Gaussian Integers form Subring of Complex Numbers

Theorem

The ring of Gaussian integers:

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

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

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:

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

$=$

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

$\ds $

$=$

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

$\ds $

$=$

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

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 $\C$.

$\blacksquare$

1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $5$: Rings: $\S 19$. Subrings: Example $33$
1970: B. Hartley and T.O. Hawkes: Rings, Modules and Linear Algebra ... (previous) ... (next): Chapter $1$: Rings - Definitions and Examples: $2$: Some examples of rings: Ring Example $5$
1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $9$: Rings: Exercise $3$