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Universal Property of Quotient Field
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I have following question about Universal Property for Quotient Field that leads me to following contradiction:
The statement is: Let $R$ be an intergal domain and $phi: R to F$ is a ring morphism. Futhermore, let $i: R to Q(R)$ be the canonic inclusion. Then there exist a unique morphism $barphi: Q(R) to F$ with $phi = barphi circ i$.
Then I consider following example: Let $R = mathbbZ, F = mathbbZ/(p)$ for a prime $p$. Let $phi: mathbbZ to mathbbZ/(p)$ be the canonic projection. Obviously, $phi neq 0$. But on the other hand the extension $barphi: mathbbQ= Q(mathbbZ) to mathbbZ/(p)$ is obviously zero map since in $mathbbQ= Q(mathbbZ)$ we have $1 = fracpp$ and therefore $barphi(1)=0$. Where is the error in my reasonings?
abstract-algebra ring-theory
asked 1 hour ago
KarlPeter
5051313
add a comment |Â
up vote
2
down vote
favorite
I have following question about Universal Property for Quotient Field that leads me to following contradiction:
The statement is: Let $R$ be an intergal domain and $phi: R to F$ is a ring morphism. Futhermore, let $i: R to Q(R)$ be the canonic inclusion. Then there exist a unique morphism $barphi: Q(R) to F$ with $phi = barphi circ i$.
Then I consider following example: Let $R = mathbbZ, F = mathbbZ/(p)$ for a prime $p$. Let $phi: mathbbZ to mathbbZ/(p)$ be the canonic projection. Obviously, $phi neq 0$. But on the other hand the extension $barphi: mathbbQ= Q(mathbbZ) to mathbbZ/(p)$ is obviously zero map since in $mathbbQ= Q(mathbbZ)$ we have $1 = fracpp$ and therefore $barphi(1)=0$. Where is the error in my reasonings?
abstract-algebra ring-theory
asked 1 hour ago
KarlPeter
5051313
It can be made to work as long as your restrict to the subring of rationals writable with denominator coprime to $,p,,$ as I explain in my answer.
– Bill Dubuque
36 mins ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have following question about Universal Property for Quotient Field that leads me to following contradiction:
The statement is: Let $R$ be an intergal domain and $phi: R to F$ is a ring morphism. Futhermore, let $i: R to Q(R)$ be the canonic inclusion. Then there exist a unique morphism $barphi: Q(R) to F$ with $phi = barphi circ i$.
Then I consider following example: Let $R = mathbbZ, F = mathbbZ/(p)$ for a prime $p$. Let $phi: mathbbZ to mathbbZ/(p)$ be the canonic projection. Obviously, $phi neq 0$. But on the other hand the extension $barphi: mathbbQ= Q(mathbbZ) to mathbbZ/(p)$ is obviously zero map since in $mathbbQ= Q(mathbbZ)$ we have $1 = fracpp$ and therefore $barphi(1)=0$. Where is the error in my reasonings?
abstract-algebra ring-theory
asked 1 hour ago
KarlPeter
5051313
I have following question about Universal Property for Quotient Field that leads me to following contradiction:
The statement is: Let $R$ be an intergal domain and $phi: R to F$ is a ring morphism. Futhermore, let $i: R to Q(R)$ be the canonic inclusion. Then there exist a unique morphism $barphi: Q(R) to F$ with $phi = barphi circ i$.
Then I consider following example: Let $R = mathbbZ, F = mathbbZ/(p)$ for a prime $p$. Let $phi: mathbbZ to mathbbZ/(p)$ be the canonic projection. Obviously, $phi neq 0$. But on the other hand the extension $barphi: mathbbQ= Q(mathbbZ) to mathbbZ/(p)$ is obviously zero map since in $mathbbQ= Q(mathbbZ)$ we have $1 = fracpp$ and therefore $barphi(1)=0$. Where is the error in my reasonings?
abstract-algebra ring-theory
abstract-algebra ring-theory
asked 1 hour ago
KarlPeter
5051313
asked 1 hour ago
KarlPeter
5051313
asked 1 hour ago
KarlPeter
5051313
asked 1 hour ago
KarlPeter
5051313
asked 1 hour ago
KarlPeter
5051313
5051313
It can be made to work as long as your restrict to the subring of rationals writable with denominator coprime to $,p,,$ as I explain in my answer.
– Bill Dubuque
36 mins ago
add a comment |Â
It can be made to work as long as your restrict to the subring of rationals writable with denominator coprime to $,p,,$ as I explain in my answer.
