Mixing
$mathfrak p$ be a prime ideal of a commutative ring $R$ , then $R_mathfrak p/mathfrak pR_mathfrak p$ is the...
$begingroup$
Let $mathfrak p$ be a prime ideal of a commutative ring $R$ with unity , then how to show that the field $R_mathfrak p/mathfrak pR_mathfrak p$ is isomorphic with the field of fractions of the integral domain $R/mathfrak p$ ?
I can show that $R_mathfrak p/mathfrak pR_mathfrak p cong (R/mathfrak p)_mathfrak p$ as $R_mathfrak p$ modules ... but I am not sure whether this implies the result I am asking or not
abstract-algebra ring-theory commutative-algebra maximal-and-prime-ideals localization
edited Jan 26 at 7:18
Andrews
1,2691421
asked Sep 4 '17 at 15:24
useruser
1
$endgroup$
add a comment |
$begingroup$
Let $mathfrak p$ be a prime ideal of a commutative ring $R$ with unity , then how to show that the field $R_mathfrak p/mathfrak pR_mathfrak p$ is isomorphic with the field of fractions of the integral domain $R/mathfrak p$ ?
I can show that $R_mathfrak p/mathfrak pR_mathfrak p cong (R/mathfrak p)_mathfrak p$ as $R_mathfrak p$ modules ... but I am not sure whether this implies the result I am asking or not
abstract-algebra ring-theory commutative-algebra maximal-and-prime-ideals localization
edited Jan 26 at 7:18
Andrews
1,2691421
asked Sep 4 '17 at 15:24
useruser
1
$endgroup$
1
$begingroup$
Does this help? math.stackexchange.com/questions/502448/…
$endgroup$
– Paul K
Sep 4 '17 at 15:34
$begingroup$
@PaulK : Not as it is ... please see my edit
$endgroup$
– user
Sep 4 '17 at 15:39
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$(R/P)_P$ is the field of fractions of $R/P$!
$endgroup$
– Bernard
Sep 4 '17 at 15:42
1
$begingroup$
@Bernard : True .. but the isomorphism I get is an $R_P$ module isomorphism not a ring isomorphism ...
$endgroup$
– user
Sep 4 '17 at 15:44
1
$begingroup$
It's also a ring homomorphism.
$endgroup$
– Bernard
Sep 4 '17 at 16:02
add a comment |
$begingroup$
Let $mathfrak p$ be a prime ideal of a commutative ring $R$ with unity , then how to show that the field $R_mathfrak p/mathfrak pR_mathfrak p$ is isomorphic with the field of fractions of the integral domain $R/mathfrak p$ ?
I can show that $R_mathfrak p/mathfrak pR_mathfrak p cong (R/mathfrak p)_mathfrak p$ as $R_mathfrak p$ modules ... but I am not sure whether this implies the result I am asking or not
abstract-algebra ring-theory commutative-algebra maximal-and-prime-ideals localization
edited Jan 26 at 7:18
Andrews
1,2691421
asked Sep 4 '17 at 15:24
useruser
1
$endgroup$
Let $mathfrak p$ be a prime ideal of a commutative ring $R$ with unity , then how to show that the field $R_mathfrak p/mathfrak pR_mathfrak p$ is isomorphic with the field of fractions of the integral domain $R/mathfrak p$ ?
I can show that $R_mathfrak p/mathfrak pR_mathfrak p cong (R/mathfrak p)_mathfrak p$ as $R_mathfrak p$ modules ... but I am not sure whether this implies the result I am asking or not
abstract-algebra ring-theory commutative-algebra maximal-and-prime-ideals localization
abstract-algebra ring-theory commutative-algebra maximal-and-prime-ideals localization
edited Jan 26 at 7:18
Andrews
1,2691421
asked Sep 4 '17 at 15:24
useruser
1
edited Jan 26 at 7:18
Andrews
1,2691421
asked Sep 4 '17 at 15:24
useruser
1
edited Jan 26 at 7:18
Andrews
1,2691421
edited Jan 26 at 7:18
Andrews
1,2691421
edited Jan 26 at 7:18
Andrews
1,2691421
1,2691421
asked Sep 4 '17 at 15:24
useruser
1
asked Sep 4 '17 at 15:24
useruser
1
asked Sep 4 '17 at 15:24
useruser
1
1
1
$begingroup$
Does this help? math.stackexchange.com/questions/502448/…
$endgroup$
– Paul K
Sep 4 '17 at 15:34
$begingroup$
@PaulK : Not as it is ... please see my edit
$endgroup$
– user
Sep 4 '17 at 15:39
$begingroup$
$(R/P)_P$ is the field of fractions of $R/P$!
