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Ring of Fractions Isomorphic to Subring of Quotient Field.
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Let $R$ be an integral domain and let $D$ be a nonempty subset of $R$ that is closed under multiplication. Prove that the ring of fractions $D^-1R$ is isomorphic to a subring of the quotient field of $R$
Proof:
Let $F$ be $R$'s quotient field, and let $iota : D to F$ be defined by $iota(d) = de/e$, where $e$ can be taken to be any element in $D$. Since this is map is an injective homomorphism, there is an injective homomorphism $phi : D^-1R to F$ such that $phi_D = iota$. The map being injective implies that $D^-1R$ is isomorphic to $phi(D^-1R)$, a subring of $F$.
Does this sound right?
EDIT:
Note that I am using the following theorem in my proof:
Theorem 15: Let $R$ be a commutative ring, $D subseteq R$ nonempty multiplicative subset without $0$ or any zero divisors. Then there is a commutative unital ring $Q$ such that $R$ is a subring of it and every element of $D$ is a unit in $Q$. The ring $Q$ has the following additional properties:
(1) every element of $Q$ is of the form $rd^-1$ for some $r in R$ and $d in D$. In particular, if $D=R-0$, then $Q$ is a field
(2) Let $S$ be any commutative unital ring and let $varphi : R to S$ be any injective ring homomorphism such that $varphi(D)$ is contained in the units of $S$. Then there is an injective homomorphism $phi : Q to S$ such that $phi|_R = varphi$.
I am using (2), in particular, in concluding that $phi$ exists.
abstract-algebra integration ring-theory
asked Jul 16 at 17:46
user193319
2,0742719
add a comment |Â
up vote
0
down vote
favorite
Let $R$ be an integral domain and let $D$ be a nonempty subset of $R$ that is closed under multiplication. Prove that the ring of fractions $D^-1R$ is isomorphic to a subring of the quotient field of $R$
Proof:
Let $F$ be $R$'s quotient field, and let $iota : D to F$ be defined by $iota(d) = de/e$, where $e$ can be taken to be any element in $D$. Since this is map is an injective homomorphism, there is an injective homomorphism $phi : D^-1R to F$ such that $phi_D = iota$. The map being injective implies that $D^-1R$ is isomorphic to $phi(D^-1R)$, a subring of $F$.
Does this sound right?
EDIT:
Note that I am using the following theorem in my proof:
Theorem 15: Let $R$ be a commutative ring, $D subseteq R$ nonempty multiplicative subset without $0$ or any zero divisors. Then there is a commutative unital ring $Q$ such that $R$ is a subring of it and every element of $D$ is a unit in $Q$. The ring $Q$ has the following additional properties:
(1) every element of $Q$ is of the form $rd^-1$ for some $r in R$ and $d in D$. In particular, if $D=R-0$, then $Q$ is a field
(2) Let $S$ be any commutative unital ring and let $varphi : R to S$ be any injective ring homomorphism such that $varphi(D)$ is contained in the units of $S$. Then there is an injective homomorphism $phi : Q to S$ such that $phi|_R = varphi$.
I am using (2), in particular, in concluding that $phi$ exists.
abstract-algebra integration ring-theory
asked Jul 16 at 17:46
user193319
2,0742719
It is necessary to require $0notin D$.
– Fabio Lucchini
Jul 16 at 18:21
@FabioLucchini Thanks. I'll be sure to pass this on to Dummit and Foote.
– user193319
Jul 16 at 19:48
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $R$ be an integral domain and let $D$ be a nonempty subset of $R$ that is closed under multiplication. Prove that the ring of fractions $D^-1R$ is isomorphic to a subring of the quotient field of $R$
Proof:
Let $F$ be $R$'s quotient field, and let $iota : D to F$ be defined by $iota(d) = de/e$, where $e$ can be taken to be any element in $D$. Since this is map is an injective homomorphism, there is an injective homomorphism $phi : D^-1R to F$ such that $phi_D = iota$. The map being injective implies that $D^-1R$ is isomorphic to $phi(D^-1R)$, a subring of $F$.
Does this sound right?
