Abstract Algebra – Proving Gaussian Integers are an Integral Domain

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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.

Best Answer

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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