What does it mean for localization to be same if multiplicative set is different? $T^{-1}R = S^{-1}R$?
$begingroup$
If R is a ring and p is a prime ideal, I was told that somehow $Frac(R/p)$ was the localisation $(R setminus p)^{-1}R/p$. I thought it was more like the localisation $(R/p setminus{p})^{-1}R/p$ but apparently these are the same. But why? What exactly should I verify? (I am new to all this localization business).
Also, in general what does it mean for $T^{-1}R$ and $S^{-1}R$ to be the same if $T$ and $S$ are different?
commutative-algebra localization
asked Dec 19 '18 at 11:54
roi_saumonroi_saumon
62938
$endgroup$
add a comment |
$begingroup$
If R is a ring and p is a prime ideal, I was told that somehow $Frac(R/p)$ was the localisation $(R setminus p)^{-1}R/p$. I thought it was more like the localisation $(R/p setminus{p})^{-1}R/p$ but apparently these are the same. But why? What exactly should I verify? (I am new to all this localization business).
Also, in general what does it mean for $T^{-1}R$ and $S^{-1}R$ to be the same if $T$ and $S$ are different?
commutative-algebra localization
asked Dec 19 '18 at 11:54
roi_saumonroi_saumon
62938
$endgroup$
$begingroup$
I guess they are isomorphic.
$endgroup$
– user370967
Dec 19 '18 at 12:33
add a comment |
$begingroup$
If R is a ring and p is a prime ideal, I was told that somehow $Frac(R/p)$ was the localisation $(R setminus p)^{-1}R/p$. I thought it was more like the localisation $(R/p setminus{p})^{-1}R/p$ but apparently these are the same. But why? What exactly should I verify? (I am new to all this localization business).
Also, in general what does it mean for $T^{-1}R$ and $S^{-1}R$ to be the same if $T$ and $S$ are different?
commutative-algebra localization
asked Dec 19 '18 at 11:54
roi_saumonroi_saumon
62938
$endgroup$
If R is a ring and p is a prime ideal, I was told that somehow $Frac(R/p)$ was the localisation $(R setminus p)^{-1}R/p$. I thought it was more like the localisation $(R/p setminus{p})^{-1}R/p$ but apparently these are the same. But why? What exactly should I verify? (I am new to all this localization business).
Also, in general what does it mean for $T^{-1}R$ and $S^{-1}R$ to be the same if $T$ and $S$ are different?
commutative-algebra localization
commutative-algebra localization
asked Dec 19 '18 at 11:54
roi_saumonroi_saumon
62938
asked Dec 19 '18 at 11:54
roi_saumonroi_saumon
62938
asked Dec 19 '18 at 11:54
roi_saumonroi_saumon
62938
asked Dec 19 '18 at 11:54
roi_saumonroi_saumon
62938
asked Dec 19 '18 at 11:54
roi_saumonroi_saumon
62938
62938
$begingroup$
I guess they are isomorphic.
$endgroup$
– user370967
Dec 19 '18 at 12:33
add a comment |
$begingroup$
I guess they are isomorphic.
$endgroup$
– user370967
Dec 19 '18 at 12:33
$begingroup$
I guess they are isomorphic.
$endgroup$
– user370967
Dec 19 '18 at 12:33
$begingroup$
I guess they are isomorphic.
$endgroup$
– user370967
Dec 19 '18 at 12:33
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Maybe $(Rsetminus p)^{-1}(R/p)$ is confusing since $Rsetminus p$ is not a subset of $R/p$? It would indeed be better to write $(overline{Rsetminus p})^{-1}(R/p)$, where the overline means taking classes modulo $p$. Alternatively, the latter can be written as $((R/p)setminus{bar 0})^{-1}(R/p)$ which coïncides with your suggestion.
Your question as to when two localizations $T^{-1}R$ and $S^{-1}R$ are isomorphic is a nice exercise: let $tilde S$ be the saturation of a multiplicative subset $S$ of $R$ defined by
$$
tilde S={rin Rmid exists tin Rcolon rtin S}.
$$
In words, $tilde S$ is the subset of all divisors of the elements of $S$.
