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!

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.
