Gaussian Integers form Subring of Complex Numbers

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Theorem

The ring of Gaussian integers:

$\struct {\Z \sqbrk i, +, \times}$

forms a subring of the set of complex numbers $\C$.


Proof

From Complex Numbers form Field, $\C$ forms a field.

By definition, a field is a ring.

Thus it is possible to use the Subring Test.


We note that $\Z \sqbrk i$ is not empty, as (for example) $0 + 0 i \in \Z \sqbrk i$.


Let $a + b i, c + d i \in \Z \sqbrk i$.

Then we have $-\paren {c + d i} = -c - d i$, and so:

\(\displaystyle \paren {a + b i} + \paren {-\paren {c + d i} }\) \(=\) \(\displaystyle \paren {a + b i} + \paren {- c - d i}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {a + \paren {-c} } + \paren {b + \paren {-d} } i\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {a - c} + \paren {b - d} i\)

We have that $a, b, c, d \in \Z$ and $\Z$ is an integral domain, therefore by definition a ring.

So it follows that $a - c \in \Z$ and $b - d \in \Z$, and hence:

$\paren {a - c} + \paren {b - d} i \in \Z \sqbrk i$


Now consider $\paren {a + b i} \paren {c + d i}$.

By the definition of complex multiplication, we have:

$\paren {a + b i} \paren {c + d i} = \paren {a c - b d} + \paren {a d + b c} i$


As $a, b, c, d \in \Z$ and $\Z$ is a ring, it follows that:

$a c - b d \in \Z$ and $ad + bc \in \Z$

So:

$\paren {a + b i} \paren {c + d i} \in \Z \sqbrk i$


So by the Subring Test, $\Z \sqbrk i$ is a subring of $\C$.

$\blacksquare$


Also see


Sources