EDIT:
Note that I am using the following theorem in my proof:
Theorem 15: Let $R$ be a commutative ring, $D subseteq R$ nonempty multiplicative subset without $0$ or any zero divisors. Then there is a commutative unital ring $Q$ such that $R$ is a subring of it and every element of $D$ is a unit in $Q$. The ring $Q$ has the following additional properties:
(1) every element of $Q$ is of the form $rd^-1$ for some $r in R$ and $d in D$. In particular, if $D=R-0$, then $Q$ is a field
(2) Let $S$ be any commutative unital ring and let $varphi : R to S$ be any injective ring homomorphism such that $varphi(D)$ is contained in the units of $S$. Then there is an injective homomorphism $phi : Q to S$ such that $phi|_R = varphi$.
I am using (2), in particular, in concluding that $phi$ exists.
abstract-algebra integration ring-theory
asked Jul 16 at 17:46
user193319
2,0742719
Let $R$ be an integral domain and let $D$ be a nonempty subset of $R$ that is closed under multiplication. Prove that the ring of fractions $D^-1R$ is isomorphic to a subring of the quotient field of $R$
Proof:
Let $F$ be $R$'s quotient field, and let $iota : D to F$ be defined by $iota(d) = de/e$, where $e$ can be taken to be any element in $D$. Since this is map is an injective homomorphism, there is an injective homomorphism $phi : D^-1R to F$ such that $phi_D = iota$. The map being injective implies that $D^-1R$ is isomorphic to $phi(D^-1R)$, a subring of $F$.
Does this sound right?
EDIT:
Note that I am using the following theorem in my proof:
Theorem 15: Let $R$ be a commutative ring, $D subseteq R$ nonempty multiplicative subset without $0$ or any zero divisors. Then there is a commutative unital ring $Q$ such that $R$ is a subring of it and every element of $D$ is a unit in $Q$. The ring $Q$ has the following additional properties:
(1) every element of $Q$ is of the form $rd^-1$ for some $r in R$ and $d in D$. In particular, if $D=R-0$, then $Q$ is a field
(2) Let $S$ be any commutative unital ring and let $varphi : R to S$ be any injective ring homomorphism such that $varphi(D)$ is contained in the units of $S$. Then there is an injective homomorphism $phi : Q to S$ such that $phi|_R = varphi$.
I am using (2), in particular, in concluding that $phi$ exists.
abstract-algebra integration ring-theory
asked Jul 16 at 17:46
user193319
2,0742719
edited Jul 17 at 12:13
asked Jul 16 at 17:46
user193319
2,0742719
asked Jul 16 at 17:46
user193319
2,0742719
asked Jul 16 at 17:46
user193319
2,0742719
2,0742719
It is necessary to require $0notin D$.
– Fabio Lucchini
Jul 16 at 18:21
@FabioLucchini Thanks. I'll be sure to pass this on to Dummit and Foote.
– user193319
Jul 16 at 19:48
add a comment |Â
It is necessary to require $0notin D$.
– Fabio Lucchini
Jul 16 at 18:21
@FabioLucchini Thanks. I'll be sure to pass this on to Dummit and Foote.
– user193319
Jul 16 at 19:48
It is necessary to require $0notin D$.
– Fabio Lucchini
Jul 16 at 18:21
It is necessary to require $0notin D$.
– Fabio Lucchini
Jul 16 at 18:21
@FabioLucchini Thanks. I'll be sure to pass this on to Dummit and Foote.
– user193319
Jul 16 at 19:48
@FabioLucchini Thanks. I'll be sure to pass this on to Dummit and Foote.
– user193319
Jul 16 at 19:48
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
It's okay but depending on your level I think you should give more justification to the assertions you state. Also, we need to assume $D$ does not contain $0$. The map $i:D to F$ should just be the composition $D to R to F$ where the last map is $rin R mapsto r/1 in F$. The image of $D$ in $F$ consists of non-zero elements (proof: if $d/1=0$ in $F$ then $dr=0$ for some nonzero $r in R$ so d=0 since R is a domain), and hence since $F$ is a field, the image of $D$ consists of invertible elements in $F$. Hence by universal property, we get a map $i:D^-1R to F$ sending $r/d in D^-1R mapsto r/d in F$. If $r/d in F$ is $0$, then $rs=0$ for nonzero $s in R$, so $r=0$. So the map is injective.
answered Jul 16 at 18:29
usr0192
1,039311
1
Note that it's not required $1in D$.
