Why must a domain be nonzero? I recently stumbled across the following definition:

A ring $\mathcal R$ is called domain, if $\mathcal R$ is nonzero and $0$ is the only
  zero divisor in $\mathcal R$.

I'm wondering why the condition $\mathcal R$ is nonzero is even necessary, since for $\mathcal R = 0, 0$ already is no zero divisor $(\not\exists y\in \mathcal R, y \not= 0: 0\cdot y = 0)$. So the zero ring wouldn't be a domain without the extra part in the definition anyway.
 What am I missing?
 A: With the usual definition, it's convenient to have that an ideal in a commutative ring $I \subset R$ is a prime ideal iff $R/I$ is an integral domain. We don't want to consider the whole ring as a prime ideal for the similar reasons that we don't want $1$ to be a prime number. (And of course we want integral domains to be the commutative domains.) It would also mess up algebraic geometry if the whole ring was a prime ideal.
More general theorems break down if we allow the zero ring as an integral domain: integral domains always have a field of fractions that we can embed them into, but for the zero ring, the only possibility is the zero ring itself. Now you could say that just shifts the question to why the zero ring is not a field, but there are multiple reasons for that: 


*

*The set of non-zero elements is empty, so it doesn't form a group.

*For example linear algebra over the zero ring looks very different (and boring) compared to other fields

*In a classification that allows one to write certain rings as a product over fields or something similar (e.g. the Artin-Wedderburn theorem) we would explicitly need to exclude the zero ring to retain uniqueness.


See also too simple to be simple
A: There are ways in which one can think of a ring as having "components" — domains are the rings that have one component, and the zero ring is the ring that has zero components.
For example, every ring $R$ has a total ring of fractions $Q(R)$: the localization formed by inverting the multiplicative subset of non-zero divisors.
In the case that $R$ is a Noetherian reduced ring, then $Q(R)$ is a product of finitely many fields, and:


*

*$R$ is the zero ring if and only if $Q(R)$ is a product of zero fields (i.e. is the zero ring)

*$R$ is a domain if and only if $Q(R)$ is a product of one field (i.e. is a field)

