What's the rationale for requiring that a field be a *non-trivial* ring? The title pretty much says it all.
Of course, one answer (IMO unsatisfactory) to such questions goes something like "a definition is a definition, period."  In my experience, mathematical definitions are rarely completely arbitrary.  Therefore I figure there must be a good reason for insisting that a field be a non-trivial ring (among other things), but it's not obvious to me.  I realize that a trivial ring would make for a very boring field, but it is also a rather boring ring, and yet it is not disallowed as such.
EDIT: I found a hint of a rationale in the statement that "the zero ring ... does not behave like a finite field," but I could not find in the source for this assertion exactly how the zero ring fails to behave like a finite field.
 A: Good question, and I don't have a complete answer. But here's a partial answer.
It is a general principle that if we can get away with using just identities (i.e. universally quantified equations), that's great, because it means we end up with a variety. The axioms of ring-theory fall into this category, which is why the category of rings is so well-behaved.
Failing that, if we can get away with using just quasi-identities, that's still pretty great, because we'll end up with a quasi-variety. By a quasi-identity, I mean a universally quantified axiom of the following form, where all the Greek letters represent equations.
$$\varphi_0 \wedge \cdots \wedge \varphi_{n-1} \rightarrow \psi$$
Cancellative semigroups fall into this category, which is why the category of cancellative semigroups is also quite nice.
However, suppose we really, really need either logical OR $(\vee)$ or logical negation $(\neg)$ for one of our axioms. This happens, for example, with integral domains; we typically assume either of the following.

*

*$xy=0 \rightarrow x=0 \vee y=0$

*$a=0 \vee (ax = ay \rightarrow x=y)$
It also happens with fields:
$$\forall x(x=0 \vee \exists y(xy=1))$$
Anyway, the point is this. If we need either $\vee$ or $\neg$ to axiomatize a class of algebraic structures, then we may as well include some non-degeneracy axioms, like $0 \neq 1,$ because frankly, our category of models already sucks, and few more non-triviality axioms aren't going to make it any suckier.
I think that is why fields (integral domains, etc.) are typically given non-triviality axioms, while rings (groups, etc.) are not.
A: The zero ring not only cannot be a field, it cannot even be either a local ring or an integral domain.  If it were either of those things, the ideal $R\subset R$ would need to be prime for any unital commutative ring $R$, which is awkward and pointless, creating unnecessary difficulty everywhere.
This is pretty much the same reason that $1$ is not considered a prime number.
There are hypothetical advantages to a field that includes into every other field, but these are not realized in practice by the zero ring.  In a typical examination of the "Field with one element", one looks at something that isn't a ring at all, such as the monoid $(\{0,1\},\times)$.
A: At some level, this is just a extension of vadim123's answer. A good definition has three properties 


*

*it covers a good class of "useful cases"

*it can be used to prove interesting and non-trivial theorems

*it (hopefully) has a high-level motivation. 


Consider the notion of a field. It certainly satisfies 1 and 2 ($\bar{\mathbf Q}$, $\mathbf R$, $\mathbf C$, $\mathbf F_q\ldots$ being good examples of 1, and Galois theory being a good example of 2). Now consider 3. One would like a field to be a "simple commutative ring," i.e. a ring for which any module is a direct sum of copies of irreducibles, and such that there is (up to isomorphism) only one irreducible. The trivial ring satisfies this, but consider the definitions of "simple group" and "simple Lie algebra." Simple Lie algebras are not allowed to be trivial, and simple groups are tacitly assumed to be nontrivial. In other words, if you're going to study a field $k$, you want to do things like: 


*

*Consider varieties over $k$

*Look at Galois extensions of $k$

*Do linear algebra over $k$


For all of these, allowing the trivial ring to be a field makes most interesting theorems have an additional assumption "$k$ is nontrivial." 
For example:


*

*we want there to actually be nontrivial varieties over $k$

*either $k$ should be algebraically closed (meaning one can do geometry over it) or there is interesting arithmetic coming e.g. from abelian varieties over $k$

*we want a theorem like: "the $K$-theory of the category of vector spaces over $k$ is $\mathbf Z$"