$endgroup$
– Bernard
Sep 4 '17 at 15:42
1
$begingroup$
@Bernard : True .. but the isomorphism I get is an $R_P$ module isomorphism not a ring isomorphism ...
$endgroup$
– user
Sep 4 '17 at 15:44
1
$begingroup$
It's also a ring homomorphism.
$endgroup$
– Bernard
Sep 4 '17 at 16:02
add a comment |
1
$begingroup$
Does this help? math.stackexchange.com/questions/502448/…
$endgroup$
– Paul K
Sep 4 '17 at 15:34
$begingroup$
@PaulK : Not as it is ... please see my edit
$endgroup$
– user
Sep 4 '17 at 15:39
$begingroup$
$(R/P)_P$ is the field of fractions of $R/P$!
$endgroup$
– Bernard
Sep 4 '17 at 15:42
1
$begingroup$
@Bernard : True .. but the isomorphism I get is an $R_P$ module isomorphism not a ring isomorphism ...
$endgroup$
– user
Sep 4 '17 at 15:44
1
$begingroup$
It's also a ring homomorphism.
$endgroup$
– Bernard
Sep 4 '17 at 16:02
1
1
$begingroup$
Does this help? math.stackexchange.com/questions/502448/…
$endgroup$
– Paul K
Sep 4 '17 at 15:34
$begingroup$
Does this help? math.stackexchange.com/questions/502448/…
$endgroup$
– Paul K
Sep 4 '17 at 15:34
$begingroup$
@PaulK : Not as it is ... please see my edit
$endgroup$
– user
Sep 4 '17 at 15:39
$begingroup$
@PaulK : Not as it is ... please see my edit
$endgroup$
– user
Sep 4 '17 at 15:39
$begingroup$
$(R/P)_P$ is the field of fractions of $R/P$!
$endgroup$
– Bernard
Sep 4 '17 at 15:42
$begingroup$
$(R/P)_P$ is the field of fractions of $R/P$!
$endgroup$
– Bernard
Sep 4 '17 at 15:42
1
1
$begingroup$
@Bernard : True .. but the isomorphism I get is an $R_P$ module isomorphism not a ring isomorphism ...
$endgroup$
– user
Sep 4 '17 at 15:44
$begingroup$
@Bernard : True .. but the isomorphism I get is an $R_P$ module isomorphism not a ring isomorphism ...
$endgroup$
– user
Sep 4 '17 at 15:44
1
1
$begingroup$
It's also a ring homomorphism.
$endgroup$
– Bernard
Sep 4 '17 at 16:02
$begingroup$
It's also a ring homomorphism.
$endgroup$
– Bernard
Sep 4 '17 at 16:02
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There's an obvious map $fcolon R_Pto text{Frac}(R/P)$, given by $a/bmapsto (a+P)/(b+P)$. You need to check:
- If $a/bin R_P$, then $f(a/b)in text{Frac}(R/P)$, i.e., $b+Pneq 0$ in $R/P$.
- If $a/b = c/d$ in $R_P$, then $f(a/b) = f(c/d)$.
- $f$ is a homomorphism.
To show that $f$ induces an isomorphism $R_P/PR_Pcong text{Frac}(R/P)$, you just need to show that the kernel of $f$ is $PR_P$. Well, suppose $a/b$ is in the kernel. Then $f(a/b) = (a+P)/(b+P)$ is equal to $0$ in $text{Frac}(R/P)$, so $a+P$ is equal to $0$ in $R/P$. So $ain P$, and $a/b = a(1/b) in PR_P$. Conversely, if $a/bin PR_P$, then it is equivalent to $p/b'$ for some $pin P$ and $b'notin P$. And $f(p/b') = (p+P)/(b'+P) = 0$, so $a/b$ is in the kernel (you've already checked that $f$ is well-defined, so it suffices to look at just one representative of its equivalence class).