– Bill Dubuque
36 mins ago
It can be made to work as long as your restrict to the subring of rationals writable with denominator coprime to $,p,,$ as I explain in my answer.
– Bill Dubuque
36 mins ago
It can be made to work as long as your restrict to the subring of rationals writable with denominator coprime to $,p,,$ as I explain in my answer.
– Bill Dubuque
36 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
The ring homomorphisms $phi:Rto F$ is required to be injective (and $F$ is required to be a field).
See here for example.
More generally, a ring homomorphism $phi:Rto F$ extends to a ring homomorphism $Q(R)to F$ if and only if $phi$ maps every non-zero element of $R$ into a unit in $F$.
answered 1 hour ago
Fabio Lucchini
6,88511126
Technically true but not the best way to view it conceptually. Rather. the point is that the map must send denominators to unis since they are units in the fraction field. Viewing it this way shows how to workaround the obstruction, as I explain in my answer.
– Bill Dubuque
31 mins ago
add a comment |Â
up vote
2
down vote
Your statement of the universal property is incorrect. It only applies when the homomorphism $phi:Rto F$ is injective.
answered 1 hour ago
Eric Wofsey
170k12198316
add a comment |Â
up vote
0
down vote
The theorem requires that the map sends denominators into units, which fails for denominators that are multiples of $p$. We can work around this if we restrict to fractions having denominators coprime to $p$. Then, by a similar proof, the natural ring map of $,Bbb Z,$ into $,Bbb Z/p,$ extends to the subring of $,Bbb Q,$ of fractions writable with denominator coprime to $p$, via $,a/b mapsto ab^-1pmod p.,$ This implies that it is valid to use such fractions when working in $,Bbb Z/p.,$ You can find many examples of such modular fraction arithmetic in my posts here.
answered 54 mins ago
Bill Dubuque
204k29189614
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The ring homomorphisms $phi:Rto F$ is required to be injective (and $F$ is required to be a field).
See here for example.
More generally, a ring homomorphism $phi:Rto F$ extends to a ring homomorphism $Q(R)to F$ if and only if $phi$ maps every non-zero element of $R$ into a unit in $F$.
answered 1 hour ago
Fabio Lucchini
6,88511126
Technically true but not the best way to view it conceptually. Rather. the point is that the map must send denominators to unis since they are units in the fraction field. Viewing it this way shows how to workaround the obstruction, as I explain in my answer.
– Bill Dubuque
31 mins ago
add a comment |Â
up vote
4
down vote
accepted
The ring homomorphisms $phi:Rto F$ is required to be injective (and $F$ is required to be a field).
See here for example.
More generally, a ring homomorphism $phi:Rto F$ extends to a ring homomorphism $Q(R)to F$ if and only if $phi$ maps every non-zero element of $R$ into a unit in $F$.
answered 1 hour ago
Fabio Lucchini
6,88511126
Technically true but not the best way to view it conceptually. Rather. the point is that the map must send denominators to unis since they are units in the fraction field. Viewing it this way shows how to workaround the obstruction, as I explain in my answer.
– Bill Dubuque
31 mins ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The ring homomorphisms $phi:Rto F$ is required to be injective (and $F$ is required to be a field).
See here for example.
More generally, a ring homomorphism $phi:Rto F$ extends to a ring homomorphism $Q(R)to F$ if and only if $phi$ maps every non-zero element of $R$ into a unit in $F$.
answered 1 hour ago
Fabio Lucchini
6,88511126
The ring homomorphisms $phi:Rto F$ is required to be injective (and $F$ is required to be a field).
See here for example.
More generally, a ring homomorphism $phi:Rto F$ extends to a ring homomorphism $Q(R)to F$ if and only if $phi$ maps every non-zero element of $R$ into a unit in $F$.
answered 1 hour ago
Fabio Lucchini
6,88511126
edited 1 hour ago
answered 1 hour ago
Fabio Lucchini
6,88511126
answered 1 hour ago
Fabio Lucchini
6,88511126
answered 1 hour ago
Fabio Lucchini
6,88511126
6,88511126
Technically true but not the best way to view it conceptually. Rather. the point is that the map must send denominators to unis since they are units in the fraction field. Viewing it this way shows how to workaround the obstruction, as I explain in my answer.
– Bill Dubuque
31 mins ago
add a comment |Â
Technically true but not the best way to view it conceptually. Rather. the point is that the map must send denominators to unis since they are units in the fraction field. Viewing it this way shows how to workaround the obstruction, as I explain in my answer.