One can prove that $tilde S$ is again multiplicative and that the natural morphism from $S^{-1}R$ to $tilde S^{-1} R$ is an isomorphism. Also, $tilde S$ is exactly the subset of elements of $R$ whose image in $S^{-1}R$ is invertible. With that in hand, it is easy to prove that the localizations $S^{-1}R$ and $T^{-1}R$ are isomorphic if and only if $tilde S=tilde T$. One has to note that by isomorphism of localizations I mean here that there is an isomorphism between $S^{-1}R$ and $T^{-1}R$ that is the identity on the image of $R$. Without that precision, it may well be that the rings $S^{-1}R$ and $T^{-1}R$ are abstractly isomorphic without $tilde S=tilde T$. An example of the latter are the rings $mathbf Z[X,Y]_X$ and $mathbf Z[X,Y]_Y$, where $R_r$ denotes the localization of the ring $R$ with respect to the multiplicative subset ${1,r,r^2,ldots}$.
answered Dec 19 '18 at 13:42
Johannes HuismanJohannes Huisman
3,146617
$endgroup$
$begingroup$
It is not too confusing that $R setminus p$ is not a subset since this is a localisation of modules where $R/p$ is seen as an R-module. But why can you just take the classes module p and say it is the same as $(R setminus p)^{-1}(R/p)$?
$endgroup$
– roi_saumon
Dec 19 '18 at 16:01
1
$begingroup$
OK, I didn't get it that you thought of $R/p$ as an $R$-module. There are, indeed, two localizations of $R/p$: the localization of $R/p$ as an $R$-module by $Rsetminus p$, and the localization of the ring $R/p$ by its multiplicative subset $(R/p)setminus{0}$. It is clear that there is a map from the former into the latter. This map is an isomorphism of $R$-modules.
$endgroup$
– Johannes Huisman
Dec 19 '18 at 16:31
$begingroup$
Is the morphism $frac{[r]_p}{t} mapsto frac{[r]_p}{[t]_p}$? Why is it enough to show that if $t-hat{t} in p$ then $frac{[r]_p}{t}-frac{[r]_p}{hat{t}}$
$endgroup$
– roi_saumon
Dec 19 '18 at 17:06
1
$begingroup$
Yes, that is the morphism. I don't understand your second question.
$endgroup$
– Johannes Huisman
Dec 19 '18 at 17:08
1
$begingroup$
I did not say that that was enough to prove; just follow the definitions: surjectivity of the map from $(Rsetminus p)^{-1}R/p$ to $(R/psetminus{0})^{-1}R/p$ is clear. As for injectivity, it suffices to prove that the kernel is trivial. Suppose that $frac rt=0$ in $(R/psetminus{0})^{-1}R/p$. This means that there is $sin Rsetminus p$ such that $sr=0$ in $R/p$. Hence, $srin p$ and $rin p$ since $snotin p$. This implies that $frac rt=0$ in $(Rsetminus p)^{-1}R/p$, and proves injectivity.
$endgroup$
– Johannes Huisman
Dec 19 '18 at 20:27
|
show 2 more comments
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$begingroup$
Maybe $(Rsetminus p)^{-1}(R/p)$ is confusing since $Rsetminus p$ is not a subset of $R/p$? It would indeed be better to write $(overline{Rsetminus p})^{-1}(R/p)$, where the overline means taking classes modulo $p$. Alternatively, the latter can be written as $((R/p)setminus{bar 0})^{-1}(R/p)$ which coïncides with your suggestion.
Your question as to when two localizations $T^{-1}R$ and $S^{-1}R$ are isomorphic is a nice exercise: let $tilde S$ be the saturation of a multiplicative subset $S$ of $R$ defined by
$$
tilde S={rin Rmid exists tin Rcolon rtin S}.
$$
In words, $tilde S$ is the subset of all divisors of the elements of $S$.
One can prove that $tilde S$ is again multiplicative and that the natural morphism from $S^{-1}R$ to $tilde S^{-1} R$ is an isomorphism. Also, $tilde S$ is exactly the subset of elements of $R$ whose image in $S^{-1}R$ is invertible. With that in hand, it is easy to prove that the localizations $S^{-1}R$ and $T^{-1}R$ are isomorphic if and only if $tilde S=tilde T$. One has to note that by isomorphism of localizations I mean here that there is an isomorphism between $S^{-1}R$ and $T^{-1}R$ that is the identity on the image of $R$. Without that precision, it may well be that the rings $S^{-1}R$ and $T^{-1}R$ are abstractly isomorphic without $tilde S=tilde T$. An example of the latter are the rings $mathbf Z[X,Y]_X$ and $mathbf Z[X,Y]_Y$, where $R_r$ denotes the localization of the ring $R$ with respect to the multiplicative subset ${1,r,r^2,ldots}$.
answered Dec 19 '18 at 13:42
Johannes HuismanJohannes Huisman
3,146617
$endgroup$
$begingroup$
It is not too confusing that $R setminus p$ is not a subset since this is a localisation of modules where $R/p$ is seen as an R-module. But why can you just take the classes module p and say it is the same as $(R setminus p)^{-1}(R/p)$?
