Show that a ring of fractions and a quotient ring are isomorphic [duplicate]
$begingroup$
This question already has an answer here:
Localisation is isomorphic to a quotient of polynomial ring
5 answers
For a commutative ring $R$ with $1neq 0$ and a nonzerodivisor $r in R$, let $S$ be the set
$S={r^nmid nin mathbb{Z}, ngeq 0}$ and denote $S^{-1}R=Rleft[frac{1}{r}right]$.
Prove that there is a ring isomorphism $$Rleft[frac{1}{r}right]cong frac{R[x]}{(rx -1)}.$$
I'm thinking maybe I can find a homomorphism from $R[x]$ to $Rleft[frac{1}{r}right]$ that has kernel $(rx-1)$, and then use the first isomorphism theorem. Is this the right approach?
abstract-algebra ring-theory
edited Dec 6 '18 at 22:15
Bernard
120k740116
asked Dec 6 '18 at 21:58
WesleyWesley
518313
$endgroup$
marked as duplicate by André 3000, darij grinberg, user10354138, Brahadeesh, KReiser Dec 7 '18 at 8:22
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Localisation is isomorphic to a quotient of polynomial ring
5 answers
For a commutative ring $R$ with $1neq 0$ and a nonzerodivisor $r in R$, let $S$ be the set
$S={r^nmid nin mathbb{Z}, ngeq 0}$ and denote $S^{-1}R=Rleft[frac{1}{r}right]$.
Prove that there is a ring isomorphism $$Rleft[frac{1}{r}right]cong frac{R[x]}{(rx -1)}.$$
I'm thinking maybe I can find a homomorphism from $R[x]$ to $Rleft[frac{1}{r}right]$ that has kernel $(rx-1)$, and then use the first isomorphism theorem. Is this the right approach?
abstract-algebra ring-theory
edited Dec 6 '18 at 22:15
Bernard
120k740116
asked Dec 6 '18 at 21:58
WesleyWesley
518313
$endgroup$
marked as duplicate by André 3000, darij grinberg, user10354138, Brahadeesh, KReiser Dec 7 '18 at 8:22
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
Yes, evaluation homomorphism can do the trick $phi(p(x))=pleft(frac{1}{r}right)$.
$endgroup$
– Anurag A
Dec 6 '18 at 22:07
$begingroup$
Please search the site before asking a new question. This has been asked many, many times before: 1, 2, 3, 4 to name a few.
$endgroup$
– André 3000
Dec 6 '18 at 23:31
add a comment |
$begingroup$
This question already has an answer here:
Localisation is isomorphic to a quotient of polynomial ring
5 answers
For a commutative ring $R$ with $1neq 0$ and a nonzerodivisor $r in R$, let $S$ be the set
$S={r^nmid nin mathbb{Z}, ngeq 0}$ and denote $S^{-1}R=Rleft[frac{1}{r}right]$.
Prove that there is a ring isomorphism $$Rleft[frac{1}{r}right]cong frac{R[x]}{(rx -1)}.$$
I'm thinking maybe I can find a homomorphism from $R[x]$ to $Rleft[frac{1}{r}right]$ that has kernel $(rx-1)$, and then use the first isomorphism theorem. Is this the right approach?
abstract-algebra ring-theory
edited Dec 6 '18 at 22:15
Bernard
120k740116
asked Dec 6 '18 at 21:58
WesleyWesley
518313
$endgroup$
This question already has an answer here:
Localisation is isomorphic to a quotient of polynomial ring
5 answers
For a commutative ring $R$ with $1neq 0$ and a nonzerodivisor $r in R$, let $S$ be the set
$S={r^nmid nin mathbb{Z}, ngeq 0}$ and denote $S^{-1}R=Rleft[frac{1}{r}right]$.
Prove that there is a ring isomorphism $$Rleft[frac{1}{r}right]cong frac{R[x]}{(rx -1)}.$$
I'm thinking maybe I can find a homomorphism from $R[x]$ to $Rleft[frac{1}{r}right]$ that has kernel $(rx-1)$, and then use the first isomorphism theorem. Is this the right approach?
