Multiple of Divisor in Integral Domain Divides Multiple
Jump to navigation
Jump to search
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 $b \times c$.
Corollary
Let $a, b, c \in \Z$.
Let:
- $a \divides b$
where $\divides$ denotes divisibility.
Then:
- $a c \divides b c$
Proof
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.
$\blacksquare$
Sources
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $6$: Polynomials and Euclidean Rings: $\S 26$. Divisibility: Theorem $49 \ \text{(ii)}$