$endgroup$
– roi_saumon
Dec 19 '18 at 16:01
1
$begingroup$
OK, I didn't get it that you thought of $R/p$ as an $R$-module. There are, indeed, two localizations of $R/p$: the localization of $R/p$ as an $R$-module by $Rsetminus p$, and the localization of the ring $R/p$ by its multiplicative subset $(R/p)setminus{0}$. It is clear that there is a map from the former into the latter. This map is an isomorphism of $R$-modules.
$endgroup$
– Johannes Huisman
Dec 19 '18 at 16:31
$begingroup$
Is the morphism $frac{[r]_p}{t} mapsto frac{[r]_p}{[t]_p}$? Why is it enough to show that if $t-hat{t} in p$ then $frac{[r]_p}{t}-frac{[r]_p}{hat{t}}$
$endgroup$
– roi_saumon
Dec 19 '18 at 17:06
1
$begingroup$
Yes, that is the morphism. I don't understand your second question.
$endgroup$
– Johannes Huisman
Dec 19 '18 at 17:08
1
$begingroup$
I did not say that that was enough to prove; just follow the definitions: surjectivity of the map from $(Rsetminus p)^{-1}R/p$ to $(R/psetminus{0})^{-1}R/p$ is clear. As for injectivity, it suffices to prove that the kernel is trivial. Suppose that $frac rt=0$ in $(R/psetminus{0})^{-1}R/p$. This means that there is $sin Rsetminus p$ such that $sr=0$ in $R/p$. Hence, $srin p$ and $rin p$ since $snotin p$. This implies that $frac rt=0$ in $(Rsetminus p)^{-1}R/p$, and proves injectivity.
$endgroup$
– Johannes Huisman
Dec 19 '18 at 20:27
|
show 2 more comments
$begingroup$
Maybe $(Rsetminus p)^{-1}(R/p)$ is confusing since $Rsetminus p$ is not a subset of $R/p$? It would indeed be better to write $(overline{Rsetminus p})^{-1}(R/p)$, where the overline means taking classes modulo $p$. Alternatively, the latter can be written as $((R/p)setminus{bar 0})^{-1}(R/p)$ which coïncides with your suggestion.
Your question as to when two localizations $T^{-1}R$ and $S^{-1}R$ are isomorphic is a nice exercise: let $tilde S$ be the saturation of a multiplicative subset $S$ of $R$ defined by
$$
tilde S={rin Rmid exists tin Rcolon rtin S}.
$$
In words, $tilde S$ is the subset of all divisors of the elements of $S$.
One can prove that $tilde S$ is again multiplicative and that the natural morphism from $S^{-1}R$ to $tilde S^{-1} R$ is an isomorphism. Also, $tilde S$ is exactly the subset of elements of $R$ whose image in $S^{-1}R$ is invertible. With that in hand, it is easy to prove that the localizations $S^{-1}R$ and $T^{-1}R$ are isomorphic if and only if $tilde S=tilde T$. One has to note that by isomorphism of localizations I mean here that there is an isomorphism between $S^{-1}R$ and $T^{-1}R$ that is the identity on the image of $R$. Without that precision, it may well be that the rings $S^{-1}R$ and $T^{-1}R$ are abstractly isomorphic without $tilde S=tilde T$. An example of the latter are the rings $mathbf Z[X,Y]_X$ and $mathbf Z[X,Y]_Y$, where $R_r$ denotes the localization of the ring $R$ with respect to the multiplicative subset ${1,r,r^2,ldots}$.
answered Dec 19 '18 at 13:42
Johannes HuismanJohannes Huisman
3,146617
$endgroup$
$begingroup$
It is not too confusing that $R setminus p$ is not a subset since this is a localisation of modules where $R/p$ is seen as an R-module. But why can you just take the classes module p and say it is the same as $(R setminus p)^{-1}(R/p)$?
$endgroup$
– roi_saumon
Dec 19 '18 at 16:01
1
$begingroup$
OK, I didn't get it that you thought of $R/p$ as an $R$-module. There are, indeed, two localizations of $R/p$: the localization of $R/p$ as an $R$-module by $Rsetminus p$, and the localization of the ring $R/p$ by its multiplicative subset $(R/p)setminus{0}$. It is clear that there is a map from the former into the latter. This map is an isomorphism of $R$-modules.