There's a more category-theoretic way to do this exercise, where you check that $R_P/PR_P$ and $text{Frac}(R/P)$ both satisfy the universal property that a ring homomorphism from this ring to $S$ is uniquely determined by a ring homomorphism $Rto S$ which maps every element of $P$ to $0$ and every element of $Rbackslash P$ to a unit, and hence they must be isomorphic (by Yoneda's Lemma). But it sounds like you might benefit more at this stage from working through the details of the more "concrete" proof.
answered Sep 4 '17 at 15:42
Alex KruckmanAlex Kruckman
28.1k32658
$endgroup$
1
$begingroup$
+1 I also think that OP will be benefited from "working through the details of the more "concrete" proof.".
$endgroup$
– Krish
Sep 4 '17 at 15:53
$begingroup$
@Alex Kruckman : why $a=pa'$ ?
$endgroup$
– user
Sep 4 '17 at 16:25
$begingroup$
@users Because that's how you multiply in the localization! If $a/b = (p/1)(a'/b')$, then $a = pa'$ and $b = 1b'$.
$endgroup$
– Alex Kruckman
Sep 4 '17 at 17:32
$begingroup$
@AlexKruckman : But $a/b=pa'/b'$ implies $s(b' a - bpa')=0 $ for some $s notin P$ ...
$endgroup$
– user
Sep 4 '17 at 17:35
$begingroup$
So $sb'a = sbpa'in P$. Since $snotin P$, and $b'notin P$, and $P$ is prime, $ain P$. But ok, maybe I've been a little sloppy in what I wrote above, I"ll rewrite that paragraph.
$endgroup$
– Alex Kruckman
Sep 4 '17 at 17:44
add a comment |
$begingroup$
Since localization preserves exact sequences we have that the exact sequence
$$ 0 to mathfrak p to R to R/mathfrak p to 0$$
becomes
$$0 to mathfrak p R_{mathfrak p} to R_{mathfrak p} to (R/mathfrak p)_{mathfrak p} to 0$$
Hence $(R/mathfrak p)_{mathfrak p}=R_{mathfrak p}/mathfrak pR_mathfrak p$. From this you can conclude that it is the field of fractions since $R/mathfrak p$ is a domain and localizing by $mathfrak p$ means inverting all nonzero elements (there are no nonzero divisor in an integral domain).
answered Sep 4 '17 at 15:45
SC30SC30
43028
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2
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this proof is better
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– Sky
Sep 4 '17 at 15:47
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Better of what?
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– SC30
Sep 4 '17 at 15:47
2
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This proof gives an isomorphism of modules - is there a way to see automatically that it gives an isomorphism of rings as well?
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– Alex Kruckman
Sep 4 '17 at 15:48
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@SC30 but you still need to explain $(R/p)_p$ is isomorphic to the fraction field!better we know this localization.
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– Sky
Sep 4 '17 at 15:51
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@SC30 : Like Alex Kruckman , I am also tempted to comment that this only shows module isomorphism , which I already know as stated in the question
$endgroup$
– user
Sep 4 '17 at 15:51
|
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2 Answers
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2 Answers
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$begingroup$
There's an obvious map $fcolon R_Pto text{Frac}(R/P)$, given by $a/bmapsto (a+P)/(b+P)$. You need to check:
- If $a/bin R_P$, then $f(a/b)in text{Frac}(R/P)$, i.e., $b+Pneq 0$ in $R/P$.
- If $a/b = c/d$ in $R_P$, then $f(a/b) = f(c/d)$.
- $f$ is a homomorphism.
To show that $f$ induces an isomorphism $R_P/PR_Pcong text{Frac}(R/P)$, you just need to show that the kernel of $f$ is $PR_P$. Well, suppose $a/b$ is in the kernel. Then $f(a/b) = (a+P)/(b+P)$ is equal to $0$ in $text{Frac}(R/P)$, so $a+P$ is equal to $0$ in $R/P$. So $ain P$, and $a/b = a(1/b) in PR_P$. Conversely, if $a/bin PR_P$, then it is equivalent to $p/b'$ for some $pin P$ and $b'notin P$. And $f(p/b') = (p+P)/(b'+P) = 0$, so $a/b$ is in the kernel (you've already checked that $f$ is well-defined, so it suffices to look at just one representative of its equivalence class).
