Intersection over all maximal ideals of localizations
$begingroup$
Let $A$ be an integral domain. Show that $$bigcap limits_{mathfrak{m}}A_{mathfrak{m}}=A,$$ where the intersection is taken over all maximal ideals $mathfrak{m}$ in $A$.
P.S. This is quite well-known problem in commutative algebra and I have found many topics in MSE which contains its solution. So please do not duplicate my question by the the following reason:
But my question is not about the solution of this problem. How these object could be equal if the LHS and RHS have elements of different structure. More prcisely, the elements of $A_{mathfrak{m}}$ are equivalence classes denoted by $dfrac{a}{s}$, where $ain A$ and $snotin mathfrak{m}$.
Can anyone explain it in a rigorous way, please?
abstract-algebra ring-theory commutative-algebra
asked Dec 16 '18 at 18:47
ZFRZFR
5,26831440
$endgroup$
add a comment |
$begingroup$
Let $A$ be an integral domain. Show that $$bigcap limits_{mathfrak{m}}A_{mathfrak{m}}=A,$$ where the intersection is taken over all maximal ideals $mathfrak{m}$ in $A$.
P.S. This is quite well-known problem in commutative algebra and I have found many topics in MSE which contains its solution. So please do not duplicate my question by the the following reason:
But my question is not about the solution of this problem. How these object could be equal if the LHS and RHS have elements of different structure. More prcisely, the elements of $A_{mathfrak{m}}$ are equivalence classes denoted by $dfrac{a}{s}$, where $ain A$ and $snotin mathfrak{m}$.
Can anyone explain it in a rigorous way, please?
abstract-algebra ring-theory commutative-algebra
asked Dec 16 '18 at 18:47
ZFRZFR
5,26831440
$endgroup$
4
$begingroup$
Each $A_mathfrak{m}$ has an embedding in $K$, the field of fractions of $A$.
$endgroup$
– Lord Shark the Unknown
Dec 16 '18 at 18:48
add a comment |
$begingroup$
Let $A$ be an integral domain. Show that $$bigcap limits_{mathfrak{m}}A_{mathfrak{m}}=A,$$ where the intersection is taken over all maximal ideals $mathfrak{m}$ in $A$.
P.S. This is quite well-known problem in commutative algebra and I have found many topics in MSE which contains its solution. So please do not duplicate my question by the the following reason:
But my question is not about the solution of this problem. How these object could be equal if the LHS and RHS have elements of different structure. More prcisely, the elements of $A_{mathfrak{m}}$ are equivalence classes denoted by $dfrac{a}{s}$, where $ain A$ and $snotin mathfrak{m}$.
Can anyone explain it in a rigorous way, please?
abstract-algebra ring-theory commutative-algebra
asked Dec 16 '18 at 18:47
ZFRZFR
5,26831440
$endgroup$
Let $A$ be an integral domain. Show that $$bigcap limits_{mathfrak{m}}A_{mathfrak{m}}=A,$$ where the intersection is taken over all maximal ideals $mathfrak{m}$ in $A$.
P.S. This is quite well-known problem in commutative algebra and I have found many topics in MSE which contains its solution. So please do not duplicate my question by the the following reason:
But my question is not about the solution of this problem. How these object could be equal if the LHS and RHS have elements of different structure. More prcisely, the elements of $A_{mathfrak{m}}$ are equivalence classes denoted by $dfrac{a}{s}$, where $ain A$ and $snotin mathfrak{m}$.
Can anyone explain it in a rigorous way, please?
abstract-algebra ring-theory commutative-algebra
abstract-algebra ring-theory commutative-algebra
asked Dec 16 '18 at 18:47
ZFRZFR
5,26831440
asked Dec 16 '18 at 18:47
ZFRZFR
5,26831440
asked Dec 16 '18 at 18:47
ZFRZFR
5,26831440
asked Dec 16 '18 at 18:47
ZFRZFR
5,26831440
asked Dec 16 '18 at 18:47
ZFRZFR
5,26831440
5,26831440
4
$begingroup$
Each $A_mathfrak{m}$ has an embedding in $K$, the field of fractions of $A$.
