What Is a Quotient Ring of the Gaussian Integers?

The Gaussian integers are the complex numbers with integer real and imaginary parts: $\mathbb{Z}[i]=\{a+bi:a,b\in\mathbb{Z}\}$, where $i^2=-1$. A quotient ring is formed by declaring certain elements equivalent, but the result depends on which ideal you quotient by. Thus, “the quotient ring of the Gaussian integers” is not one specific ring until the ideal is given.

Because $\mathbb{Z}[i]$ is a principal ideal domain, every ideal is generated by one Gaussian integer $\alpha$. The usual quotient therefore has the form $\mathbb{Z}[i]/(\alpha)$.

What the quotient means

The ideal $(\alpha)$ consists of all Gaussian-integer multiples of $\alpha$:

$$(\alpha)=\{\alpha\beta:\beta\in\mathbb{Z}[i]\}.$$

In the quotient $\mathbb{Z}[i]/(\alpha)$, two Gaussian integers $z$ and $w$ represent the same element precisely when their difference belongs to this ideal:

$$z\sim w \quad\Longleftrightarrow\quad z-w\in(\alpha),$$

or, equivalently, when $\alpha$ divides $z-w$ in $\mathbb{Z}[i]$. The elements of the quotient are these equivalence classes, written $z+(\alpha)$. Addition and multiplication are performed using representatives: $(z+(\alpha))+(w+(\alpha))=(z+w)+(\alpha)$, and likewise for multiplication.

This is the Gaussian-integer version of reducing ordinary integers modulo $n$. In $\mathbb{Z}/n\mathbb{Z}$, two integers are equivalent when their difference is divisible by $n$; here the divisor can be a complex number with integer coordinates.

How many elements are in the quotient?

For $\alpha=a+bi\ne0$, define its norm by

$$N(\alpha)=a^2+b^2.$$

Then $\mathbb{Z}[i]/(\alpha)$ has exactly $N(\alpha)$ elements. One way to see the count is to view multiplication by $a+bi$ as a linear transformation on the integer-coordinate lattice. It sends $(x,y)$ to $(ax-by,bx+ay)$, whose matrix has determinant $a^2+b^2$. The absolute determinant is the index of the image lattice, which is the number of distinct quotient classes.

For example, $N(2+i)=2^2+1^2=5$, so $\mathbb{Z}[i]/(2+i)$ has five elements. By contrast, $\mathbb{Z}[i]/(2)$ has $N(2)=4$ elements.

This count assumes $\alpha\ne0$. If $\alpha=0$, the quotient by $(0)$ is $\mathbb{Z}[i]$ itself and is infinite. If $\alpha$ is a unit, then $(\alpha)=\mathbb{Z}[i]$, so the quotient has just one element.

Worked example: quotient by (2)

In $\mathbb{Z}[i]/(2)$, coefficients are reduced modulo 2. Every class has one of the four representatives $0$, $1$, $i$, or $1+i$. Also, in this quotient, $i^2=-1\equiv1$ because $-1\equiv1\pmod 2$.

Equivalently, write the ring as $\mathbb{F}_2[x]/(x^2+1)$, identifying $x$ with $i$. In the field $\mathbb{F}_2$, $x^2+1=(x+1)^2$, so this is not a field: it contains the nonzero element $1+i$ whose square is zero:

$$(1+i)^2=1+2i+i^2=0\quad\text{in }\mathbb{Z}[i]/(2).$$

A nonzero element whose power is zero is called nilpotent. Its presence shows that the quotient has zero divisors and cannot be a field.

Worked example: quotient by (2 + i)

Now consider $\mathbb{Z}[i]/(2+i)$. In the quotient, $2+i=0$, so $i=-2$. Since the quotient has five elements, calculations can be made in $\mathbb{F}_5$. The map

$$a+bi\longmapsto a-2b\pmod 5$$

respects addition and multiplication and sends $i$ to $-2$. This is consistent with $i^2=-1$, since $(-2)^2=4\equiv-1\pmod 5$. It gives an isomorphism $\mathbb{Z}[i]/(2+i)\cong\mathbb{F}_5$, so this quotient is a field.

When is the quotient a field?

For a commutative ring, the quotient by an ideal is a field exactly when that ideal is maximal. In $\mathbb{Z}[i]$, every ideal is principal, so $\mathbb{Z}[i]/(\alpha)$ is a field exactly when $\alpha$ is a Gaussian prime (a nonzero nonunit that is irreducible).

A useful caution: a prime norm is enough to show that $\alpha$ is a Gaussian prime, but it is not necessary. For instance, $3$ is a Gaussian prime even though $N(3)=9$. Up to multiplication by units, Gaussian primes are $1+i$, rational primes $p$ with $p\equiv3\pmod4$, and numbers $a+bi$ whose norm is a rational prime. This classification can help decide whether a particular quotient is a field.

Common mistakes

• Leaving out the ideal. The notation $\mathbb{Z}[i]/I$ needs a specified ideal $I$; different choices give different rings.
• Treating the modulus like an ordinary integer. Divisibility is in $\mathbb{Z}[i]$, and the modulus may have both real and imaginary parts.
• Assuming every finite quotient is a field. A finite quotient can have zero divisors, as $\mathbb{Z}[i]/(2)$ demonstrates.
• Using the norm as the field test. The norm counts elements when the generator is nonzero. A prime norm guarantees a field, but composite norm alone does not rule one out; check whether the generator is a Gaussian prime.