There's a more category-theoretic way to do this exercise, where you check that $R_P/PR_P$ and $text{Frac}(R/P)$ both satisfy the universal property that a ring homomorphism from this ring to $S$ is uniquely determined by a ring homomorphism $Rto S$ which maps every element of $P$ to $0$ and every element of $Rbackslash P$ to a unit, and hence they must be isomorphic (by Yoneda's Lemma). But it sounds like you might benefit more at this stage from working through the details of the more "concrete" proof.
answered Sep 4 '17 at 15:42
Alex KruckmanAlex Kruckman
28.1k32658
$endgroup$
1
$begingroup$
+1 I also think that OP will be benefited from "working through the details of the more "concrete" proof.".
$endgroup$
– Krish
Sep 4 '17 at 15:53
$begingroup$
@Alex Kruckman : why $a=pa'$ ?
$endgroup$
– user
Sep 4 '17 at 16:25
$begingroup$
@users Because that's how you multiply in the localization! If $a/b = (p/1)(a'/b')$, then $a = pa'$ and $b = 1b'$.
$endgroup$
– Alex Kruckman
Sep 4 '17 at 17:32
$begingroup$
@AlexKruckman : But $a/b=pa'/b'$ implies $s(b' a - bpa')=0 $ for some $s notin P$ ...
$endgroup$
– user
Sep 4 '17 at 17:35
$begingroup$
So $sb'a = sbpa'in P$. Since $snotin P$, and $b'notin P$, and $P$ is prime, $ain P$. But ok, maybe I've been a little sloppy in what I wrote above, I"ll rewrite that paragraph.
$endgroup$
– Alex Kruckman
Sep 4 '17 at 17:44
add a comment |
$begingroup$
There's an obvious map $fcolon R_Pto text{Frac}(R/P)$, given by $a/bmapsto (a+P)/(b+P)$. You need to check:
- If $a/bin R_P$, then $f(a/b)in text{Frac}(R/P)$, i.e., $b+Pneq 0$ in $R/P$.
- If $a/b = c/d$ in $R_P$, then $f(a/b) = f(c/d)$.
- $f$ is a homomorphism.
To show that $f$ induces an isomorphism $R_P/PR_Pcong text{Frac}(R/P)$, you just need to show that the kernel of $f$ is $PR_P$. Well, suppose $a/b$ is in the kernel. Then $f(a/b) = (a+P)/(b+P)$ is equal to $0$ in $text{Frac}(R/P)$, so $a+P$ is equal to $0$ in $R/P$. So $ain P$, and $a/b = a(1/b) in PR_P$. Conversely, if $a/bin PR_P$, then it is equivalent to $p/b'$ for some $pin P$ and $b'notin P$. And $f(p/b') = (p+P)/(b'+P) = 0$, so $a/b$ is in the kernel (you've already checked that $f$ is well-defined, so it suffices to look at just one representative of its equivalence class).
There's a more category-theoretic way to do this exercise, where you check that $R_P/PR_P$ and $text{Frac}(R/P)$ both satisfy the universal property that a ring homomorphism from this ring to $S$ is uniquely determined by a ring homomorphism $Rto S$ which maps every element of $P$ to $0$ and every element of $Rbackslash P$ to a unit, and hence they must be isomorphic (by Yoneda's Lemma). But it sounds like you might benefit more at this stage from working through the details of the more "concrete" proof.
answered Sep 4 '17 at 15:42
Alex KruckmanAlex Kruckman
28.1k32658
$endgroup$
1
$begingroup$
+1 I also think that OP will be benefited from "working through the details of the more "concrete" proof.".
$endgroup$
– Krish
Sep 4 '17 at 15:53
$begingroup$
@Alex Kruckman : why $a=pa'$ ?
$endgroup$
– user
Sep 4 '17 at 16:25
$begingroup$
@users Because that's how you multiply in the localization! If $a/b = (p/1)(a'/b')$, then $a = pa'$ and $b = 1b'$.
$endgroup$
– Alex Kruckman
Sep 4 '17 at 17:32
$begingroup$
@AlexKruckman : But $a/b=pa'/b'$ implies $s(b' a - bpa')=0 $ for some $s notin P$ ...
$endgroup$
– user
Sep 4 '17 at 17:35
$begingroup$
So $sb'a = sbpa'in P$. Since $snotin P$, and $b'notin P$, and $P$ is prime, $ain P$. But ok, maybe I've been a little sloppy in what I wrote above, I"ll rewrite that paragraph.