You could say that adding the nontriviality hypothesis to the definition of a field is arbitrary and ad-hoc, but I respond that all definitions are (to some degree) arbitrary and ad-hoc. Mathematicians are interested in what interesting or useful theorems you can prove, and are generally only interested in "better" definitions if they a) make theorems cleaner or b) make proofs nicer. Allowing the trivial ring to be a field accomplishes neither of these. 
Finally, note that there is a (not entirely rigorous) theory of a "field with one element," but it is much more sophisticated than simply saying "the field with one element is the trivial ring." For exampe:


*

*A vector space over $\mathbf F_1$ is  pointed set

*$\operatorname{GL}(n,\mathbf F_1)=S_n$

*The Riemann hypothesis "should" follow from considering $\operatorname{Spec}(\mathbf Z)$ as a curve over $\mathbf F_1$. 


None of these work if we do the naive thing and set $\mathbf F_1=0$. 
A: If it's not part of the definition, then almost every proof that uses that definition will need a special case to handle the trivial ring.  To eliminate that wasteful and annoying step, many people write the definition to exclude the trivial ring.
A: It is primarily a matter of convenience that fields (and sometimes domains) are required by definition to satisfy $1\ne 0,\,$ i.e. the trivial ring $\{0\}$ is excluded. There are various motivations for such. Without this convention many definitions and theorems would require cumbersome exceptions to handle trivial degenerate cases. Also in domains and fields it often proves very convenient to assume that one has a nonzero element available. This permits proofs by contradiction to conclude by deducing $\,1 = 0,\,$ or that $\,0\,$ is invertible, or equivalent field "absurdities". More importantly, it implies that the  unit group of a domain is nonempty, so unit groups always exist. It would be very inconvenient to have to always add the proviso (except if $\;\rm R = 0)$ to the ubiquitous arguments involving units and unit groups. More generally it is worth emphasizing that the usual rules for equational logic are not complete for empty structures. That is why groups and other algebraic structures are always axiomatized to prevent nonempty structures (see this sci.math thread for further details).  
Remark $ $ Perhaps worth mention is that there are some examples on MSE of just how confusing things can get if one starts reasoning in the zero ring, esp. in a proof by contradiction.  For one example of such, see the long discussion in comments to my answer here, where it took a surprisingly long time to convince some readers that a reinterpretation of one of Rudin's proofs  (by contradiction) was actually valid, and had a natural interpretation in the trivial ring (where $0/0 = 1$). See also the closely related discussion in this comment thread. If those discussions aren't enought to convince one of the pedagogical difficulties with such, then I suspect nothing will! $ $  Note: when reading those threads, be sure to click on "show more comments" so you can read the entire discussion (the top voted comments are skewed by the misunderstandings in the earlier part of the discussions).
A: I think of the requirement $0\neq1$ for fields as a consequence of the desires for (1) fields to be integral domains and (2) integral domains to satisfy $0\neq1$. Of course that just shifts the question to: Why do I want integral domains to have $0\neq1$?  That desire comes from my (quite general) inclination to treat the empty set just like other finite sets. Here's how that's relevant: The key clause in the definition of integral domains (besides the axioms for commutative rings with unit) is that if $xy=0$ then $x=0$ or $y=0$. It follows immediately by induction that, if a product of $n\geq 2$ factors is $0$, then there must be a $0$ among those factors. The same holds trivially for $n=1$ under the obvious convention that the product of a single factor is that factor.  So I'd like it to hold also for $n=0$.  Now the product of no factors is $1$, so what I want is that, if $1=0$, then there is a $0$ in the (empty) set of factors.  As there is no $0$ (or anything else) in the empty set, I infer that $1\neq0$.
(The same underlying idea explains why I don't regard the integer $1$ as prime and why I want lattices to have top and bottom elements.)
A: And Bourbaki said (Algebra I, Chapter 1, §9, definition 1)  
"A ring $K$ is called a field if it does not consist only of $0$ and every non-zero element of $K$ is invertible"
and fields did not consist only of $0$.  
Bourbaki  saw that the rings were good, and he separated the fields from the other rings.
He  called the fields with commutative multiplication "commutative fields" and non-commutative fields he called "skew fields" .
And there were domains  and there were division rings-the first day.
Edit: For miscreants, heretics, apostates, schismatics, infidels and other iconoclasts
Since your ilk might not know: this answer was shamefully plagiarized from lines 3,4,5 of this text.