$endgroup$
– Lord Shark the Unknown
Dec 16 '18 at 18:48
add a comment |
4
$begingroup$
Each $A_mathfrak{m}$ has an embedding in $K$, the field of fractions of $A$.
$endgroup$
– Lord Shark the Unknown
Dec 16 '18 at 18:48
4
4
$begingroup$
Each $A_mathfrak{m}$ has an embedding in $K$, the field of fractions of $A$.
$endgroup$
– Lord Shark the Unknown
Dec 16 '18 at 18:48
$begingroup$
Each $A_mathfrak{m}$ has an embedding in $K$, the field of fractions of $A$.
$endgroup$
– Lord Shark the Unknown
Dec 16 '18 at 18:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To elaborate on Lord Shark's comment: if $A$ is an integral domain, then $A$ embeds into any localization $Ato S^{-1}A,$ provided the multiplicative set $S$ does not contain $0.$ Moreover, if $Ssubseteq Tsubseteq A$ are both multiplicative subsets of $A,$ then we have natural maps
$$
Ato S^{-1}Ato T^{-1}A,
$$
and the composition $Ato T^{-1}A$ is simply the localization map (the map $S^{-1}Ato T^{-1}A$ may either be thought of as a localization of $S^{-1}A$ or as the map induced by the universal property of localization).
An integral domain has a maximal multiplicative subset not containing $0$: namely, $Asetminus{0}.$ The localization of $A$ by this multiplicative subset is the field of fractions $K$ of $A.$ So, if $mathfrak{m}$ is a maximal ideal of $A,$ we have natural embeddings
$$
Ato A_mathfrak{m}to K.
$$
Inside of $K,$ we may take the intersection of [the images of] all the $A_mathfrak{m}$ and compare that with [the image of] $A,$ and one way to make the theorem precise is that these two subsets of $K$ are equal. Since $Ato K$ is an isomorphism of $A$ onto its image (and the same is true of each $A_mathfrak{m}to K$), it is reasonable to say that $bigcap_mathfrak{m} A_mathfrak{m} = A$ (or at least that $bigcap_mathfrak{m} A_mathfrak{m} cong A$).
answered Dec 16 '18 at 20:35
StahlStahl
16.7k43455
$endgroup$
$begingroup$
That's what I want to read! Thanks a lot for that! To be precise, the element of $A$ and $A_{mathfrak{m}}$ are different, right? But we can identify $ain A$ with $frac{a}{1}$ in $A_{mathfrak{m}}$. This is my understanding.
$endgroup$
– ZFR
Dec 16 '18 at 20:40
$begingroup$
@ZFR Right; technically, $A$ and $A_mathfrak{m}$ share no elements (as $A_mathfrak{m}$ is typically defined as equivalence classes of ordered pairs $(a,s)in Atimes(Asetminusmathfrak{m})$), but as there are natural inclusions $Ato A_mathfrak{m}$ and $A_mathfrak{m}to K,$ we often identify both $A$ and $A_mathfrak{m}$ with their isomorphic images in $K$ to ease notation. (It would be somewhat obnoxious to always be spelling out that we really want to look at $iota_K(A)$ and $iota_K(A_mathfrak{m})$, especially since it really doesn't add much content.)
$endgroup$
– Stahl
Dec 16 '18 at 20:45
$begingroup$
And as you say, one can (and often does) also identify $A$ with its image in $A_mathfrak{m}$ under the natural map $amapsto frac{a}{1} := [(a,1)]$.
$endgroup$
– Stahl
Dec 16 '18 at 20:46
$begingroup$
The most natural thing here is to identify $A$ with ${1}^{-1} A$, which is $A$ localized at the minimal multiplicative subset.