$endgroup$
– Alex Kruckman
Sep 4 '17 at 17:44
add a comment |
$begingroup$
There's an obvious map $fcolon R_Pto text{Frac}(R/P)$, given by $a/bmapsto (a+P)/(b+P)$. You need to check:
- If $a/bin R_P$, then $f(a/b)in text{Frac}(R/P)$, i.e., $b+Pneq 0$ in $R/P$.
- If $a/b = c/d$ in $R_P$, then $f(a/b) = f(c/d)$.
- $f$ is a homomorphism.
To show that $f$ induces an isomorphism $R_P/PR_Pcong text{Frac}(R/P)$, you just need to show that the kernel of $f$ is $PR_P$. Well, suppose $a/b$ is in the kernel. Then $f(a/b) = (a+P)/(b+P)$ is equal to $0$ in $text{Frac}(R/P)$, so $a+P$ is equal to $0$ in $R/P$. So $ain P$, and $a/b = a(1/b) in PR_P$. Conversely, if $a/bin PR_P$, then it is equivalent to $p/b'$ for some $pin P$ and $b'notin P$. And $f(p/b') = (p+P)/(b'+P) = 0$, so $a/b$ is in the kernel (you've already checked that $f$ is well-defined, so it suffices to look at just one representative of its equivalence class).
There's a more category-theoretic way to do this exercise, where you check that $R_P/PR_P$ and $text{Frac}(R/P)$ both satisfy the universal property that a ring homomorphism from this ring to $S$ is uniquely determined by a ring homomorphism $Rto S$ which maps every element of $P$ to $0$ and every element of $Rbackslash P$ to a unit, and hence they must be isomorphic (by Yoneda's Lemma). But it sounds like you might benefit more at this stage from working through the details of the more "concrete" proof.
answered Sep 4 '17 at 15:42
Alex KruckmanAlex Kruckman
28.1k32658
$endgroup$
There's an obvious map $fcolon R_Pto text{Frac}(R/P)$, given by $a/bmapsto (a+P)/(b+P)$. You need to check:
- If $a/bin R_P$, then $f(a/b)in text{Frac}(R/P)$, i.e., $b+Pneq 0$ in $R/P$.
- If $a/b = c/d$ in $R_P$, then $f(a/b) = f(c/d)$.
- $f$ is a homomorphism.
To show that $f$ induces an isomorphism $R_P/PR_Pcong text{Frac}(R/P)$, you just need to show that the kernel of $f$ is $PR_P$. Well, suppose $a/b$ is in the kernel. Then $f(a/b) = (a+P)/(b+P)$ is equal to $0$ in $text{Frac}(R/P)$, so $a+P$ is equal to $0$ in $R/P$. So $ain P$, and $a/b = a(1/b) in PR_P$. Conversely, if $a/bin PR_P$, then it is equivalent to $p/b'$ for some $pin P$ and $b'notin P$. And $f(p/b') = (p+P)/(b'+P) = 0$, so $a/b$ is in the kernel (you've already checked that $f$ is well-defined, so it suffices to look at just one representative of its equivalence class).
There's a more category-theoretic way to do this exercise, where you check that $R_P/PR_P$ and $text{Frac}(R/P)$ both satisfy the universal property that a ring homomorphism from this ring to $S$ is uniquely determined by a ring homomorphism $Rto S$ which maps every element of $P$ to $0$ and every element of $Rbackslash P$ to a unit, and hence they must be isomorphic (by Yoneda's Lemma). But it sounds like you might benefit more at this stage from working through the details of the more "concrete" proof.
answered Sep 4 '17 at 15:42
Alex KruckmanAlex Kruckman
28.1k32658
edited Sep 4 '17 at 17:46
answered Sep 4 '17 at 15:42
Alex KruckmanAlex Kruckman
28.1k32658
answered Sep 4 '17 at 15:42
Alex KruckmanAlex Kruckman
28.1k32658
answered Sep 4 '17 at 15:42
Alex KruckmanAlex Kruckman
28.1k32658
28.1k32658
1
$begingroup$
+1 I also think that OP will be benefited from "working through the details of the more "concrete" proof.".
