---
tag: 2SQBC82310
kind: proof
name: Proof of cancellation in an integral domain
msc: [13G05]
proves: 0GJM24EXP9
uses: [09S3PFBEPM, 0GYGHKX834, 0894J980RK]
sources:
  - url: https://en.wikipedia.org/wiki/Integral_domain
    locator: Properties
---

## Proof

1. Let $R$ be an [[integral domain|09S3PFBEPM]], and let $a, b, c \in R$ with $a \neq 0$ and $a b = a c$.
2. Subtract $a c$ from both sides. This gives $a b - a c = 0$.
3. By the distributive law of the [[ring|0894J980RK]] $R$, $a (b - c) + a c = a (b - c + c) = a b$. So $a (b - c) = a b - a c = 0$.
4. In an integral domain, a product is $0$ only if one factor is $0$. Since $a \neq 0$, we get $b - c = 0$.
5. Adding $c$ to both sides gives $b = c$.

## Idea

Move everything to one side and factor out $a$. The product $a (b - c)$ is zero, and an integral domain has no zero divisors. So the factor $b - c$ must vanish, even though $a$ need not have an inverse.