Example of an integral domain which is not a field I just proved that a commutative ring $R$ is an integral domain iff $R$ is isomorphic to a subring of a field. 

My question is why can't $R$ be a field with these conditions? Aren't we satisfying all of the field's properties?

Thanks.
 A: No, subring of a field does not satisfy all the field's axioms. Namely, the problem is twofold: the subring doesn't have to contain $1$ and even when it does, there is trouble with inverses.
Let $1\in R\subseteq \mathbb F$, where $\mathbb F$ is a field and let $0\neq r\in R$. Sure, $r$ is invertible in $\mathbb F$, but what guarantees that $r^{-1}\in R$? Well, nothing. See the answer by Famke where they give example of $\mathbb Z\subseteq \mathbb Q$.
A: A yet another example are polynomials $R[x]$ over a field $R$. While (obviously) all other properties are fine, there are much less inverses than there are required for a field.
Now formal power series offer somewhat a remedy...
A: For a counter-example, 
let's have a look at $\mathbb{Z} \subseteq \mathbb{Q}$.
Here $\mathbb{Z}$ is an integral domain which is not a field;
also you can check that $\mathbb{Z}$ 
is a sub-ring of the field of rational numbers $\mathbb{Q}$.  

Note that $\mathbb{Z}$ satisfies all of the field's properties;  except the property which concerns the
existence of multiplicative inverses for non-zero elements. 
For example $2^{-1} \notin \mathbb{Z}$ .