$endgroup$
– Krish
Sep 4 '17 at 15:53
$begingroup$
@Alex Kruckman : why $a=pa'$ ?
$endgroup$
– user
Sep 4 '17 at 16:25
$begingroup$
@users Because that's how you multiply in the localization! If $a/b = (p/1)(a'/b')$, then $a = pa'$ and $b = 1b'$.
$endgroup$
– Alex Kruckman
Sep 4 '17 at 17:32
$begingroup$
@AlexKruckman : But $a/b=pa'/b'$ implies $s(b' a - bpa')=0 $ for some $s notin P$ ...
$endgroup$
– user
Sep 4 '17 at 17:35
$begingroup$
So $sb'a = sbpa'in P$. Since $snotin P$, and $b'notin P$, and $P$ is prime, $ain P$. But ok, maybe I've been a little sloppy in what I wrote above, I"ll rewrite that paragraph.
$endgroup$
– Alex Kruckman
Sep 4 '17 at 17:44
add a comment |
1
$begingroup$
+1 I also think that OP will be benefited from "working through the details of the more "concrete" proof.".
$endgroup$
– Krish
Sep 4 '17 at 15:53
$begingroup$
@Alex Kruckman : why $a=pa'$ ?
$endgroup$
– user
Sep 4 '17 at 16:25
$begingroup$
@users Because that's how you multiply in the localization! If $a/b = (p/1)(a'/b')$, then $a = pa'$ and $b = 1b'$.
$endgroup$
– Alex Kruckman
Sep 4 '17 at 17:32
$begingroup$
@AlexKruckman : But $a/b=pa'/b'$ implies $s(b' a - bpa')=0 $ for some $s notin P$ ...
$endgroup$
– user
Sep 4 '17 at 17:35
$begingroup$
So $sb'a = sbpa'in P$. Since $snotin P$, and $b'notin P$, and $P$ is prime, $ain P$. But ok, maybe I've been a little sloppy in what I wrote above, I"ll rewrite that paragraph.
$endgroup$
– Alex Kruckman
Sep 4 '17 at 17:44
1
1
$begingroup$
+1 I also think that OP will be benefited from "working through the details of the more "concrete" proof.".
$endgroup$
– Krish
Sep 4 '17 at 15:53
$begingroup$
+1 I also think that OP will be benefited from "working through the details of the more "concrete" proof.".
$endgroup$
– Krish
Sep 4 '17 at 15:53
$begingroup$
@Alex Kruckman : why $a=pa'$ ?
$endgroup$
– user
Sep 4 '17 at 16:25
$begingroup$
@Alex Kruckman : why $a=pa'$ ?
$endgroup$
– user
Sep 4 '17 at 16:25
$begingroup$
@users Because that's how you multiply in the localization! If $a/b = (p/1)(a'/b')$, then $a = pa'$ and $b = 1b'$.
$endgroup$
– Alex Kruckman
Sep 4 '17 at 17:32
$begingroup$
@users Because that's how you multiply in the localization! If $a/b = (p/1)(a'/b')$, then $a = pa'$ and $b = 1b'$.
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– Alex Kruckman
Sep 4 '17 at 17:32
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@AlexKruckman : But $a/b=pa'/b'$ implies $s(b' a - bpa')=0 $ for some $s notin P$ ...
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– user
Sep 4 '17 at 17:35
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@AlexKruckman : But $a/b=pa'/b'$ implies $s(b' a - bpa')=0 $ for some $s notin P$ ...
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– user
Sep 4 '17 at 17:35
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So $sb'a = sbpa'in P$. Since $snotin P$, and $b'notin P$, and $P$ is prime, $ain P$. But ok, maybe I've been a little sloppy in what I wrote above, I"ll rewrite that paragraph.
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– Alex Kruckman
Sep 4 '17 at 17:44
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So $sb'a = sbpa'in P$. Since $snotin P$, and $b'notin P$, and $P$ is prime, $ain P$. But ok, maybe I've been a little sloppy in what I wrote above, I"ll rewrite that paragraph.
