Multiple of Divisor in Integral Domain Divides Multiple

Theorem

Let $\struct {D, +, \times}$ be an integral domain.

Let $a, b, c \in D$.

Let $a \divides b$, where $\divides$ denotes divisibility.

Then $a \times c$ is a divisor of $b \times c$.

Let $a, b, c \in \Z$.

Let:

$a \divides b$

where $\divides$ denotes divisibility.

Then:

$a c \divides b c$

By definition, if $a \divides b$ then $\exists d \in D: a \times d = b$.

Then $\paren {a \times d} \times c = b \times c$, that is:

$\paren {a \times c} \times d = b \times c$

which follows because $\times$ is commutative and associative in an integral domain.

Hence the result.

$\blacksquare$

1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $6$: Polynomials and Euclidean Rings: $\S 26$. Divisibility: Theorem $49 \ \text{(ii)}$