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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.

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    $\begingroup$ 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 $0\cdot a=0$) $\endgroup$ – Asaf Rosemarin Jul 16 at 11:21
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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 $1+1=0$.

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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.

In fact, this ring is even a field, since the only non-zero element has itself as a multiplicative inverse.

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    $\begingroup$ +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). $\endgroup$ – Stahl Jul 16 at 11:49

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