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– Alex Kruckman
Sep 4 '17 at 17:44
add a comment |
$begingroup$
Since localization preserves exact sequences we have that the exact sequence
$$ 0 to mathfrak p to R to R/mathfrak p to 0$$
becomes
$$0 to mathfrak p R_{mathfrak p} to R_{mathfrak p} to (R/mathfrak p)_{mathfrak p} to 0$$
Hence $(R/mathfrak p)_{mathfrak p}=R_{mathfrak p}/mathfrak pR_mathfrak p$. From this you can conclude that it is the field of fractions since $R/mathfrak p$ is a domain and localizing by $mathfrak p$ means inverting all nonzero elements (there are no nonzero divisor in an integral domain).
answered Sep 4 '17 at 15:45
SC30SC30
43028
$endgroup$
2
$begingroup$
this proof is better
$endgroup$
– Sky
Sep 4 '17 at 15:47
$begingroup$
Better of what?
$endgroup$
– SC30
Sep 4 '17 at 15:47
2
$begingroup$
This proof gives an isomorphism of modules - is there a way to see automatically that it gives an isomorphism of rings as well?
$endgroup$
– Alex Kruckman
Sep 4 '17 at 15:48
$begingroup$
@SC30 but you still need to explain $(R/p)_p$ is isomorphic to the fraction field!better we know this localization.
$endgroup$
– Sky
Sep 4 '17 at 15:51
$begingroup$
@SC30 : Like Alex Kruckman , I am also tempted to comment that this only shows module isomorphism , which I already know as stated in the question
$endgroup$
– user
Sep 4 '17 at 15:51
|
show 2 more comments
$begingroup$
Since localization preserves exact sequences we have that the exact sequence
$$ 0 to mathfrak p to R to R/mathfrak p to 0$$
becomes
$$0 to mathfrak p R_{mathfrak p} to R_{mathfrak p} to (R/mathfrak p)_{mathfrak p} to 0$$
Hence $(R/mathfrak p)_{mathfrak p}=R_{mathfrak p}/mathfrak pR_mathfrak p$. From this you can conclude that it is the field of fractions since $R/mathfrak p$ is a domain and localizing by $mathfrak p$ means inverting all nonzero elements (there are no nonzero divisor in an integral domain).
answered Sep 4 '17 at 15:45
SC30SC30
43028
$endgroup$
2
$begingroup$
this proof is better
$endgroup$
– Sky
Sep 4 '17 at 15:47
$begingroup$
Better of what?
$endgroup$
– SC30
Sep 4 '17 at 15:47
2
$begingroup$
This proof gives an isomorphism of modules - is there a way to see automatically that it gives an isomorphism of rings as well?
$endgroup$
– Alex Kruckman
Sep 4 '17 at 15:48
$begingroup$
@SC30 but you still need to explain $(R/p)_p$ is isomorphic to the fraction field!better we know this localization.
$endgroup$
– Sky
Sep 4 '17 at 15:51
$begingroup$
@SC30 : Like Alex Kruckman , I am also tempted to comment that this only shows module isomorphism , which I already know as stated in the question
$endgroup$
– user
Sep 4 '17 at 15:51
|
show 2 more comments
$begingroup$
Since localization preserves exact sequences we have that the exact sequence
$$ 0 to mathfrak p to R to R/mathfrak p to 0$$
becomes
$$0 to mathfrak p R_{mathfrak p} to R_{mathfrak p} to (R/mathfrak p)_{mathfrak p} to 0$$
Hence $(R/mathfrak p)_{mathfrak p}=R_{mathfrak p}/mathfrak pR_mathfrak p$. From this you can conclude that it is the field of fractions since $R/mathfrak p$ is a domain and localizing by $mathfrak p$ means inverting all nonzero elements (there are no nonzero divisor in an integral domain).
answered Sep 4 '17 at 15:45
SC30SC30
43028
$endgroup$
Since localization preserves exact sequences we have that the exact sequence
$$ 0 to mathfrak p to R to R/mathfrak p to 0$$
becomes
$$0 to mathfrak p R_{mathfrak p} to R_{mathfrak p} to (R/mathfrak p)_{mathfrak p} to 0$$
Hence $(R/mathfrak p)_{mathfrak p}=R_{mathfrak p}/mathfrak pR_mathfrak p$. From this you can conclude that it is the field of fractions since $R/mathfrak p$ is a domain and localizing by $mathfrak p$ means inverting all nonzero elements (there are no nonzero divisor in an integral domain).
answered Sep 4 '17 at 15:45
SC30SC30
43028
edited Sep 4 '17 at 15:50
answered Sep 4 '17 at 15:45
SC30SC30
43028
answered Sep 4 '17 at 15:45
SC30SC30
43028
answered Sep 4 '17 at 15:45
SC30SC30
43028
43028
2
$begingroup$
this proof is better
$endgroup$
– Sky
Sep 4 '17 at 15:47
$begingroup$
Better of what?
