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We have the following Theorem: A non-zero commutative ring is an integral domain if and only if for all $a$,$b$ $\neq 0$ $\implies ab \neq 0$.

Now, we need to prove that the Gaussian integers form an integral domain.

Proof: Let $\Bbb Z[i]$ denote the Gaussian Integers, which is a commutative ring. Take $z,w \in \Bbb Z[i]$ s.t: $z,w \neq 0$ and $z = a + ib$, $w = c + id$.

Then, $zw = (ac - bd) + (ad + bc)i \in \Bbb Z[i]$. Since the elements of $\Bbb Z[i]$ are non-zero $\implies zw \neq 0$. QED.

I am wondering if this is correct? Thanks.

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    $\begingroup$ Why should not $ac-bd$ and $ad+bc$ be $0$? Use the fact that $\Bbb Z[i]\subseteq\Bbb C$ which is a field $\endgroup$ Commented Mar 28, 2016 at 0:33
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    $\begingroup$ If the first paragraph is a theorem about integral domains, what is your definition of an integral domain? $\endgroup$
    – rschwieb
    Commented Mar 28, 2016 at 0:35
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    $\begingroup$ This is usually the definition of an integral domain, not a theorem. $\endgroup$ Commented Mar 28, 2016 at 0:36
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    $\begingroup$ Hint to what is probably the simplest proof: If $zw = 0$, then $z\overline{z} w\overline{w}= 0$ as well. But for every nonzero $u \in \mathbb{Z}\left[i\right]$, the complex number $u\overline{u}$ is actually a positive integer (why?). When is the product of two positive integers zero? $\endgroup$ Commented Mar 28, 2016 at 1:13

2 Answers 2

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So the problem with your proof is that you assume that $ad-bc\ne 0$ and $ad+bc\ne 0$. This is not supported by your argument at all. However, if you note that $\Bbb Z[i]\subseteq\Bbb C$ you can use polar coordinates, write $z=r_1e^{i\theta_1}, w=r_2e^{i\theta_2}$ with $r_1, r_2>0$. Then their product is

$$zw= r_1r_2e^{i(\theta_1+\theta_2)}$$

which has absolute value $r_1r_2$. Since $r_1,r_2$ are real numbers which are not-zero, they multiple to a non-zero value. But if $|zw|\ne 0$ clearly $zw\ne 0$.

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  • $\begingroup$ Why does $Z[i] \subseteq C$ allow use to use polar coordinates and write $z = r e^{i \theta}$? Where does the formula come from? $\endgroup$
    – Javier
    Commented Mar 28, 2016 at 0:41
  • $\begingroup$ @Javier previous classes. That's how $\Bbb C$ is defined as pairs of real numbers and using the famous $e^{i\theta}=\cos\theta+i\sin\theta$ $\endgroup$ Commented Mar 28, 2016 at 0:46
  • $\begingroup$ Ah, I see. Thank you so much! $\endgroup$
    – Javier
    Commented Mar 28, 2016 at 0:50
  • $\begingroup$ You proved that the product of two nonzero complex numbers is a nonzero complex number, and therefore the result requested by the OP follows as a special case of your proof. But is there a way to prove OP's claim by using only that $\mathbb{Z}$ is an integral domain, or perhaps another way that doesn't use the property that $\mathbb{C}$ is a field? $\endgroup$
    – Junglemath
    Commented Apr 21, 2019 at 2:40
  • $\begingroup$ @junglemath there are often plenty of ways to prove things, this is no different. You'll find answers with as many different proofs of the exact same fact quite uncommon. Mostly we try to keep things down to 1 $\endgroup$ Commented Apr 21, 2019 at 2:43
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Alternative; I'd use the fact that norm (or absolute value) of Gaussian numbers, $\mathbb Z[i]$, is multiplicative.

$$0 \not =a,b \in \mathbb Z[i] \implies |a|^2 \not = 0, |b|^2 \not=0$$ $$ |a|^2|b|^2 = |ab|^2 \not= 0 \implies ab \not=0$$

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