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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.

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  • $\begingroup$ You should separate each question, giving each its own post. Also, use the tag (proof-verification). $\endgroup$ – Omnomnomnom Jul 18 '14 at 20:40
  • $\begingroup$ Done. I will take care in the future about separating the question. I guess, I just gave in while I was writing. or may be I will just do that right now as well $\endgroup$ – MathMan Jul 18 '14 at 20:43
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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!

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There are a few issues here, but good effort.

First of all, it might clarify things to refer to elements of the field of fractions via their representative ordered pairs, rather than as fractions. I know that you have explained your notation, but when you write elements of the fraction field in this way, they look like elements of $F$ which may lead to confusion.

Your proof that $\Phi$ is a ring homomorphism looks good, although you could spell out explicitly how the rules of addition/multiplication in the field of fractions are applied.

Injectivity (one-to-one): you have to show that $\Phi(x)=\Phi(y) \implies x=y$. This is equivalent to proving that if $(x,1)=(y,1)$ then $x=y$.

Surjectivity (onto): you need to show that for any element $(x,y)$ in the field of fractions, there is some element $p \in F$ so that $\Phi(p)=(x,y)$. (This is where it becomes important not to refer to elements of the quotient as fractions -- you're going to need to divide to get the element $p$!).

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  • $\begingroup$ I'm glad I looked at Kevin's post, if only after the fact -- as he points out, you also need to check that $1$ is sent to $1$. $\endgroup$ – vociferous_rutabaga Jul 18 '14 at 20:54

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