$endgroup$
– Slade
Dec 16 '18 at 21:30
add a comment |
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$begingroup$
To elaborate on Lord Shark's comment: if $A$ is an integral domain, then $A$ embeds into any localization $Ato S^{-1}A,$ provided the multiplicative set $S$ does not contain $0.$ Moreover, if $Ssubseteq Tsubseteq A$ are both multiplicative subsets of $A,$ then we have natural maps
$$
Ato S^{-1}Ato T^{-1}A,
$$
and the composition $Ato T^{-1}A$ is simply the localization map (the map $S^{-1}Ato T^{-1}A$ may either be thought of as a localization of $S^{-1}A$ or as the map induced by the universal property of localization).
An integral domain has a maximal multiplicative subset not containing $0$: namely, $Asetminus{0}.$ The localization of $A$ by this multiplicative subset is the field of fractions $K$ of $A.$ So, if $mathfrak{m}$ is a maximal ideal of $A,$ we have natural embeddings
$$
Ato A_mathfrak{m}to K.
$$
Inside of $K,$ we may take the intersection of [the images of] all the $A_mathfrak{m}$ and compare that with [the image of] $A,$ and one way to make the theorem precise is that these two subsets of $K$ are equal. Since $Ato K$ is an isomorphism of $A$ onto its image (and the same is true of each $A_mathfrak{m}to K$), it is reasonable to say that $bigcap_mathfrak{m} A_mathfrak{m} = A$ (or at least that $bigcap_mathfrak{m} A_mathfrak{m} cong A$).
answered Dec 16 '18 at 20:35
StahlStahl
16.7k43455
$endgroup$
$begingroup$
That's what I want to read! Thanks a lot for that! To be precise, the element of $A$ and $A_{mathfrak{m}}$ are different, right? But we can identify $ain A$ with $frac{a}{1}$ in $A_{mathfrak{m}}$. This is my understanding.
$endgroup$
– ZFR
Dec 16 '18 at 20:40
$begingroup$
@ZFR Right; technically, $A$ and $A_mathfrak{m}$ share no elements (as $A_mathfrak{m}$ is typically defined as equivalence classes of ordered pairs $(a,s)in Atimes(Asetminusmathfrak{m})$), but as there are natural inclusions $Ato A_mathfrak{m}$ and $A_mathfrak{m}to K,$ we often identify both $A$ and $A_mathfrak{m}$ with their isomorphic images in $K$ to ease notation. (It would be somewhat obnoxious to always be spelling out that we really want to look at $iota_K(A)$ and $iota_K(A_mathfrak{m})$, especially since it really doesn't add much content.)
$endgroup$
– Stahl
Dec 16 '18 at 20:45
$begingroup$
And as you say, one can (and often does) also identify $A$ with its image in $A_mathfrak{m}$ under the natural map $amapsto frac{a}{1} := [(a,1)]$.
$endgroup$
– Stahl
Dec 16 '18 at 20:46
$begingroup$
The most natural thing here is to identify $A$ with ${1}^{-1} A$, which is $A$ localized at the minimal multiplicative subset.
$endgroup$
– Slade
Dec 16 '18 at 21:30
add a comment |
$begingroup$
To elaborate on Lord Shark's comment: if $A$ is an integral domain, then $A$ embeds into any localization $Ato S^{-1}A,$ provided the multiplicative set $S$ does not contain $0.$ Moreover, if $Ssubseteq Tsubseteq A$ are both multiplicative subsets of $A,$ then we have natural maps
$$
Ato S^{-1}Ato T^{-1}A,
$$
and the composition $Ato T^{-1}A$ is simply the localization map (the map $S^{-1}Ato T^{-1}A$ may either be thought of as a localization of $S^{-1}A$ or as the map induced by the universal property of localization).
