mathfrak{p}}) cong lbrace mathfrak{q} in text{Spec}(R);| mathfrak{q} subset mathfrak{p} rbrace, $ and I am trying to figure out whether this is true or not, but I am stuck trying to show that there is an injective map $ alpha : text{Spec}(R_{mathfrak{p}}) longrightarrow lbrace mathfrak{q} in text{Spec}(R);| mathfrak{q} subset mathfrak{p} rbrace.$
I suspect that there is a way to show this by exploiting the fact that any (prime) ideal of $ text{Spec}(R_{p}) $ cannot contain units of $ R_{mathfrak{p}} $, but I can't find a way through.
Any assistance would be much appreciated.
commutative-algebra
asked Jan 11 at 2:53
Addled StudentAddled Student
749
$endgroup$
add a comment |
$begingroup$
Suppose $ R $ is a ring, and $ mathfrak{p} in text{Spec}(R). $
I have been told that $ text{Spec}(R_{mathfrak{p}}) cong lbrace mathfrak{q} in text{Spec}(R);| mathfrak{q} subset mathfrak{p} rbrace, $ and I am trying to figure out whether this is true or not, but I am stuck trying to show that there is an injective map $ alpha : text{Spec}(R_{mathfrak{p}}) longrightarrow lbrace mathfrak{q} in text{Spec}(R);| mathfrak{q} subset mathfrak{p} rbrace.$
I suspect that there is a way to show this by exploiting the fact that any (prime) ideal of $ text{Spec}(R_{p}) $ cannot contain units of $ R_{mathfrak{p}} $, but I can't find a way through.
Any assistance would be much appreciated.
commutative-algebra
asked Jan 11 at 2:53
Addled StudentAddled Student
749
$endgroup$
Suppose $ R $ is a ring, and $ mathfrak{p} in text{Spec}(R). $
I have been told that $ text{Spec}(R_{mathfrak{p}}) cong lbrace mathfrak{q} in text{Spec}(R);| mathfrak{q} subset mathfrak{p} rbrace, $ and I am trying to figure out whether this is true or not, but I am stuck trying to show that there is an injective map $ alpha : text{Spec}(R_{mathfrak{p}}) longrightarrow lbrace mathfrak{q} in text{Spec}(R);| mathfrak{q} subset mathfrak{p} rbrace.$
I suspect that there is a way to show this by exploiting the fact that any (prime) ideal of $ text{Spec}(R_{p}) $ cannot contain units of $ R_{mathfrak{p}} $, but I can't find a way through.
Any assistance would be much appreciated.
commutative-algebra
commutative-algebra
asked Jan 11 at 2:53
Addled StudentAddled Student
749
asked Jan 11 at 2:53
Addled StudentAddled Student
749
asked Jan 11 at 2:53
Addled StudentAddled Student
749
asked Jan 11 at 2:53
Addled StudentAddled Student
749
asked Jan 11 at 2:53
Addled StudentAddled Student
749
749
add a comment |
add a comment |
1 Answer
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$begingroup$
In a word, yes.
In general, for a multiplicatively closed subset $S$ of $R$,
there is a natural correspondence between the prime ideals of
$S^{-1}R$ and the prime ideals of $R$ disjoint from $S$. Your
example is the case where $S=R-newcommand{fp}{mathfrak{p}}
newcommand{fq}{mathfrak{q}}fp$.
The prime ideals of $S^{-1}R$ all have the form $S^{-1}fq$
where $fqsubseteq fp$ is a prime ideal of $S$. It is straightforward
to check that these $S^{-1}fq$ are prime in $S^{-1}R$. Conversely,
let $Q$ be a prime ideal of $S^{-1}R$. Then
$fq={ain R:a/1in Q}$ is a prime ideal of $R$, being the inverse image
of $Q$ under the map $phi:amapsto a/1$ from $R$ to $S^{-1}R$.
If $b/sin Q$, then $b/1=(b/s)(s/1)in Q$ so $binfq$ and so $Q=S^{-1}fq$
etc.
Texts on commutative algebra will have more details.
Your $alpha$ is $Qmapstophi^{-1}(Q)$.
answered Jan 11 at 4:20
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
$begingroup$
Thanks for this. I think I see your argument. It's quite nice. Arguments like this make the effort of trying to understand these things worth it in the end.
$endgroup$
– Addled Student
Jan 20 at 4:29
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
In a word, yes.
In general, for a multiplicatively closed subset $S$ of $R$,
there is a natural correspondence between the prime ideals of
$S^{-1}R$ and the prime ideals of $R$ disjoint from $S$. Your
example is the case where $S=R-newcommand{fp}{mathfrak{p}}
newcommand{fq}{mathfrak{q}}fp$.
The prime ideals of $S^{-1}R$ all have the form $S^{-1}fq$
where $fqsubseteq fp$ is a prime ideal of $S$. It is straightforward
to check that these $S^{-1}fq$ are prime in $S^{-1}R$. Conversely,
let $Q$ be a prime ideal of $S^{-1}R$. Then
$fq={ain R:a/1in Q}$ is a prime ideal of $R$, being the inverse image
of $Q$ under the map $phi:amapsto a/1$ from $R$ to $S^{-1}R$.
If $b/sin Q$, then $b/1=(b/s)(s/1)in Q$ so $binfq$ and so $Q=S^{-1}fq$
etc.