This question already has an answer here:
Localisation is isomorphic to a quotient of polynomial ring
5 answers
abstract-algebra ring-theory
abstract-algebra ring-theory
edited Dec 6 '18 at 22:15
Bernard
120k740116
asked Dec 6 '18 at 21:58
WesleyWesley
518313
edited Dec 6 '18 at 22:15
Bernard
120k740116
asked Dec 6 '18 at 21:58
WesleyWesley
518313
edited Dec 6 '18 at 22:15
Bernard
120k740116
edited Dec 6 '18 at 22:15
Bernard
120k740116
edited Dec 6 '18 at 22:15
Bernard
120k740116
120k740116
asked Dec 6 '18 at 21:58
WesleyWesley
518313
asked Dec 6 '18 at 21:58
WesleyWesley
518313
asked Dec 6 '18 at 21:58
WesleyWesley
518313
518313
marked as duplicate by André 3000, darij grinberg, user10354138, Brahadeesh, KReiser Dec 7 '18 at 8:22
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by André 3000, darij grinberg, user10354138, Brahadeesh, KReiser Dec 7 '18 at 8:22
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
Yes, evaluation homomorphism can do the trick $phi(p(x))=pleft(frac{1}{r}right)$.
$endgroup$
– Anurag A
Dec 6 '18 at 22:07
$begingroup$
Please search the site before asking a new question. This has been asked many, many times before: 1, 2, 3, 4 to name a few.
$endgroup$
– André 3000
Dec 6 '18 at 23:31
add a comment |
1
$begingroup$
Yes, evaluation homomorphism can do the trick $phi(p(x))=pleft(frac{1}{r}right)$.
$endgroup$
– Anurag A
Dec 6 '18 at 22:07
$begingroup$
Please search the site before asking a new question. This has been asked many, many times before: 1, 2, 3, 4 to name a few.
$endgroup$
– André 3000
Dec 6 '18 at 23:31
1
1
$begingroup$
Yes, evaluation homomorphism can do the trick $phi(p(x))=pleft(frac{1}{r}right)$.
$endgroup$
– Anurag A
Dec 6 '18 at 22:07
$begingroup$
Yes, evaluation homomorphism can do the trick $phi(p(x))=pleft(frac{1}{r}right)$.
$endgroup$
– Anurag A
Dec 6 '18 at 22:07
$begingroup$
Please search the site before asking a new question. This has been asked many, many times before: 1, 2, 3, 4 to name a few.
$endgroup$
– André 3000
Dec 6 '18 at 23:31
$begingroup$
Please search the site before asking a new question. This has been asked many, many times before: 1, 2, 3, 4 to name a few.
$endgroup$
– André 3000
Dec 6 '18 at 23:31
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Tricky is proving the kernel $K = (rx!-!1).,$ A simple way: if $fin K$ then by nonmonic division
$$ r^n f(x) = (rx!-!1),q(x) + r', {rm for} r'in R, nin Bbb N$$
Evaluating at $, x = 1/r,$ shows $,r'! = 0,$ so $,rx!-!1mid r^n f,Rightarrow,rx!-!1mid f,,$ by $,(rx!-!1,r) = (1);,$ more explicitly $,rx!-!1mid rg,Rightarrow, rx!-!1mid g = x(rg)-(rx!-!1)g$.
Remark $ $ See this answer for another proof and further discussion. If you already know basic properties of localizations then see also the linked dupe for ways to exploit these proeprties.
answered Dec 6 '18 at 23:59
Bill DubuqueBill Dubuque
210k29192643
$endgroup$
add a comment |
$begingroup$
You should note that $frac{1}{r}$ has the property $frac{1}{r}cdot r=1$. That is, $frac{1}{r}$ is not just an indeterminate object like $x$. Conceptually, this isomorphism should be easy to construct. Send $sin R$ to itself for all $s$. Send $xmapsto frac{1}{r}$. This is clearly a surjective homomorphism
$$ R[x]xrightarrow{phi} S^{-1}R.$$
Calculate $kerphi$. You will see that it is $(rx-1)$. Conclude using the first isomorphism theorem.
answered Dec 6 '18 at 22:08
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.2k41641
$endgroup$
$begingroup$
Why is the map clearly a surjection?
$endgroup$
– Wesley
Dec 6 '18 at 22:18
$begingroup$
Let $ain R$ and $r^kin S$. Is it surjective because $phi(ax^k) = frac{a}{r^k}$, and the latter term is a general term of the ring of fractions?
$endgroup$
– Wesley
Dec 6 '18 at 22:26
$begingroup$
Yes that is one way to see it.
