Let R1 and R2 be integral domains with quotient fields F1 and F2 respectively. If phi: R1 -> R2 is a ring isomorphism, show that phi extends to an isomorphism phi hat : F1 -> F2. Here extends means that phi hat(a) = phi (a) for all a in R1 (Hint: under the givien assumptions, there is really only one way to define phi hat; of course you still have to prove that phi hat is well-defined and is an isomorphism)
Note: Since I don't have Math software, I use "phi hat" just notation "phi with ^".© BrainMass Inc. brainmass.com June 3, 2020, 5:41 pm ad1c9bdddf
First, we give a definition of phi-hat: F1->F2.
For any a/b in F1, where a,b are in R1, we define phi-hat(a/b)=phi(a)/phi(b)
Then for any a in R, we know a=a/1. So phi-hat(a)=phi-hat(a/1)=phi(a)/phi(1)=phi(a).
This phi-hat is well-defined because phi(a)/phi(b) is really in F2.
Now we show that phi-hat is a ...
This is a proof regarding ring isomorphisms.