Sum of Integer Ideals is Greatest Common Divisor
From ProofWiki
Theorem
Let $\left({m}\right)$ and $\left({n}\right)$ be ideals of the integers $\Z$.
Let $\left({d}\right) = \left({m}\right) + \left({n}\right)$.
Then $d = \gcd \left\{{m, n}\right\}$.
Proof
By Sum of Ideals is an Ideal we have that $\left({d}\right) = \left({m}\right) + \left({n}\right)$ is an ideal of $\Z$.
By Ring of Integers is Principal Ideal Domain we have that $\left({m}\right)$, $\left({n}\right)$ and $\left({d}\right)$ are all necessarily principal ideals.
By Subrings of the Integers we have that:
- $\left({m}\right) = m \Z, \left({n}\right) = n \Z$
Thus:
- $\left({d}\right) = \left({m}\right) + \left({n}\right) = \left\{{x \in \Z: \exists a, b \in \Z: x = a m + b n}\right\}$
That is, $\left({d}\right)$ is the set of all integer combinations of $m$ and $n$.
The result follows by Bézout's Lemma.
$\blacksquare$
Sources
- C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 5.21$: Example $38$