$endgroup$
– Antonios-Alexandros Robotis
Dec 6 '18 at 22:50
$begingroup$
It's clear that $(rx-1) subseteq ker(phi)$, but how would I show that $ker(phi) subseteq (rx-1)$? If I could use the division algorithm I could see how, but I can't see how to do it without the division algorithm.
$endgroup$
– Wesley
Dec 6 '18 at 23:14
1
$begingroup$
@Wesley One simple way is to use the nonmonic division algorithm - see my answer.
$endgroup$
– Bill Dubuque
Dec 7 '18 at 0:02
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Tricky is proving the kernel $K = (rx!-!1).,$ A simple way: if $fin K$ then by nonmonic division
$$ r^n f(x) = (rx!-!1),q(x) + r', {rm for} r'in R, nin Bbb N$$
Evaluating at $, x = 1/r,$ shows $,r'! = 0,$ so $,rx!-!1mid r^n f,Rightarrow,rx!-!1mid f,,$ by $,(rx!-!1,r) = (1);,$ more explicitly $,rx!-!1mid rg,Rightarrow, rx!-!1mid g = x(rg)-(rx!-!1)g$.
Remark $ $ See this answer for another proof and further discussion. If you already know basic properties of localizations then see also the linked dupe for ways to exploit these proeprties.
answered Dec 6 '18 at 23:59
Bill DubuqueBill Dubuque
210k29192643
$endgroup$
add a comment |
$begingroup$
Tricky is proving the kernel $K = (rx!-!1).,$ A simple way: if $fin K$ then by nonmonic division
$$ r^n f(x) = (rx!-!1),q(x) + r', {rm for} r'in R, nin Bbb N$$
Evaluating at $, x = 1/r,$ shows $,r'! = 0,$ so $,rx!-!1mid r^n f,Rightarrow,rx!-!1mid f,,$ by $,(rx!-!1,r) = (1);,$ more explicitly $,rx!-!1mid rg,Rightarrow, rx!-!1mid g = x(rg)-(rx!-!1)g$.
Remark $ $ See this answer for another proof and further discussion. If you already know basic properties of localizations then see also the linked dupe for ways to exploit these proeprties.
answered Dec 6 '18 at 23:59
Bill DubuqueBill Dubuque
210k29192643
$endgroup$
add a comment |
$begingroup$
Tricky is proving the kernel $K = (rx!-!1).,$ A simple way: if $fin K$ then by nonmonic division
$$ r^n f(x) = (rx!-!1),q(x) + r', {rm for} r'in R, nin Bbb N$$
Evaluating at $, x = 1/r,$ shows $,r'! = 0,$ so $,rx!-!1mid r^n f,Rightarrow,rx!-!1mid f,,$ by $,(rx!-!1,r) = (1);,$ more explicitly $,rx!-!1mid rg,Rightarrow, rx!-!1mid g = x(rg)-(rx!-!1)g$.
Remark $ $ See this answer for another proof and further discussion. If you already know basic properties of localizations then see also the linked dupe for ways to exploit these proeprties.
answered Dec 6 '18 at 23:59
Bill DubuqueBill Dubuque
210k29192643
$endgroup$
Tricky is proving the kernel $K = (rx!-!1).,$ A simple way: if $fin K$ then by nonmonic division
$$ r^n f(x) = (rx!-!1),q(x) + r', {rm for} r'in R, nin Bbb N$$
Evaluating at $, x = 1/r,$ shows $,r'! = 0,$ so $,rx!-!1mid r^n f,Rightarrow,rx!-!1mid f,,$ by $,(rx!-!1,r) = (1);,$ more explicitly $,rx!-!1mid rg,Rightarrow, rx!-!1mid g = x(rg)-(rx!-!1)g$.
Remark $ $ See this answer for another proof and further discussion. If you already know basic properties of localizations then see also the linked dupe for ways to exploit these proeprties.
answered Dec 6 '18 at 23:59
Bill DubuqueBill Dubuque
210k29192643
edited Dec 7 '18 at 0:22
answered Dec 6 '18 at 23:59
Bill DubuqueBill Dubuque
210k29192643
answered Dec 6 '18 at 23:59
Bill DubuqueBill Dubuque
210k29192643
answered Dec 6 '18 at 23:59
Bill DubuqueBill Dubuque
210k29192643
210k29192643
add a comment |
add a comment |
$begingroup$
You should note that $frac{1}{r}$ has the property $frac{1}{r}cdot r=1$. That is, $frac{1}{r}$ is not just an indeterminate object like $x$. Conceptually, this isomorphism should be easy to construct. Send $sin R$ to itself for all $s$. Send $xmapsto frac{1}{r}$. This is clearly a surjective homomorphism
$$ R[x]xrightarrow{phi} S^{-1}R.$$
Calculate $kerphi$. You will see that it is $(rx-1)$. Conclude using the first isomorphism theorem.
answered Dec 6 '18 at 22:08
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.2k41641
$endgroup$
$begingroup$
Why is the map clearly a surjection?