– Bill Dubuque
31 mins ago
Technically true but not the best way to view it conceptually. Rather. the point is that the map must send denominators to unis since they are units in the fraction field. Viewing it this way shows how to workaround the obstruction, as I explain in my answer.
– Bill Dubuque
31 mins ago
Technically true but not the best way to view it conceptually. Rather. the point is that the map must send denominators to unis since they are units in the fraction field. Viewing it this way shows how to workaround the obstruction, as I explain in my answer.
– Bill Dubuque
31 mins ago
add a comment |Â
up vote
2
down vote
Your statement of the universal property is incorrect. It only applies when the homomorphism $phi:Rto F$ is injective.
answered 1 hour ago
Eric Wofsey
170k12198316
add a comment |Â
up vote
2
down vote
Your statement of the universal property is incorrect. It only applies when the homomorphism $phi:Rto F$ is injective.
answered 1 hour ago
Eric Wofsey
170k12198316
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Your statement of the universal property is incorrect. It only applies when the homomorphism $phi:Rto F$ is injective.
answered 1 hour ago
Eric Wofsey
170k12198316
Your statement of the universal property is incorrect. It only applies when the homomorphism $phi:Rto F$ is injective.
answered 1 hour ago
Eric Wofsey
170k12198316
answered 1 hour ago
Eric Wofsey
170k12198316
answered 1 hour ago
Eric Wofsey
170k12198316
answered 1 hour ago
Eric Wofsey
170k12198316
170k12198316
add a comment |Â
add a comment |Â
up vote
0
down vote
The theorem requires that the map sends denominators into units, which fails for denominators that are multiples of $p$. We can work around this if we restrict to fractions having denominators coprime to $p$. Then, by a similar proof, the natural ring map of $,Bbb Z,$ into $,Bbb Z/p,$ extends to the subring of $,Bbb Q,$ of fractions writable with denominator coprime to $p$, via $,a/b mapsto ab^-1pmod p.,$ This implies that it is valid to use such fractions when working in $,Bbb Z/p.,$ You can find many examples of such modular fraction arithmetic in my posts here.
answered 54 mins ago
Bill Dubuque
204k29189614
add a comment |Â
up vote
0
down vote
The theorem requires that the map sends denominators into units, which fails for denominators that are multiples of $p$. We can work around this if we restrict to fractions having denominators coprime to $p$. Then, by a similar proof, the natural ring map of $,Bbb Z,$ into $,Bbb Z/p,$ extends to the subring of $,Bbb Q,$ of fractions writable with denominator coprime to $p$, via $,a/b mapsto ab^-1pmod p.,$ This implies that it is valid to use such fractions when working in $,Bbb Z/p.,$ You can find many examples of such modular fraction arithmetic in my posts here.
answered 54 mins ago
Bill Dubuque
204k29189614
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The theorem requires that the map sends denominators into units, which fails for denominators that are multiples of $p$. We can work around this if we restrict to fractions having denominators coprime to $p$. Then, by a similar proof, the natural ring map of $,Bbb Z,$ into $,Bbb Z/p,$ extends to the subring of $,Bbb Q,$ of fractions writable with denominator coprime to $p$, via $,a/b mapsto ab^-1pmod p.,$ This implies that it is valid to use such fractions when working in $,Bbb Z/p.,$ You can find many examples of such modular fraction arithmetic in my posts here.
answered 54 mins ago
Bill Dubuque
204k29189614
The theorem requires that the map sends denominators into units, which fails for denominators that are multiples of $p$. We can work around this if we restrict to fractions having denominators coprime to $p$. Then, by a similar proof, the natural ring map of $,Bbb Z,$ into $,Bbb Z/p,$ extends to the subring of $,Bbb Q,$ of fractions writable with denominator coprime to $p$, via $,a/b mapsto ab^-1pmod p.,$ This implies that it is valid to use such fractions when working in $,Bbb Z/p.,$ You can find many examples of such modular fraction arithmetic in my posts here.
answered 54 mins ago
Bill Dubuque
204k29189614
edited 49 mins ago
answered 54 mins ago
Bill Dubuque
204k29189614
answered 54 mins ago
Bill Dubuque
204k29189614
answered 54 mins ago
Bill Dubuque
204k29189614
204k29189614
add a comment |Â
add a comment |Â
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It can be made to work as long as your restrict to the subring of rationals writable with denominator coprime to $,p,,$ as I explain in my answer.
– Bill Dubuque
36 mins ago