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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    up vote
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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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      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











      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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          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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            up vote
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            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.






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              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























                up vote
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                accepted







                up vote
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                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













                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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