Finite Set of Elements in Principal Ideal Domain has GCD
Theorem
Let $\struct {D, +, \circ}$ be a principal ideal domain.
Let $a_1, a_2, \dotsc, a_n$ be non-zero elements of $D$.
Then $a_1, a_2, \dotsc, a_n$ all have a greatest common divisor.
Proof
Let $0_D$ and $1_D$ be the zero and unity respectively of $D$.
Let $J$ be the set of all linear combinations in $D$ of $\set {a_1, a_2, \dotsc, a_n}$.
From Set of Linear Combinations of Finite Set of Elements of Principal Ideal Domain is Principal Ideal:
- $J = \ideal x$
for some $x \in D$, where $\ideal x$ denotes the principal ideal generated by $x$.
We have that each $a_i$ can be expressed as a linear combination of $\set {a_1, a_2, \dotsc, a_n}$:
- $a_i = 0_D a_1 + 0_D a_2 + \dotsb + 1_D a_i + \dotsb + 0_D a_n$
Thus:
- $\forall i \in \set {0, 1, \dotsc, n}: a_i \in J$
and so by definition of $J$:
- $\forall i \in \set {0, 1, \dotsc, n}: a_i = t_i x$
for some $t_i \in D$.
Thus $x$ is a common divisor of $a_1, a_2, \dotsc, a_n$.
As $x \in \ideal x = J$, we have:
- $x = c_1 a_1 + c_2 a_2 + \dotsb + c_n a_n$
for some $c_1, c_2, \dotsc, c_n \in D$.
Thus every common divisor of $a_1, a_2, \dotsc, a_n$ also is a divisor of $x$.
Thus $x$ is a greatest common divisor of $a_1, a_2, \dotsc, a_n$.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 62.4$ Factorization in an integral domain: $\text{(i)}$