Is every commutative ring contained in a field? I imagine the answer to this question is very simple, but I haven't been able to locate it.
Can every commutative ring be imbedded in a field? This seems very plausible to me, for it seems we could just take the "closure" under inverses. 
Thanks!
 A: Let $R$ be a commutative ring. Define $\bar Q(R)=R\times (R-\{0\})$ as the set of pairs $(r,s)$ with $r,s\in R$ and $s\not=0$. We define the equivalence relation
$$(r,s)\sim(r',s')\quad:\Leftrightarrow\quad rs'=sr'.$$
We define $Q(R):=\bar Q(R)/\!\!\sim$ and give it a ring structure (denoting by $x$ the equivalence class that contains $x$) via
$$(r,s)+(r',s')=(rs'+sr',ss'),\qquad (r,s)\cdot (r',s')=(rr',ss').$$
This turns out to be well-defined if and only if $R$ was an integral domain. Then $Q(R)$ will be a field $-$ the field of fractions. Think of $(r,s)\in Q(R)$ as the fraction $r/s$ or the element $rs^{-1}$. $R$ can be embedded into $Q(R)$ via  $r\mapsto(rx,x)$ for arbitrary $x$. The zero and unity of $Q(R)$ are $(0,x)$ and $(x,x)$ respectively. A non-zero element $(r,s)$ has the inverse $(s,r)$.
A: 
A commutative ring can be embedded in a field iff it is an integral domain.

Indeed, if a ring can be embedded in a field then it cannot have zero divisors because fields cannot have zero divisors.
Conversely, every integral domain can be embedded in a field, namely, its field of fractions.
A: No. Every nonzero element of a field is invertible; commutative rings can contain zero divisors.
The "closure" operation you're talking about is localization. In a (commutative) domain $A$, localizing at $0$ gives you the field of fractions $\operatorname{Frac}(A)$ of $A$, and the canonical map $A \to \operatorname{Frac}(A)$ is an embedding. In general, the localization is a more complicated ring than just a field.