$endgroup$
– Johannes Huisman
Dec 19 '18 at 16:31
$begingroup$
Is the morphism $frac{[r]_p}{t} mapsto frac{[r]_p}{[t]_p}$? Why is it enough to show that if $t-hat{t} in p$ then $frac{[r]_p}{t}-frac{[r]_p}{hat{t}}$
$endgroup$
– roi_saumon
Dec 19 '18 at 17:06
1
$begingroup$
Yes, that is the morphism. I don't understand your second question.
$endgroup$
– Johannes Huisman
Dec 19 '18 at 17:08
1
$begingroup$
I did not say that that was enough to prove; just follow the definitions: surjectivity of the map from $(Rsetminus p)^{-1}R/p$ to $(R/psetminus{0})^{-1}R/p$ is clear. As for injectivity, it suffices to prove that the kernel is trivial. Suppose that $frac rt=0$ in $(R/psetminus{0})^{-1}R/p$. This means that there is $sin Rsetminus p$ such that $sr=0$ in $R/p$. Hence, $srin p$ and $rin p$ since $snotin p$. This implies that $frac rt=0$ in $(Rsetminus p)^{-1}R/p$, and proves injectivity.
$endgroup$
– Johannes Huisman
Dec 19 '18 at 20:27
|
show 2 more comments
$begingroup$
Maybe $(Rsetminus p)^{-1}(R/p)$ is confusing since $Rsetminus p$ is not a subset of $R/p$? It would indeed be better to write $(overline{Rsetminus p})^{-1}(R/p)$, where the overline means taking classes modulo $p$. Alternatively, the latter can be written as $((R/p)setminus{bar 0})^{-1}(R/p)$ which coïncides with your suggestion.
Your question as to when two localizations $T^{-1}R$ and $S^{-1}R$ are isomorphic is a nice exercise: let $tilde S$ be the saturation of a multiplicative subset $S$ of $R$ defined by
$$
tilde S={rin Rmid exists tin Rcolon rtin S}.
$$
In words, $tilde S$ is the subset of all divisors of the elements of $S$.
One can prove that $tilde S$ is again multiplicative and that the natural morphism from $S^{-1}R$ to $tilde S^{-1} R$ is an isomorphism. Also, $tilde S$ is exactly the subset of elements of $R$ whose image in $S^{-1}R$ is invertible. With that in hand, it is easy to prove that the localizations $S^{-1}R$ and $T^{-1}R$ are isomorphic if and only if $tilde S=tilde T$. One has to note that by isomorphism of localizations I mean here that there is an isomorphism between $S^{-1}R$ and $T^{-1}R$ that is the identity on the image of $R$. Without that precision, it may well be that the rings $S^{-1}R$ and $T^{-1}R$ are abstractly isomorphic without $tilde S=tilde T$. An example of the latter are the rings $mathbf Z[X,Y]_X$ and $mathbf Z[X,Y]_Y$, where $R_r$ denotes the localization of the ring $R$ with respect to the multiplicative subset ${1,r,r^2,ldots}$.
answered Dec 19 '18 at 13:42
Johannes HuismanJohannes Huisman
3,146617
$endgroup$
Maybe $(Rsetminus p)^{-1}(R/p)$ is confusing since $Rsetminus p$ is not a subset of $R/p$? It would indeed be better to write $(overline{Rsetminus p})^{-1}(R/p)$, where the overline means taking classes modulo $p$. Alternatively, the latter can be written as $((R/p)setminus{bar 0})^{-1}(R/p)$ which coïncides with your suggestion.
Your question as to when two localizations $T^{-1}R$ and $S^{-1}R$ are isomorphic is a nice exercise: let $tilde S$ be the saturation of a multiplicative subset $S$ of $R$ defined by
$$
tilde S={rin Rmid exists tin Rcolon rtin S}.
$$
In words, $tilde S$ is the subset of all divisors of the elements of $S$.