An integral domain has a maximal multiplicative subset not containing $0$: namely, $Asetminus{0}.$ The localization of $A$ by this multiplicative subset is the field of fractions $K$ of $A.$ So, if $mathfrak{m}$ is a maximal ideal of $A,$ we have natural embeddings
$$
Ato A_mathfrak{m}to K.
$$
Inside of $K,$ we may take the intersection of [the images of] all the $A_mathfrak{m}$ and compare that with [the image of] $A,$ and one way to make the theorem precise is that these two subsets of $K$ are equal. Since $Ato K$ is an isomorphism of $A$ onto its image (and the same is true of each $A_mathfrak{m}to K$), it is reasonable to say that $bigcap_mathfrak{m} A_mathfrak{m} = A$ (or at least that $bigcap_mathfrak{m} A_mathfrak{m} cong A$).
answered Dec 16 '18 at 20:35
StahlStahl
16.7k43455
$endgroup$
$begingroup$
That's what I want to read! Thanks a lot for that! To be precise, the element of $A$ and $A_{mathfrak{m}}$ are different, right? But we can identify $ain A$ with $frac{a}{1}$ in $A_{mathfrak{m}}$. This is my understanding.
$endgroup$
– ZFR
Dec 16 '18 at 20:40
$begingroup$
@ZFR Right; technically, $A$ and $A_mathfrak{m}$ share no elements (as $A_mathfrak{m}$ is typically defined as equivalence classes of ordered pairs $(a,s)in Atimes(Asetminusmathfrak{m})$), but as there are natural inclusions $Ato A_mathfrak{m}$ and $A_mathfrak{m}to K,$ we often identify both $A$ and $A_mathfrak{m}$ with their isomorphic images in $K$ to ease notation. (It would be somewhat obnoxious to always be spelling out that we really want to look at $iota_K(A)$ and $iota_K(A_mathfrak{m})$, especially since it really doesn't add much content.)
$endgroup$
– Stahl
Dec 16 '18 at 20:45
$begingroup$
And as you say, one can (and often does) also identify $A$ with its image in $A_mathfrak{m}$ under the natural map $amapsto frac{a}{1} := [(a,1)]$.
$endgroup$
– Stahl
Dec 16 '18 at 20:46
$begingroup$
The most natural thing here is to identify $A$ with ${1}^{-1} A$, which is $A$ localized at the minimal multiplicative subset.
$endgroup$
– Slade
Dec 16 '18 at 21:30
add a comment |
$begingroup$
To elaborate on Lord Shark's comment: if $A$ is an integral domain, then $A$ embeds into any localization $Ato S^{-1}A,$ provided the multiplicative set $S$ does not contain $0.$ Moreover, if $Ssubseteq Tsubseteq A$ are both multiplicative subsets of $A,$ then we have natural maps
$$
Ato S^{-1}Ato T^{-1}A,
$$
and the composition $Ato T^{-1}A$ is simply the localization map (the map $S^{-1}Ato T^{-1}A$ may either be thought of as a localization of $S^{-1}A$ or as the map induced by the universal property of localization).
An integral domain has a maximal multiplicative subset not containing $0$: namely, $Asetminus{0}.$ The localization of $A$ by this multiplicative subset is the field of fractions $K$ of $A.$ So, if $mathfrak{m}$ is a maximal ideal of $A,$ we have natural embeddings
$$
Ato A_mathfrak{m}to K.
$$
Inside of $K,$ we may take the intersection of [the images of] all the $A_mathfrak{m}$ and compare that with [the image of] $A,$ and one way to make the theorem precise is that these two subsets of $K$ are equal. Since $Ato K$ is an isomorphism of $A$ onto its image (and the same is true of each $A_mathfrak{m}to K$), it is reasonable to say that $bigcap_mathfrak{m} A_mathfrak{m} = A$ (or at least that $bigcap_mathfrak{m} A_mathfrak{m} cong A$).
answered Dec 16 '18 at 20:35
StahlStahl
16.7k43455
$endgroup$
To elaborate on Lord Shark's comment: if $A$ is an integral domain, then $A$ embeds into any localization $Ato S^{-1}A,$ provided the multiplicative set $S$ does not contain $0.$ Moreover, if $Ssubseteq Tsubseteq A$ are both multiplicative subsets of $A,$ then we have natural maps
$$
Ato S^{-1}Ato T^{-1}A,
$$
and the composition $Ato T^{-1}A$ is simply the localization map (the map $S^{-1}Ato T^{-1}A$ may either be thought of as a localization of $S^{-1}A$ or as the map induced by the universal property of localization).
