Extending a ring hom from $R to L$ to $K to L$, where $K$ is fraction field of domain $R$.

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Will this work as a proof?

Let $R$ be a domain and $L$ a field. Let $f : R to L$ be a ring hom. Let $K$ be the field of fractions of $R$. Then to extend $f$ to $K$ means there is a map $f^* : K to L$ such that $f^*|_R = f$. The obvious choice for $f^*$ is that it respects precisely the fractions so that $f^*(a/b) = f(a)f(b)^-1$. Let $g : K to L$ be a ring hom that also restricts to $f$ on $R$. Then $g(a/b) = g(a cdot (1/b)) = g(a) cdot g(1/b)$, but $1/b equiv b^-1$ in field $K$ so that $g(a/b) = g(a)g(b)^-1 = f(a) f(b)^-1 = f^*(a/b). square$

Thank you. I am putting this into a flash card so don't want to write down an invalid proof.

abstract-algebra proof-verification ring-theory fractions integral-domain

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asked Aug 5 at 22:57

EnjoysMath

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up vote
1
down vote

favorite

Will this work as a proof?

Let $R$ be a domain and $L$ a field. Let $f : R to L$ be a ring hom. Let $K$ be the field of fractions of $R$. Then to extend $f$ to $K$ means there is a map $f^* : K to L$ such that $f^*|_R = f$. The obvious choice for $f^*$ is that it respects precisely the fractions so that $f^*(a/b) = f(a)f(b)^-1$. Let $g : K to L$ be a ring hom that also restricts to $f$ on $R$. Then $g(a/b) = g(a cdot (1/b)) = g(a) cdot g(1/b)$, but $1/b equiv b^-1$ in field $K$ so that $g(a/b) = g(a)g(b)^-1 = f(a) f(b)^-1 = f^*(a/b). square$

Thank you. I am putting this into a flash card so don't want to write down an invalid proof.

abstract-algebra proof-verification ring-theory fractions integral-domain

share|cite|improve this question

asked Aug 5 at 22:57

EnjoysMath

8,64142154

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Will this work as a proof?

Let $R$ be a domain and $L$ a field. Let $f : R to L$ be a ring hom. Let $K$ be the field of fractions of $R$. Then to extend $f$ to $K$ means there is a map $f^* : K to L$ such that $f^*|_R = f$. The obvious choice for $f^*$ is that it respects precisely the fractions so that $f^*(a/b) = f(a)f(b)^-1$. Let $g : K to L$ be a ring hom that also restricts to $f$ on $R$. Then $g(a/b) = g(a cdot (1/b)) = g(a) cdot g(1/b)$, but $1/b equiv b^-1$ in field $K$ so that $g(a/b) = g(a)g(b)^-1 = f(a) f(b)^-1 = f^*(a/b). square$

Thank you. I am putting this into a flash card so don't want to write down an invalid proof.

abstract-algebra proof-verification ring-theory fractions integral-domain

share|cite|improve this question

asked Aug 5 at 22:57

EnjoysMath

8,64142154

Will this work as a proof?

Let $R$ be a domain and $L$ a field. Let $f : R to L$ be a ring hom. Let $K$ be the field of fractions of $R$. Then to extend $f$ to $K$ means there is a map $f^* : K to L$ such that $f^*|_R = f$. The obvious choice for $f^*$ is that it respects precisely the fractions so that $f^*(a/b) = f(a)f(b)^-1$. Let $g : K to L$ be a ring hom that also restricts to $f$ on $R$. Then $g(a/b) = g(a cdot (1/b)) = g(a) cdot g(1/b)$, but $1/b equiv b^-1$ in field $K$ so that $g(a/b) = g(a)g(b)^-1 = f(a) f(b)^-1 = f^*(a/b). square$

Thank you. I am putting this into a flash card so don't want to write down an invalid proof.

abstract-algebra proof-verification ring-theory fractions integral-domain

share|cite|improve this question

asked Aug 5 at 22:57

EnjoysMath

8,64142154

share|cite|improve this question

share|cite|improve this question

asked Aug 5 at 22:57

EnjoysMath

8,64142154

asked Aug 5 at 22:57

EnjoysMath

8,64142154

asked Aug 5 at 22:57

EnjoysMath

8,64142154

8,64142154

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add a comment | 

1 Answer
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You need to check the map is well defined. What does this mean? It means if we select two equivalent elements of the field of fractions $a/b sim c/d$ then your definition of $f(a/b)$ and $f(c/d)$ give the same answer.

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answered Aug 5 at 23:04

Daron

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

You need to check the map is well defined. What does this mean? It means if we select two equivalent elements of the field of fractions $a/b sim c/d$ then your definition of $f(a/b)$ and $f(c/d)$ give the same answer.

share|cite|improve this answer

answered Aug 5 at 23:04

Daron

4,3581923

add a comment | 

up vote
1
down vote



accepted

You need to check the map is well defined. What does this mean? It means if we select two equivalent elements of the field of fractions $a/b sim c/d$ then your definition of $f(a/b)$ and $f(c/d)$ give the same answer.

share|cite|improve this answer

answered Aug 5 at 23:04

Daron

4,3581923

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

You need to check the map is well defined. What does this mean? It means if we select two equivalent elements of the field of fractions $a/b sim c/d$ then your definition of $f(a/b)$ and $f(c/d)$ give the same answer.

share|cite|improve this answer

answered Aug 5 at 23:04

Daron

4,3581923

You need to check the map is well defined. What does this mean? It means if we select two equivalent elements of the field of fractions $a/b sim c/d$ then your definition of $f(a/b)$ and $f(c/d)$ give the same answer.

share|cite|improve this answer

answered Aug 5 at 23:04

Daron

4,3581923

share|cite|improve this answer

share|cite|improve this answer

answered Aug 5 at 23:04

Daron

4,3581923

answered Aug 5 at 23:04

Daron

4,3581923

answered Aug 5 at 23:04

Daron

4,3581923

4,3581923

add a comment | 

add a comment | 

 

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