One can prove that $tilde S$ is again multiplicative and that the natural morphism from $S^{-1}R$ to $tilde S^{-1} R$ is an isomorphism. Also, $tilde S$ is exactly the subset of elements of $R$ whose image in $S^{-1}R$ is invertible. With that in hand, it is easy to prove that the localizations $S^{-1}R$ and $T^{-1}R$ are isomorphic if and only if $tilde S=tilde T$. One has to note that by isomorphism of localizations I mean here that there is an isomorphism between $S^{-1}R$ and $T^{-1}R$ that is the identity on the image of $R$. Without that precision, it may well be that the rings $S^{-1}R$ and $T^{-1}R$ are abstractly isomorphic without $tilde S=tilde T$. An example of the latter are the rings $mathbf Z[X,Y]_X$ and $mathbf Z[X,Y]_Y$, where $R_r$ denotes the localization of the ring $R$ with respect to the multiplicative subset ${1,r,r^2,ldots}$.
answered Dec 19 '18 at 13:42
Johannes HuismanJohannes Huisman
3,146617
answered Dec 19 '18 at 13:42
Johannes HuismanJohannes Huisman
3,146617
answered Dec 19 '18 at 13:42
Johannes HuismanJohannes Huisman
3,146617
answered Dec 19 '18 at 13:42
Johannes HuismanJohannes Huisman
3,146617
3,146617
$begingroup$
It is not too confusing that $R setminus p$ is not a subset since this is a localisation of modules where $R/p$ is seen as an R-module. But why can you just take the classes module p and say it is the same as $(R setminus p)^{-1}(R/p)$?
$endgroup$
– roi_saumon
Dec 19 '18 at 16:01
1
$begingroup$
OK, I didn't get it that you thought of $R/p$ as an $R$-module. There are, indeed, two localizations of $R/p$: the localization of $R/p$ as an $R$-module by $Rsetminus p$, and the localization of the ring $R/p$ by its multiplicative subset $(R/p)setminus{0}$. It is clear that there is a map from the former into the latter. This map is an isomorphism of $R$-modules.
$endgroup$
– Johannes Huisman
Dec 19 '18 at 16:31
$begingroup$
Is the morphism $frac{[r]_p}{t} mapsto frac{[r]_p}{[t]_p}$? Why is it enough to show that if $t-hat{t} in p$ then $frac{[r]_p}{t}-frac{[r]_p}{hat{t}}$
$endgroup$
– roi_saumon
Dec 19 '18 at 17:06
1
$begingroup$
Yes, that is the morphism. I don't understand your second question.
$endgroup$
– Johannes Huisman
Dec 19 '18 at 17:08
1
$begingroup$
I did not say that that was enough to prove; just follow the definitions: surjectivity of the map from $(Rsetminus p)^{-1}R/p$ to $(R/psetminus{0})^{-1}R/p$ is clear. As for injectivity, it suffices to prove that the kernel is trivial. Suppose that $frac rt=0$ in $(R/psetminus{0})^{-1}R/p$. This means that there is $sin Rsetminus p$ such that $sr=0$ in $R/p$. Hence, $srin p$ and $rin p$ since $snotin p$. This implies that $frac rt=0$ in $(Rsetminus p)^{-1}R/p$, and proves injectivity.
$endgroup$
– Johannes Huisman
Dec 19 '18 at 20:27
|
show 2 more comments
$begingroup$
It is not too confusing that $R setminus p$ is not a subset since this is a localisation of modules where $R/p$ is seen as an R-module. But why can you just take the classes module p and say it is the same as $(R setminus p)^{-1}(R/p)$?
$endgroup$
– roi_saumon
Dec 19 '18 at 16:01
1
$begingroup$
OK, I didn't get it that you thought of $R/p$ as an $R$-module. There are, indeed, two localizations of $R/p$: the localization of $R/p$ as an $R$-module by $Rsetminus p$, and the localization of the ring $R/p$ by its multiplicative subset $(R/p)setminus{0}$. It is clear that there is a map from the former into the latter. This map is an isomorphism of $R$-modules.
$endgroup$
– Johannes Huisman
Dec 19 '18 at 16:31
$begingroup$
Is the morphism $frac{[r]_p}{t} mapsto frac{[r]_p}{[t]_p}$? Why is it enough to show that if $t-hat{t} in p$ then $frac{[r]_p}{t}-frac{[r]_p}{hat{t}}$
$endgroup$
– roi_saumon
Dec 19 '18 at 17:06
1
$begingroup$
Yes, that is the morphism. I don't understand your second question.
