Every subring of a field is a domain. Is this reciprocal? I'm reading my notes on ring theory, and we proved on class that every subring of a field is a domain.

Proof: Let $S \subseteq K$ be a subring of $K$, with $K$ a field. Let $x,y \in S$. If $xy=0$, then $xy=0$ in $K$ too, so $x=0$ or $y=0$ (because $K$ is a field).

Automatically the question that came to my mind was: Is the reciprocal true? Can we say that every domain is a subring of a field?
Thank you.
 A: Yes, every integral domain $D$ is a subring of a field. Suppose $D$ has more than one element. The construction of the field looks much like the standard formal construction of the rationals from the integers. The intuition is that the elements of the field should "behave" like fractions $\frac{a}{b}$, where $b\ne 0$. The complication is that we need to build in the fact that the fraction $\frac{a}{b}$ is the same as the fraction $\frac{ka}{kb}$. 
Let $W$ be the set of all ordered pairs $(a,b)$, where $a$ ranges over $D$ and $b$ ranges over the non-zero elements of $D$. On this set $W$, define an equivalence relation $\equiv$ by $(a,b)\equiv (c,d)$ if $ad=cb$. Let $F$ be the set of equivalence classes. 
We define addition and multiplication on $F$ in the natural way. Multiplication is simpler. The product of the equivalence classes of $(a,b)$ and $(c,d)$ is defined to be the equivalence class of $(ab,cd)$. The sum is defined to be the equivalence class of $(ad+cb, bd)$. We have to check that sum and product are well-defined. It is not hard to check that $F$ is a field under these operations. 
Technically, $D$ is not a subring of $F$. However, the mapping $\varphi$ that takes an element $x$ of $D$ to $F$ by mapping $x$ to the equivalence class of $(xb,b)$, where $b$ is non-zero, embeds $D$ isomorphically in $F$. By replacing $\varphi(x)$ by $x$ we obtain a field $F'$ that has $D$ as a subring.