$endgroup$
– SC30
Sep 4 '17 at 15:47
2
$begingroup$
This proof gives an isomorphism of modules - is there a way to see automatically that it gives an isomorphism of rings as well?
$endgroup$
– Alex Kruckman
Sep 4 '17 at 15:48
$begingroup$
@SC30 but you still need to explain $(R/p)_p$ is isomorphic to the fraction field!better we know this localization.
$endgroup$
– Sky
Sep 4 '17 at 15:51
$begingroup$
@SC30 : Like Alex Kruckman , I am also tempted to comment that this only shows module isomorphism , which I already know as stated in the question
$endgroup$
– user
Sep 4 '17 at 15:51
|
show 2 more comments
2
$begingroup$
this proof is better
$endgroup$
– Sky
Sep 4 '17 at 15:47
$begingroup$
Better of what?
$endgroup$
– SC30
Sep 4 '17 at 15:47
2
$begingroup$
This proof gives an isomorphism of modules - is there a way to see automatically that it gives an isomorphism of rings as well?
$endgroup$
– Alex Kruckman
Sep 4 '17 at 15:48
$begingroup$
@SC30 but you still need to explain $(R/p)_p$ is isomorphic to the fraction field!better we know this localization.
$endgroup$
– Sky
Sep 4 '17 at 15:51
$begingroup$
@SC30 : Like Alex Kruckman , I am also tempted to comment that this only shows module isomorphism , which I already know as stated in the question
$endgroup$
– user
Sep 4 '17 at 15:51
2
2
$begingroup$
this proof is better
$endgroup$
– Sky
Sep 4 '17 at 15:47
$begingroup$
this proof is better
$endgroup$
– Sky
Sep 4 '17 at 15:47
$begingroup$
Better of what?
$endgroup$
– SC30
Sep 4 '17 at 15:47
$begingroup$
Better of what?
$endgroup$
– SC30
Sep 4 '17 at 15:47
2
2
$begingroup$
This proof gives an isomorphism of modules - is there a way to see automatically that it gives an isomorphism of rings as well?
$endgroup$
– Alex Kruckman
Sep 4 '17 at 15:48
$begingroup$
This proof gives an isomorphism of modules - is there a way to see automatically that it gives an isomorphism of rings as well?
$endgroup$
– Alex Kruckman
Sep 4 '17 at 15:48
$begingroup$
@SC30 but you still need to explain $(R/p)_p$ is isomorphic to the fraction field!better we know this localization.
$endgroup$
– Sky
Sep 4 '17 at 15:51
$begingroup$
@SC30 but you still need to explain $(R/p)_p$ is isomorphic to the fraction field!better we know this localization.
$endgroup$
– Sky
Sep 4 '17 at 15:51
$begingroup$
@SC30 : Like Alex Kruckman , I am also tempted to comment that this only shows module isomorphism , which I already know as stated in the question
$endgroup$
– user
Sep 4 '17 at 15:51
$begingroup$
@SC30 : Like Alex Kruckman , I am also tempted to comment that this only shows module isomorphism , which I already know as stated in the question
$endgroup$
– user
Sep 4 '17 at 15:51
|
show 2 more comments
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$begingroup$
Does this help? math.stackexchange.com/questions/502448/…
$endgroup$
– Paul K
Sep 4 '17 at 15:34
$begingroup$
@PaulK : Not as it is ... please see my edit
$endgroup$
– user
Sep 4 '17 at 15:39
$begingroup$
$(R/P)_P$ is the field of fractions of $R/P$!
$endgroup$
– Bernard
Sep 4 '17 at 15:42
1
$begingroup$
@Bernard : True .. but the isomorphism I get is an $R_P$ module isomorphism not a ring isomorphism ...
$endgroup$
– user
Sep 4 '17 at 15:44
1
$begingroup$
It's also a ring homomorphism.
$endgroup$
– Bernard
Sep 4 '17 at 16:02