Prove that the cancellation law of multiplication is equivalent in a commutative ring to the assertion that the product of non-zero factors is not zero

To prove that the cancellation law of multiplication is equivalent to the assertion that the product of non-zero factors is not zero in a commutative ring, we can approach it by proving both directions.

1. Cancellation Law ⇒ Non-zero Factors:
Assume that the cancellation law of multiplication holds in a commutative ring. Let's prove that in this case, the product of non-zero factors is not zero.

Suppose we have two non-zero elements, a and b, in the commutative ring. Assume, by contradiction, that their product is zero, i.e., ab = 0. Since the commutative ring satisfies the cancellation law, we can multiply both sides of this equation by the inverse of a, which exists because a is non-zero. This yields a^(-1)ab = a^(-1)0. Simplifying the left side gives b = 0.

However, this contradicts our assumption that b is non-zero. Thus, we conclude that the product of non-zero factors cannot be zero in a commutative ring.

2. Non-zero Factors ⇒ Cancellation Law:
Assume that the product of non-zero factors is not zero in a commutative ring. Let's prove that in this case, the cancellation law of multiplication holds.

Consider three elements, a, b, and c, in the commutative ring, where a ≠ 0. Assume that ab = ac. We aim to show that b = c.

Since a ≠ 0, we know that aa^(-1) = 1, where a^(-1) is the inverse of a. We can multiply both sides of the given equation by a^(-1) on the left, yielding (a^(-1)ab) = (a^(-1)ac). By associativity of multiplication, this simplifies to (a^(-1)a)b = (a^(-1)a)c. Since (a^(-1)a) equals 1, it becomes 1b = 1c.

By the definition of the multiplicative identity, 1b = b and 1c = c. Therefore, b = c. This shows that the cancellation law of multiplication holds in the commutative ring.

Thus, we have proven that the cancellation law of multiplication is equivalent to the assertion that the product of non-zero factors is not zero in a commutative ring.