Is $0 \neq 1$ a necessary ring axiom?

Question

In the listing of the ring axioms, is $0 \neq 1$ a necessary axiom? I see it in some versions, but not others. Also, is the existence of a multiplicative identity necessary in listing the axioms? Thanks.

Answer 1

This is a bad axiom and you should not have it. The zero ring is a perfectly fine ring, and without it the category of rings fails to have some very natural constructions, such as tensor products. In algebraic geometry this would mean that affine schemes would fail to be closed under fiber products which would be very silly. You want to be able to talk about empty intersections.

I am also firmly of the opinion that rings should have multiplicative identities by default. This convention is, as far as I know, universal in fields such as algebraic geometry and algebraic topology. It is less universal in functional analysis / operator algebras.

Answer 2

As far as I am concerned, the only ring in which $0=1$ is the null ring $A={0}$. As this ring is not of interest, many books include $0 \neq 1$ as an axiom. For your second question, the existence of a multiplicative identity is a characteristic of unitary rings. Not all the rings have an identity element.

Answer 3

It is not a necessary axiom, but, as far as I know, there is no benefit in allowing such a thing. The additional structure that you would get is quite boring. Assuming that $0=1$ in $R$, we get that, for every element $a \in R$, $$a=1 \cdot a = 0 \cdot a=0,$$ so $R$ is a ring of only one element.

Answer 4

In a ring with $1$ the assumption $1\ne0$ is not necessary and, in fact, the standard modern definition allows the zero ring to exist. The way I personally make sense of this is that the zero ring is the terminal object of the category of rings with $1$, and the reason rings with $1$ should have a terminal object is that they naturally have products of every non-empty family. Therefore it's just natural that the product of the empty family should be there as well.

For the record, allowing the zero ring to exist makes the hypothesis $f(1)=1$ for homomorphisms come to its full fruition: in fact, there is no homomorphism $\{0\}\to R$ for any $R\ne \{0\}$, because it would send $1_0=0$ to $0_R\ne 1_R$.

To my knowledge, the current standard definition is that $1\ne 0$ in fields and, commonly, in integral domains.

For the second question, the nomenclature for a "ring without $1$" varies; some names in use are: ring, non-unital ring, pseudo-ring, rng. Likewise, possible terms for a ring with $1$ are: ring with 1, unital ring, ring.