Let $F$ be a field. Show that the field of quotients of $F$ is ring isomorphic to $F$.
Attempt: Let $F'$ be the field of Quotients of the field $F$.
Let $\Phi:F \rightarrow F'$ such that $\Phi(x)=x/1$
$x/1$ refers to the equivalence class containing the element $(x,1) ~;~x,1 \in F $ . The equivalence relations satisfies $a/b = c/d$ iff $ad=bc$
Operation Preservation : Then, $\Phi(x+y)= (x/1) + (y/1) = \Phi(x) + \Phi(y)$
$\Phi(xy) = (xy)/1 = (x/1) . (y/1)$
These operations are consistent with the definition of addition and multiplication defined for the elements of the field of quotient and hence, both addition and multiplication are preseerved.
One – One nature : If $x/1$ denotes the equivalence class containing $(x,1)$. Now, how do I prove the one-one nature. I have read about equivalence classes, but, don't possess much expert intuition. So, here's my attempt :
$(x,1) \in x/1 \implies (1,x) \in x/1 \implies (x,x) \in x/1$
Also, the equivalence classes are disjoint, $\implies$ No other equivalence class shall contain elements of the form $(x,1)$ or $(1,x)$ or $(x,x) \implies x/1 =y/1~~ \forall ~~x,y \in F$ . How does this $\implies x=y$ which is necessary for the one-one condition?
Onto Nature : Corresponding to any element $(x/1)$ in the quotient of field, we can find $x \in F$. Hence, the mapping is onto.
Is my attempt correct?
Thank you for your help.
Best Answer
First, make sure that you also check that $1\in F$ maps to $1\in F^\prime$, as that's also a criteria for a ring homomorphism.
To show one to one, you need to show that $\frac{x}{1} \simeq \frac{y}{1}$ implies that $x=y$. this isn't difficult based on the definition of the equivalence class. Clearly, two different elements of $F$ map to different elements of $F^\prime$. This now gives an embedding from $F$ into $F^\prime$.
More difficult, and the essence of the proof, is to show that the homomorphism is onto. This is important because it now shows that if $F$ is a field to begin with, then $F$ is in fact isomorphic to $F^\prime$ which makes sense since the "smallest" field you can embed a field into should be that field itself. Note this is the only part of the proof that requires you to use the fact that $F$ is a field.
To do this, you need to show that given any element of $y\in F^\prime$, not necessarily those with form $\frac{x}{1}$, that there exists $x\in F$ such that $\frac{x}{1} \simeq y$.
Given that $F$ is a field this should not be very difficult, but give it a shot!