An integral domain has a maximal multiplicative subset not containing $0$: namely, $Asetminus{0}.$ The localization of $A$ by this multiplicative subset is the field of fractions $K$ of $A.$ So, if $mathfrak{m}$ is a maximal ideal of $A,$ we have natural embeddings
$$
Ato A_mathfrak{m}to K.
$$
Inside of $K,$ we may take the intersection of [the images of] all the $A_mathfrak{m}$ and compare that with [the image of] $A,$ and one way to make the theorem precise is that these two subsets of $K$ are equal. Since $Ato K$ is an isomorphism of $A$ onto its image (and the same is true of each $A_mathfrak{m}to K$), it is reasonable to say that $bigcap_mathfrak{m} A_mathfrak{m} = A$ (or at least that $bigcap_mathfrak{m} A_mathfrak{m} cong A$).
answered Dec 16 '18 at 20:35
StahlStahl
16.7k43455
answered Dec 16 '18 at 20:35
StahlStahl
16.7k43455
answered Dec 16 '18 at 20:35
StahlStahl
16.7k43455
answered Dec 16 '18 at 20:35
StahlStahl
16.7k43455
16.7k43455
$begingroup$
That's what I want to read! Thanks a lot for that! To be precise, the element of $A$ and $A_{mathfrak{m}}$ are different, right? But we can identify $ain A$ with $frac{a}{1}$ in $A_{mathfrak{m}}$. This is my understanding.
$endgroup$
– ZFR
Dec 16 '18 at 20:40
$begingroup$
@ZFR Right; technically, $A$ and $A_mathfrak{m}$ share no elements (as $A_mathfrak{m}$ is typically defined as equivalence classes of ordered pairs $(a,s)in Atimes(Asetminusmathfrak{m})$), but as there are natural inclusions $Ato A_mathfrak{m}$ and $A_mathfrak{m}to K,$ we often identify both $A$ and $A_mathfrak{m}$ with their isomorphic images in $K$ to ease notation. (It would be somewhat obnoxious to always be spelling out that we really want to look at $iota_K(A)$ and $iota_K(A_mathfrak{m})$, especially since it really doesn't add much content.)
$endgroup$
– Stahl
Dec 16 '18 at 20:45
$begingroup$
And as you say, one can (and often does) also identify $A$ with its image in $A_mathfrak{m}$ under the natural map $amapsto frac{a}{1} := [(a,1)]$.
$endgroup$
– Stahl
Dec 16 '18 at 20:46
$begingroup$
The most natural thing here is to identify $A$ with ${1}^{-1} A$, which is $A$ localized at the minimal multiplicative subset.
$endgroup$
– Slade
Dec 16 '18 at 21:30
add a comment |
$begingroup$
That's what I want to read! Thanks a lot for that! To be precise, the element of $A$ and $A_{mathfrak{m}}$ are different, right? But we can identify $ain A$ with $frac{a}{1}$ in $A_{mathfrak{m}}$. This is my understanding.
$endgroup$
– ZFR
Dec 16 '18 at 20:40
$begingroup$
@ZFR Right; technically, $A$ and $A_mathfrak{m}$ share no elements (as $A_mathfrak{m}$ is typically defined as equivalence classes of ordered pairs $(a,s)in Atimes(Asetminusmathfrak{m})$), but as there are natural inclusions $Ato A_mathfrak{m}$ and $A_mathfrak{m}to K,$ we often identify both $A$ and $A_mathfrak{m}$ with their isomorphic images in $K$ to ease notation. (It would be somewhat obnoxious to always be spelling out that we really want to look at $iota_K(A)$ and $iota_K(A_mathfrak{m})$, especially since it really doesn't add much content.)