$endgroup$
– Wesley
Dec 6 '18 at 22:18
$begingroup$
Let $ain R$ and $r^kin S$. Is it surjective because $phi(ax^k) = frac{a}{r^k}$, and the latter term is a general term of the ring of fractions?
$endgroup$
– Wesley
Dec 6 '18 at 22:26
$begingroup$
Yes that is one way to see it.
$endgroup$
– Antonios-Alexandros Robotis
Dec 6 '18 at 22:50
$begingroup$
It's clear that $(rx-1) subseteq ker(phi)$, but how would I show that $ker(phi) subseteq (rx-1)$? If I could use the division algorithm I could see how, but I can't see how to do it without the division algorithm.
$endgroup$
– Wesley
Dec 6 '18 at 23:14
1
$begingroup$
@Wesley One simple way is to use the nonmonic division algorithm - see my answer.
$endgroup$
– Bill Dubuque
Dec 7 '18 at 0:02
add a comment |
$begingroup$
You should note that $frac{1}{r}$ has the property $frac{1}{r}cdot r=1$. That is, $frac{1}{r}$ is not just an indeterminate object like $x$. Conceptually, this isomorphism should be easy to construct. Send $sin R$ to itself for all $s$. Send $xmapsto frac{1}{r}$. This is clearly a surjective homomorphism
$$ R[x]xrightarrow{phi} S^{-1}R.$$
Calculate $kerphi$. You will see that it is $(rx-1)$. Conclude using the first isomorphism theorem.
answered Dec 6 '18 at 22:08
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.2k41641
$endgroup$
$begingroup$
Why is the map clearly a surjection?
$endgroup$
– Wesley
Dec 6 '18 at 22:18
$begingroup$
Let $ain R$ and $r^kin S$. Is it surjective because $phi(ax^k) = frac{a}{r^k}$, and the latter term is a general term of the ring of fractions?
$endgroup$
– Wesley
Dec 6 '18 at 22:26
$begingroup$
Yes that is one way to see it.
$endgroup$
– Antonios-Alexandros Robotis
Dec 6 '18 at 22:50
$begingroup$
It's clear that $(rx-1) subseteq ker(phi)$, but how would I show that $ker(phi) subseteq (rx-1)$? If I could use the division algorithm I could see how, but I can't see how to do it without the division algorithm.
$endgroup$
– Wesley
Dec 6 '18 at 23:14
1
$begingroup$
@Wesley One simple way is to use the nonmonic division algorithm - see my answer.
$endgroup$
– Bill Dubuque
Dec 7 '18 at 0:02
add a comment |
$begingroup$
You should note that $frac{1}{r}$ has the property $frac{1}{r}cdot r=1$. That is, $frac{1}{r}$ is not just an indeterminate object like $x$. Conceptually, this isomorphism should be easy to construct. Send $sin R$ to itself for all $s$. Send $xmapsto frac{1}{r}$. This is clearly a surjective homomorphism
$$ R[x]xrightarrow{phi} S^{-1}R.$$
Calculate $kerphi$. You will see that it is $(rx-1)$. Conclude using the first isomorphism theorem.
answered Dec 6 '18 at 22:08
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.2k41641
$endgroup$
You should note that $frac{1}{r}$ has the property $frac{1}{r}cdot r=1$. That is, $frac{1}{r}$ is not just an indeterminate object like $x$. Conceptually, this isomorphism should be easy to construct. Send $sin R$ to itself for all $s$. Send $xmapsto frac{1}{r}$. This is clearly a surjective homomorphism
$$ R[x]xrightarrow{phi} S^{-1}R.$$
Calculate $kerphi$. You will see that it is $(rx-1)$. Conclude using the first isomorphism theorem.
answered Dec 6 '18 at 22:08
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.2k41641
edited Dec 6 '18 at 22:37
answered Dec 6 '18 at 22:08
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.2k41641
answered Dec 6 '18 at 22:08
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.2k41641
answered Dec 6 '18 at 22:08
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.2k41641
10.2k41641
$begingroup$
Why is the map clearly a surjection?