$endgroup$
– Johannes Huisman
Dec 19 '18 at 17:08
1
$begingroup$
I did not say that that was enough to prove; just follow the definitions: surjectivity of the map from $(Rsetminus p)^{-1}R/p$ to $(R/psetminus{0})^{-1}R/p$ is clear. As for injectivity, it suffices to prove that the kernel is trivial. Suppose that $frac rt=0$ in $(R/psetminus{0})^{-1}R/p$. This means that there is $sin Rsetminus p$ such that $sr=0$ in $R/p$. Hence, $srin p$ and $rin p$ since $snotin p$. This implies that $frac rt=0$ in $(Rsetminus p)^{-1}R/p$, and proves injectivity.
$endgroup$
– Johannes Huisman
Dec 19 '18 at 20:27
$begingroup$
It is not too confusing that $R setminus p$ is not a subset since this is a localisation of modules where $R/p$ is seen as an R-module. But why can you just take the classes module p and say it is the same as $(R setminus p)^{-1}(R/p)$?
$endgroup$
– roi_saumon
Dec 19 '18 at 16:01
$begingroup$
It is not too confusing that $R setminus p$ is not a subset since this is a localisation of modules where $R/p$ is seen as an R-module. But why can you just take the classes module p and say it is the same as $(R setminus p)^{-1}(R/p)$?
$endgroup$
– roi_saumon
Dec 19 '18 at 16:01
1
1
$begingroup$
OK, I didn't get it that you thought of $R/p$ as an $R$-module. There are, indeed, two localizations of $R/p$: the localization of $R/p$ as an $R$-module by $Rsetminus p$, and the localization of the ring $R/p$ by its multiplicative subset $(R/p)setminus{0}$. It is clear that there is a map from the former into the latter. This map is an isomorphism of $R$-modules.
$endgroup$
– Johannes Huisman
Dec 19 '18 at 16:31
$begingroup$
OK, I didn't get it that you thought of $R/p$ as an $R$-module. There are, indeed, two localizations of $R/p$: the localization of $R/p$ as an $R$-module by $Rsetminus p$, and the localization of the ring $R/p$ by its multiplicative subset $(R/p)setminus{0}$. It is clear that there is a map from the former into the latter. This map is an isomorphism of $R$-modules.
$endgroup$
– Johannes Huisman
Dec 19 '18 at 16:31
$begingroup$
Is the morphism $frac{[r]_p}{t} mapsto frac{[r]_p}{[t]_p}$? Why is it enough to show that if $t-hat{t} in p$ then $frac{[r]_p}{t}-frac{[r]_p}{hat{t}}$
$endgroup$
– roi_saumon
Dec 19 '18 at 17:06
$begingroup$
Is the morphism $frac{[r]_p}{t} mapsto frac{[r]_p}{[t]_p}$? Why is it enough to show that if $t-hat{t} in p$ then $frac{[r]_p}{t}-frac{[r]_p}{hat{t}}$
$endgroup$
– roi_saumon
Dec 19 '18 at 17:06
1
1
$begingroup$
Yes, that is the morphism. I don't understand your second question.
$endgroup$
– Johannes Huisman
Dec 19 '18 at 17:08
$begingroup$
Yes, that is the morphism. I don't understand your second question.
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– Johannes Huisman
Dec 19 '18 at 17:08
1
1
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I did not say that that was enough to prove; just follow the definitions: surjectivity of the map from $(Rsetminus p)^{-1}R/p$ to $(R/psetminus{0})^{-1}R/p$ is clear. As for injectivity, it suffices to prove that the kernel is trivial. Suppose that $frac rt=0$ in $(R/psetminus{0})^{-1}R/p$. This means that there is $sin Rsetminus p$ such that $sr=0$ in $R/p$. Hence, $srin p$ and $rin p$ since $snotin p$. This implies that $frac rt=0$ in $(Rsetminus p)^{-1}R/p$, and proves injectivity.
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– Johannes Huisman
Dec 19 '18 at 20:27
$begingroup$
I did not say that that was enough to prove; just follow the definitions: surjectivity of the map from $(Rsetminus p)^{-1}R/p$ to $(R/psetminus{0})^{-1}R/p$ is clear. As for injectivity, it suffices to prove that the kernel is trivial. Suppose that $frac rt=0$ in $(R/psetminus{0})^{-1}R/p$. This means that there is $sin Rsetminus p$ such that $sr=0$ in $R/p$. Hence, $srin p$ and $rin p$ since $snotin p$. This implies that $frac rt=0$ in $(Rsetminus p)^{-1}R/p$, and proves injectivity.
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– Johannes Huisman
Dec 19 '18 at 20:27
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$begingroup$
I guess they are isomorphic.
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– user370967
Dec 19 '18 at 12:33