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In the definition of field, we exclude $\{0\}$ by assuming $1\ne0$ (i.e. there must be at least two elements in the field). Therefore, it is not a field. However, in the case of ring definition, even though there is an axiom of existence of multiplicative identity $1$ as in the field definition, we do not assume $1\ne0$ and call $\{0\}$ the zero ring. Is there a reason why we do not exclude $\{0\}$ from ring and treat it as a rng?

Prown
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    At first, this ring looks totally useless, and many theorems only hold for nontrivial rings. (like existence of a maximal ideal) But when you start learning algebraic geometry, and in particular scheme theory, it becomes important that the category of rings has a terminal object. Otherwise, there would be some technical problems. – Mark Jun 29 '24 at 13:34
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    There are actually authors who do require their rings to satisfy $0 \ne 1$. Here's just one example: Rings and Categories of Modules by Anderson & Fuller. Related question: https://math.stackexchange.com/questions/3917370/is-0-neq-1-a-necessary-ring-axiom. – Hans Lundmark Jun 29 '24 at 17:22

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Do you agree that every abelian group $A$ has a ring $\text{End}(A)$ of endomorphisms? Do you agree that the trivial group is an abelian group? Now, what's the ring of endomorphisms of the trivial group?

Note that the trivial group is also the zero-dimensional vector space over every field $F$, so this is equivalent to asking: what's the ring of $0 \times 0$ matrices over $F$?

Qiaochu Yuan
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If $R$ is a set with one element $\star$, then $(R,+,\cdot)$ is a ring under the operations $+:R\times R\to R$ and $\cdot:R\times R\to R$ given by $\star+\star=\star\cdot\star=\star$. If you follow the definitions of additive identity and multiplicative identity carefully, you will see that $\star=0=1$ in the ring $(R,+,\cdot)$. So that answers why we consider it to be a ring, not just a rng. It is conventional to call the ring $(R,+,\cdot)$ the zero ring and denote it by $\{0\}$, even though any one element set can be given a ring structure. (This convention is understandable in light of the fact that the zero ring is unique up to unique isomorphism, but perhaps you haven't covered ring homomorphisms and isomorphisms yet.)

As for why we don't consider the zero ring to be a field, you might want to consult this thread. In contrast to the situation with fields, there is no compelling reason to exclude the zero ring from being a ring.

Joe
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  • That is, we exclude ${0}$ from field because it would be burdensome if we think it as a field, but we don't do this for ring because it doesn't make that much hard work even if we regard the zero ring as a ring. Do I understand correctly? – Prown Jun 29 '24 at 14:19
  • @Prown: Yes, that’s correct. The zero ring simply doesn’t behave like a field, e.g. the nonzero elements of a field form a group under multiplication. But it is convenient to regard the zero ring as a a ring, though the reasons for this are perhaps a little more technical. – Joe Jun 29 '24 at 14:39
  • Thank you for answering my question! – Prown Jun 29 '24 at 14:44
  • @Prown: No worries! I'm glad I could help. Let me know if you have any more questions. – Joe Jun 29 '24 at 15:15
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    @Joe, to clarify one point: if we say that "non-zero elements of a field form a group", then, since a group has to contain a unit, that set cannot be empty. So ${0}$ could not be a field. On another hand, if we say that "non-zero elements of a field have multiplicative inverses", then ${0}$ is a field, because the condition is vacuously true. :) – paul garrett Jun 29 '24 at 16:33