# Is there any finite integral domain consisting entirely of $\mathbf 0$ and $\mathbf 1$?

I'd like to know whether there is an integral domain comprising only the additive identity $$\mathbf 0$$ and the multiplicative identity $$\mathbf 1$$. For unification, let me introduce some definitions.

Definition An algebraic system $$(R,+,\times)$$ is a ring if

$$\qquad R_1$$: $$(R,+)$$ is an abelian group.

$$\qquad R_2$$: $$\times$$ is associative.

$$\qquad R_3$$: $$\times$$ is distributive over $$+$$.

Definition If $$a,b$$ are nonzero elements of a ring $$R$$ s.t. $$a\times b=\mathbf 0$$, then $$a,b$$ are divisors of $$\mathbf 0$$.

Definition An integral domain is a commutative ring with a multiplicative identity $$\mathbf 1\not=\mathbf 0$$ and with no divisor of $$\mathbf 0$$.

To answer the question, I try to challenge $$\{\mathbf 0,\mathbf1\}$$ with the definition of an integral domain. But I have difficulty defining $$\mathbf 1 +\mathbf 1$$. What can I do next? Thank you.

• It seems hat every field is an integral domain, so you can just take $\mathbb{F}_2$. It is the only integral domain with 2 elements, since there is only one group of order 2 (and there is only one way to define multiplication as $0\cdot a=0$) – Asaf Rosemarin Jul 16 at 11:21

Any finite integral domain is a field. Hence it has $$q=p^n$$ elements, for some $$n>0$$, where $$p$$ is a prime which is the characteristic of the ring/field.
In the present case, you have the field $$\mathbf F_2=\mathbf Z/2\mathbf Z$$, and by definition $$1+1=0$$.
You already got answers giving the integral domain in question, but I want to address your difficulty with defining $$1+1$$. You basically have two choices. Either $$1+1=0$$ or $$1+1=1$$. The latter case doesn't work since then there would be no additive inverse of $$1$$: Neither $$1+0$$ nor $$1+1$$ would be $$0$$, so no additive inverse. This only leaves you with $$1+1=0$$. You can then go through all the axioms and show that with this definition, your ring is actually an integral domain: $$(R,+)$$ is an Abelian group, $$\times$$ is associative, commutative and distributes over addition, and the only non-zero element $$1$$ is not a zero-divisor.
• +1. Another way to look at the $1+1 = 1$ case: if $1 + 1 = 1,$ then $1 = 1 + 1 - 1 = 1 - 1 = 0,$ so your ring is the zero ring (hence not an integral domain nor a ring with two elements). – Stahl Jul 16 at 11:49