1
$\begingroup$

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?

Improve this question
$\endgroup$
  • 1
    $\begingroup$ That ring is contained in the field of algebraic numbers, so it is commutative, and the unity is obvious $\endgroup$ – J. W. Tanner Jul 10 '19 at 14:43
  • $\begingroup$ 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. $\endgroup$ – Fareed Abi Farraj Jul 10 '19 at 14:47
  • 1
    $\begingroup$ 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.) $\endgroup$ – user247327 Jul 10 '19 at 15:05
1
$\begingroup$

To prove that it is an integral domain, you would indeed have to prove all three. That it has a unity and is commutative is quite obvious, but that it has no zero-divisors may not be immediately obvious.

Fortunately, all three follow from the fact that it is a subring of the real numbers, which is an integral domain.

Improve this answer
$\endgroup$
  • $\begingroup$ Anyone care to explain the dowvote? $\endgroup$ – Servaes Jul 12 '19 at 19:31
2
$\begingroup$

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.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.