$endgroup$
– Wesley
Dec 6 '18 at 22:18
$begingroup$
Let $ain R$ and $r^kin S$. Is it surjective because $phi(ax^k) = frac{a}{r^k}$, and the latter term is a general term of the ring of fractions?
$endgroup$
– Wesley
Dec 6 '18 at 22:26
$begingroup$
Yes that is one way to see it.
$endgroup$
– Antonios-Alexandros Robotis
Dec 6 '18 at 22:50
$begingroup$
It's clear that $(rx-1) subseteq ker(phi)$, but how would I show that $ker(phi) subseteq (rx-1)$? If I could use the division algorithm I could see how, but I can't see how to do it without the division algorithm.
$endgroup$
– Wesley
Dec 6 '18 at 23:14
1
$begingroup$
@Wesley One simple way is to use the nonmonic division algorithm - see my answer.
$endgroup$
– Bill Dubuque
Dec 7 '18 at 0:02
add a comment |
$begingroup$
Why is the map clearly a surjection?
$endgroup$
– Wesley
Dec 6 '18 at 22:18
$begingroup$
Let $ain R$ and $r^kin S$. Is it surjective because $phi(ax^k) = frac{a}{r^k}$, and the latter term is a general term of the ring of fractions?
$endgroup$
– Wesley
Dec 6 '18 at 22:26
$begingroup$
Yes that is one way to see it.
$endgroup$
– Antonios-Alexandros Robotis
Dec 6 '18 at 22:50
$begingroup$
It's clear that $(rx-1) subseteq ker(phi)$, but how would I show that $ker(phi) subseteq (rx-1)$? If I could use the division algorithm I could see how, but I can't see how to do it without the division algorithm.
$endgroup$
– Wesley
Dec 6 '18 at 23:14
1
$begingroup$
@Wesley One simple way is to use the nonmonic division algorithm - see my answer.
$endgroup$
– Bill Dubuque
Dec 7 '18 at 0:02
$begingroup$
Why is the map clearly a surjection?
$endgroup$
– Wesley
Dec 6 '18 at 22:18
$begingroup$
Why is the map clearly a surjection?
$endgroup$
– Wesley
Dec 6 '18 at 22:18
$begingroup$
Let $ain R$ and $r^kin S$. Is it surjective because $phi(ax^k) = frac{a}{r^k}$, and the latter term is a general term of the ring of fractions?
$endgroup$
– Wesley
Dec 6 '18 at 22:26
$begingroup$
Let $ain R$ and $r^kin S$. Is it surjective because $phi(ax^k) = frac{a}{r^k}$, and the latter term is a general term of the ring of fractions?
$endgroup$
– Wesley
Dec 6 '18 at 22:26
$begingroup$
Yes that is one way to see it.
$endgroup$
– Antonios-Alexandros Robotis
Dec 6 '18 at 22:50
$begingroup$
Yes that is one way to see it.
$endgroup$
– Antonios-Alexandros Robotis
Dec 6 '18 at 22:50
$begingroup$
It's clear that $(rx-1) subseteq ker(phi)$, but how would I show that $ker(phi) subseteq (rx-1)$? If I could use the division algorithm I could see how, but I can't see how to do it without the division algorithm.
$endgroup$
– Wesley
Dec 6 '18 at 23:14
$begingroup$
It's clear that $(rx-1) subseteq ker(phi)$, but how would I show that $ker(phi) subseteq (rx-1)$? If I could use the division algorithm I could see how, but I can't see how to do it without the division algorithm.
$endgroup$
– Wesley
Dec 6 '18 at 23:14
1
1
$begingroup$
@Wesley One simple way is to use the nonmonic division algorithm - see my answer.
$endgroup$
– Bill Dubuque
Dec 7 '18 at 0:02
$begingroup$
@Wesley One simple way is to use the nonmonic division algorithm - see my answer.
$endgroup$
– Bill Dubuque
Dec 7 '18 at 0:02
add a comment |
1
$begingroup$
Yes, evaluation homomorphism can do the trick $phi(p(x))=pleft(frac{1}{r}right)$.
$endgroup$
– Anurag A
Dec 6 '18 at 22:07
$begingroup$
Please search the site before asking a new question. This has been asked many, many times before: 1, 2, 3, 4 to name a few.
$endgroup$
– André 3000
Dec 6 '18 at 23:31