$endgroup$
– Stahl
Dec 16 '18 at 20:45
$begingroup$
And as you say, one can (and often does) also identify $A$ with its image in $A_mathfrak{m}$ under the natural map $amapsto frac{a}{1} := [(a,1)]$.
$endgroup$
– Stahl
Dec 16 '18 at 20:46
$begingroup$
The most natural thing here is to identify $A$ with ${1}^{-1} A$, which is $A$ localized at the minimal multiplicative subset.
$endgroup$
– Slade
Dec 16 '18 at 21:30
$begingroup$
That's what I want to read! Thanks a lot for that! To be precise, the element of $A$ and $A_{mathfrak{m}}$ are different, right? But we can identify $ain A$ with $frac{a}{1}$ in $A_{mathfrak{m}}$. This is my understanding.
$endgroup$
– ZFR
Dec 16 '18 at 20:40
$begingroup$
That's what I want to read! Thanks a lot for that! To be precise, the element of $A$ and $A_{mathfrak{m}}$ are different, right? But we can identify $ain A$ with $frac{a}{1}$ in $A_{mathfrak{m}}$. This is my understanding.
$endgroup$
– ZFR
Dec 16 '18 at 20:40
$begingroup$
@ZFR Right; technically, $A$ and $A_mathfrak{m}$ share no elements (as $A_mathfrak{m}$ is typically defined as equivalence classes of ordered pairs $(a,s)in Atimes(Asetminusmathfrak{m})$), but as there are natural inclusions $Ato A_mathfrak{m}$ and $A_mathfrak{m}to K,$ we often identify both $A$ and $A_mathfrak{m}$ with their isomorphic images in $K$ to ease notation. (It would be somewhat obnoxious to always be spelling out that we really want to look at $iota_K(A)$ and $iota_K(A_mathfrak{m})$, especially since it really doesn't add much content.)
$endgroup$
– Stahl
Dec 16 '18 at 20:45
$begingroup$
@ZFR Right; technically, $A$ and $A_mathfrak{m}$ share no elements (as $A_mathfrak{m}$ is typically defined as equivalence classes of ordered pairs $(a,s)in Atimes(Asetminusmathfrak{m})$), but as there are natural inclusions $Ato A_mathfrak{m}$ and $A_mathfrak{m}to K,$ we often identify both $A$ and $A_mathfrak{m}$ with their isomorphic images in $K$ to ease notation. (It would be somewhat obnoxious to always be spelling out that we really want to look at $iota_K(A)$ and $iota_K(A_mathfrak{m})$, especially since it really doesn't add much content.)
$endgroup$
– Stahl
Dec 16 '18 at 20:45
$begingroup$
And as you say, one can (and often does) also identify $A$ with its image in $A_mathfrak{m}$ under the natural map $amapsto frac{a}{1} := [(a,1)]$.
$endgroup$
– Stahl
Dec 16 '18 at 20:46
$begingroup$
And as you say, one can (and often does) also identify $A$ with its image in $A_mathfrak{m}$ under the natural map $amapsto frac{a}{1} := [(a,1)]$.
$endgroup$
– Stahl
Dec 16 '18 at 20:46
$begingroup$
The most natural thing here is to identify $A$ with ${1}^{-1} A$, which is $A$ localized at the minimal multiplicative subset.
$endgroup$
– Slade
Dec 16 '18 at 21:30
$begingroup$
The most natural thing here is to identify $A$ with ${1}^{-1} A$, which is $A$ localized at the minimal multiplicative subset.
$endgroup$
– Slade
Dec 16 '18 at 21:30
add a comment |
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$begingroup$
Each $A_mathfrak{m}$ has an embedding in $K$, the field of fractions of $A$.
$endgroup$
– Lord Shark the Unknown
Dec 16 '18 at 18:48