– Fabio Lucchini
Jul 16 at 19:31
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
It's okay but depending on your level I think you should give more justification to the assertions you state. Also, we need to assume $D$ does not contain $0$. The map $i:D to F$ should just be the composition $D to R to F$ where the last map is $rin R mapsto r/1 in F$. The image of $D$ in $F$ consists of non-zero elements (proof: if $d/1=0$ in $F$ then $dr=0$ for some nonzero $r in R$ so d=0 since R is a domain), and hence since $F$ is a field, the image of $D$ consists of invertible elements in $F$. Hence by universal property, we get a map $i:D^-1R to F$ sending $r/d in D^-1R mapsto r/d in F$. If $r/d in F$ is $0$, then $rs=0$ for nonzero $s in R$, so $r=0$. So the map is injective.
answered Jul 16 at 18:29
usr0192
1,039311
1
Note that it's not required $1in D$.
– Fabio Lucchini
Jul 16 at 19:31
add a comment |Â
up vote
0
down vote
It's okay but depending on your level I think you should give more justification to the assertions you state. Also, we need to assume $D$ does not contain $0$. The map $i:D to F$ should just be the composition $D to R to F$ where the last map is $rin R mapsto r/1 in F$. The image of $D$ in $F$ consists of non-zero elements (proof: if $d/1=0$ in $F$ then $dr=0$ for some nonzero $r in R$ so d=0 since R is a domain), and hence since $F$ is a field, the image of $D$ consists of invertible elements in $F$. Hence by universal property, we get a map $i:D^-1R to F$ sending $r/d in D^-1R mapsto r/d in F$. If $r/d in F$ is $0$, then $rs=0$ for nonzero $s in R$, so $r=0$. So the map is injective.
answered Jul 16 at 18:29
usr0192
1,039311
1
Note that it's not required $1in D$.
– Fabio Lucchini
Jul 16 at 19:31
add a comment |Â
up vote
0
down vote
up vote
0
down vote
It's okay but depending on your level I think you should give more justification to the assertions you state. Also, we need to assume $D$ does not contain $0$. The map $i:D to F$ should just be the composition $D to R to F$ where the last map is $rin R mapsto r/1 in F$. The image of $D$ in $F$ consists of non-zero elements (proof: if $d/1=0$ in $F$ then $dr=0$ for some nonzero $r in R$ so d=0 since R is a domain), and hence since $F$ is a field, the image of $D$ consists of invertible elements in $F$. Hence by universal property, we get a map $i:D^-1R to F$ sending $r/d in D^-1R mapsto r/d in F$. If $r/d in F$ is $0$, then $rs=0$ for nonzero $s in R$, so $r=0$. So the map is injective.
answered Jul 16 at 18:29
usr0192
1,039311
It's okay but depending on your level I think you should give more justification to the assertions you state. Also, we need to assume $D$ does not contain $0$. The map $i:D to F$ should just be the composition $D to R to F$ where the last map is $rin R mapsto r/1 in F$. The image of $D$ in $F$ consists of non-zero elements (proof: if $d/1=0$ in $F$ then $dr=0$ for some nonzero $r in R$ so d=0 since R is a domain), and hence since $F$ is a field, the image of $D$ consists of invertible elements in $F$. Hence by universal property, we get a map $i:D^-1R to F$ sending $r/d in D^-1R mapsto r/d in F$. If $r/d in F$ is $0$, then $rs=0$ for nonzero $s in R$, so $r=0$. So the map is injective.
answered Jul 16 at 18:29
usr0192
1,039311
answered Jul 16 at 18:29
usr0192
1,039311
answered Jul 16 at 18:29
usr0192
1,039311
answered Jul 16 at 18:29
usr0192
1,039311
1,039311
1
Note that it's not required $1in D$.
– Fabio Lucchini
Jul 16 at 19:31
add a comment |Â
1
Note that it's not required $1in D$.
– Fabio Lucchini
Jul 16 at 19:31
1
1
Note that it's not required $1in D$.
– Fabio Lucchini
Jul 16 at 19:31
Note that it's not required $1in D$.
– Fabio Lucchini
Jul 16 at 19:31
add a comment |Â
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It is necessary to require $0notin D$.
– Fabio Lucchini
Jul 16 at 18:21
@FabioLucchini Thanks. I'll be sure to pass this on to Dummit and Foote.
– user193319
Jul 16 at 19:48