Integral Domains & Fields

No zero divisors; ℤ/nℤ a field if and only if n prime.

The idea

Integral domains and fields are the commutative rings in which the familiar arithmetic of $\mathbb{Z}$ and of $\mathbb{Q}$ survives. In a general ring a product can vanish without either factor vanishing: in $\mathbb{Z}/4\mathbb{Z}$, $2 \cdot 2 = 0$. Where that happens, cancellation fails — $2 \cdot 1 = 2 \cdot 3$ in $\mathbb{Z}/4\mathbb{Z}$, yet $1 \ne 3$. A nonzero element whose product with some other nonzero element is $0$ is a zero divisor.

An integral domain is a commutative ring with $1 \ne 0$ and no zero divisors: $ab = 0$ forces $a = 0$ or $b = 0$. Equivalently, we may cancel any nonzero factor, since $ax = ay$ with $a \ne 0$ gives $a(x - y) = 0$, so $x = y$.

A field is a commutative ring with $1 \ne 0$ in which every nonzero $a$ has a multiplicative inverse $a^{-1}$, so we can divide by anything nonzero. Every field is an integral domain: if $ab = 0$ and $a \ne 0$, multiplying by $a^{-1}$ gives $b = 0$. In a domain we can cancel; in a field we can also divide.

The two multiplication tables below display the contrast: $\mathbb{Z}/4\mathbb{Z}$ carries a $0$ in the interior of its table, at $2 \cdot 2$, while in the field $\mathbb{Z}/5\mathbb{Z}$ every $0$ sits on the border, where one factor is itself $0$.

Ways to work on it

• Walkthrough. Zero divisors, fields, and inverses mod a prime.
• Practice. Invert an element mod a prime.
• Hardest. Decide which ℤ/nℤ is a field.

Not sure where to start? Take the ten-question placement test.