Texts on commutative algebra will have more details.
Your $alpha$ is $Qmapstophi^{-1}(Q)$.
answered Jan 11 at 4:20
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
$begingroup$
Thanks for this. I think I see your argument. It's quite nice. Arguments like this make the effort of trying to understand these things worth it in the end.
$endgroup$
– Addled Student
Jan 20 at 4:29
add a comment |
$begingroup$
In a word, yes.
In general, for a multiplicatively closed subset $S$ of $R$,
there is a natural correspondence between the prime ideals of
$S^{-1}R$ and the prime ideals of $R$ disjoint from $S$. Your
example is the case where $S=R-newcommand{fp}{mathfrak{p}}
newcommand{fq}{mathfrak{q}}fp$.
The prime ideals of $S^{-1}R$ all have the form $S^{-1}fq$
where $fqsubseteq fp$ is a prime ideal of $S$. It is straightforward
to check that these $S^{-1}fq$ are prime in $S^{-1}R$. Conversely,
let $Q$ be a prime ideal of $S^{-1}R$. Then
$fq={ain R:a/1in Q}$ is a prime ideal of $R$, being the inverse image
of $Q$ under the map $phi:amapsto a/1$ from $R$ to $S^{-1}R$.
If $b/sin Q$, then $b/1=(b/s)(s/1)in Q$ so $binfq$ and so $Q=S^{-1}fq$
etc.
Texts on commutative algebra will have more details.
Your $alpha$ is $Qmapstophi^{-1}(Q)$.
answered Jan 11 at 4:20
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
$begingroup$
Thanks for this. I think I see your argument. It's quite nice. Arguments like this make the effort of trying to understand these things worth it in the end.
$endgroup$
– Addled Student
Jan 20 at 4:29
add a comment |
$begingroup$
In a word, yes.
In general, for a multiplicatively closed subset $S$ of $R$,
there is a natural correspondence between the prime ideals of
$S^{-1}R$ and the prime ideals of $R$ disjoint from $S$. Your
example is the case where $S=R-newcommand{fp}{mathfrak{p}}
newcommand{fq}{mathfrak{q}}fp$.
The prime ideals of $S^{-1}R$ all have the form $S^{-1}fq$
where $fqsubseteq fp$ is a prime ideal of $S$. It is straightforward
to check that these $S^{-1}fq$ are prime in $S^{-1}R$. Conversely,
let $Q$ be a prime ideal of $S^{-1}R$. Then
$fq={ain R:a/1in Q}$ is a prime ideal of $R$, being the inverse image
of $Q$ under the map $phi:amapsto a/1$ from $R$ to $S^{-1}R$.
If $b/sin Q$, then $b/1=(b/s)(s/1)in Q$ so $binfq$ and so $Q=S^{-1}fq$
etc.
Texts on commutative algebra will have more details.
Your $alpha$ is $Qmapstophi^{-1}(Q)$.
answered Jan 11 at 4:20
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
In a word, yes.
In general, for a multiplicatively closed subset $S$ of $R$,
there is a natural correspondence between the prime ideals of
$S^{-1}R$ and the prime ideals of $R$ disjoint from $S$. Your
example is the case where $S=R-newcommand{fp}{mathfrak{p}}
newcommand{fq}{mathfrak{q}}fp$.
The prime ideals of $S^{-1}R$ all have the form $S^{-1}fq$
where $fqsubseteq fp$ is a prime ideal of $S$. It is straightforward
to check that these $S^{-1}fq$ are prime in $S^{-1}R$. Conversely,
let $Q$ be a prime ideal of $S^{-1}R$. Then
$fq={ain R:a/1in Q}$ is a prime ideal of $R$, being the inverse image
of $Q$ under the map $phi:amapsto a/1$ from $R$ to $S^{-1}R$.
If $b/sin Q$, then $b/1=(b/s)(s/1)in Q$ so $binfq$ and so $Q=S^{-1}fq$
etc.
Texts on commutative algebra will have more details.
Your $alpha$ is $Qmapstophi^{-1}(Q)$.
answered Jan 11 at 4:20
Lord Shark the UnknownLord Shark the Unknown
108k1162136
answered Jan 11 at 4:20
Lord Shark the UnknownLord Shark the Unknown
108k1162136
answered Jan 11 at 4:20
Lord Shark the UnknownLord Shark the Unknown
108k1162136
answered Jan 11 at 4:20
Lord Shark the UnknownLord Shark the Unknown
108k1162136
108k1162136
$begingroup$
Thanks for this. I think I see your argument. It's quite nice. Arguments like this make the effort of trying to understand these things worth it in the end.
$endgroup$
– Addled Student
Jan 20 at 4:29
add a comment |
$begingroup$
Thanks for this. I think I see your argument. It's quite nice. Arguments like this make the effort of trying to understand these things worth it in the end.
$endgroup$
– Addled Student
Jan 20 at 4:29
$begingroup$
Thanks for this. I think I see your argument. It's quite nice. Arguments like this make the effort of trying to understand these things worth it in the end.
$endgroup$
– Addled Student
Jan 20 at 4:29
$begingroup$
Thanks for this. I think I see your argument. It's quite nice. Arguments like this make the effort of trying to understand these things worth it in the end.
$endgroup$
– Addled Student
Jan 20 at 4:29
add a comment |
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