# How to prove a set is an integral domain?

Let's say I have the ring: $$\mathbb{Z}[\sqrt{2}]=\{a+b\sqrt{2}\mid a,b\in\mathbb{Z}\}$$.

Now the question asked is to prove whether or not this ring is an integral domain.

By definition: "An integral domain is a commutative ring with unity and no zero-divisor"

From this I would try to prove three things:

1. ring is commutative
2. ring has a unity
3. ring has no zero-divisors

Yet on multiple occasions and multiple examples only the last one (no zero-divisor is proven), why is this the case? Is there no need to prove the other two?

• That ring is contained in the field of algebraic numbers, so it is commutative, and the unity is obvious – J. W. Tanner Jul 10 at 14:43
• I think when the first 2 are left not proven, then either it is in the given, or it is considered trivial and easy. But if 1. or 2. is not satisfied then the ring will NOT be an integral domain. – Fareed Abi Farraj Jul 10 at 14:47
• By the way, you want to prove statements or theorems, not "proof" them. "Proof" in this sense is a noun, not a verb. (I had to say "in this sense" because you can "proof" bread or whiskey, etc.) – user247327 Jul 10 at 15:05

These "multiple occasions" may be implicitly applying the subring test, i.e. $$\,\Bbb Z[\sqrt 2] \subseteq \Bbb R\,$$ contains $$\,1_{\Bbb R}$$ and is easily verified to be closed under subtraction and multiplication hence it is a subring of $$\,\Bbb R.\,$$ Furthermore, it is clear that nontrivial subrings of domains remain domains (since the inference$$\,a,b\neq0\,\Rightarrow\, ab\neq 0$$ necessarily remains true in every subring containing $$\,a,b)$$
Remark  In fact, by the general definition of ring adjunction, $$\,\Bbb Z[\sqrt 2]\subseteq\Bbb R$$ is the intersection of all subrings of $$\Bbb R$$ containing $$\,\Bbb Z\,$$ and $$\,\sqrt 2\,$$ so it is a domain, being a nontrivial intersection of domains.
Such inferences are quite common in algebra, because we define (minimally) "generated" structures via such intersections, and the (universal) logical form of the axioms of many algebraic structures makes it immediately clear that they are closed under intersections and subalgebras (e.g. the $$\rm\color{#c00}{universal}$$ form ring axiom $$\,\color{#c00}{\forall x,y:}\ x+y = y+x).\,$$ This is made more precise when studies the relationship between syntax and semantics in universal algebra and model theory.