The field of fractions of an integral domain is a field 
Proposition: The field of fractions of an integral domain is a field

I founded the above proposition from Wikipedia. 
"In abstract algebra, the field of fractions or field of quotients of an integral domain is the smallest field in which it can be embedded" (http://en.wikipedia.org/wiki/Field_of_fractions)
How is this true?
Can you help me prove this one is true?
 A: This will probably be easiest to see by the universal property that the field of fractions satisfies.
Let $R$ be an integral domain.  The field of fractions can be described as a pair $(Frac(R),f)$, where $Frac(R)$ is a field, and $f:R\to Frac(R)$ an embedding, satisfying the following: if $k$ is another field and $g:R\to k$ is another embedding, then there exists an embedding $h:Frac(R)\to k$ such that the $g=h\circ f$.
From this definition, can you see why $Frac(R)$ is the smallest field containing $R$?  Also, under this definition, $Frac(R)$ is automatically a field, because we define it to be so.  We have yet to show that any such structure exists, but by the universal property, if it does exist, it is unique up to unique isomorphism.  The usual construction of such an object mimics the construction of $\mathbb{Q}$ from $\mathbb{Z}$.
A: What the proposition precisely means is the following. Assume that $D$ is an integral domain. Assume that you are given an injective ring homomorphism
$f\colon D\to F$, where $F$ is some field. Let $i\colon D\to Q(D)$ be the embedding of $D$
into its field of fractions $Q(D)$ given by the mapping $i(d)=d/1$ for all $d\in D$.
Then there exists a unique homomorphism of fields $\phi\colon Q(D)\to F$ such that $\phi\circ i=f$.
To prove this you can simply first guess that we must have
$$
\phi\left(\frac a b\right)=\frac{f(a)}{f(b)}
$$
for all $a,b\in D$, $b\neq0$. It will be easy to show that this is a homomorphism of
fields, IF you can FIRST show that it is well-defined. In other words, if
$$
\frac a b\qquad \text{and}\qquad \frac{a'}{b'}
$$
represent the same element in $Q(D)$, then you should show that this implies that
$$
\frac{f(a)}{f(b)}=\frac{f(a')}{f(b')}
$$
in the field $F$. Leaving that to you (do ask, if you get stuck).
Exercise #1: Why was it necessary to assume that $f$ is injective?
Exercise #2: The claim that "$Q(D)$ is the smallest field..." has the natural interpretation here coming from the fact that $\phi$ is also injective. Why is that necessarily the case?
