Making definitions so that $\{0\}$ is not a field. I have a follow-up question to my question here: Are there any rings where $0$ has a multiplicative inverse?
Here it was made clear that in the trivial ring $\{0\}$ the element $0$ does have a multiplicative inverse. From some of the linked answers and comment(Are there any rings where $0$ has a multiplicative inverse?) it was pointed out that this is one of the reasons that some people don't allow rings with $1=0$ (I think). 
I can see that if one defines a field as a commutative ring where all non-zero elements have multiplicative inverses, then $\{0\}$ is a field.
My question is if the following is correct way to avoid having $\{0\}$ be a field by simply defining a field as a commutative ring with $R^\times = R\setminus \{0\}$. (Here $R^\times$ is the set of units). If this is the definition of a field, then for the trivial (commutative) ring $\{0\}$ we would have
$$
R^\times = \{0\} \neq R\setminus \{0\}.
$$
Did I do that right?
 A: To my mind, the cleanest definition is:

Definition: A field is a commutative ring with exactly two quotients (namely, the field itself and $0$), or equivalently with exactly two ideals.

Said another way, fields are the simple objects in commutative rings. The zero ring $\{ 0 \}$ is still a ring (and you really want it in your theory, because it's the terminal object in the category of rings), it's just not a field by this definition because it has only one quotient / only one ideal. 
The nLab calls this general phenomenon too simple to be simple. Here are some other definitions along these lines:


*

*A simple group is a group with exactly two quotient groups, or equivalently exactly two normal subgroups. The trivial group $\{ 1 \}$ is not simple. 

*A connected graph is a graph with exactly one connected component (where a connected component of a graph is an equivalence class under the equivalence relation generated by being connected by an edge). The empty graph is not connected. 

*A prime is a positive integer with exactly two divisors. $1$ is not a prime.


(The reason for the flip between one and two is that some of these definitions describe "connected components" directly and some of these definitions describe, in some sense, the lattice of subsets of connected components.) 
There are also some interesting things known about which definitions of a field are and aren't equivalent constructively; for example, some of the obvious definitions don't apply constructively to $\mathbb{R}$! 
A: There are more handy definitions based on $R-\{0\}$ where $R$ is a ring with identity:


*

*trivial ring if $R-\{0\}=\varnothing$.

*ring without zero-divisors if $R-\{0\}$  is a monoid.

*integral domain if $R-\{0\}$  is a commutative monoid.

*division ring if $R-\{0\}$ is a group.

*field if $R-\{0\}$  is a commutative group.

A: One of the field axioms is called the non triviality axiom and asks that $1 \neq 0$
For your question, this would certainly work but it is much easier to just ask that $0 \neq 1$, but in the end, it doesn't matter which definition we use, as long as they are equivalent (which is the case here)
