It is a well-known fact that a unique factorization domain is a principal ideal domain, in which all ideals are principal ideals. [EDIT: I got dyslexic on this one, should've said something along the lines of "a principal ideal domain is a unique factorization domain."]
But what happens if one tries to generate a non-principal ideal in a principal ideal domain? My hunch is that it always winds up eventually leading back to the whole ring itself. Some of the "questions that may already have your answer" seem to validate my hunch, but buried under layers of abstraction, so it's entirely possible I have misunderstood them.
So to make it more concrete, I thought of trying it in $\mathbb{Z}$. I thought of $\langle 4, 7 \rangle$, which doesn't look like a principal ideal. This would consist of numbers of the form $4a + 7b$, with $\{a, b\} \in \mathbb{Z}$. But:
- $4 \times 2 + 7 \times (-1) = 1$
- $4 \times 4 + 7 \times (-2) = 2$
- $4 \times 6 + 7 \times (-3) = 3$
- $4 \times 10 + 7 \times (-5) = 5$
- $4 \times 5 + 7 \times (-2) = 6$
Clearly, with the right choices of $a$ and $b$, every $n \in \mathbb{Z}$ can be obtained and therefore $\langle 4, 7 \rangle = \langle 1 \rangle = \mathbb{Z}$.
Does this always happen in $\mathbb{Z}$? And if that's the case, might there be some other PID (e.g., $\mathbb{Z}[\sqrt{-2}]$) in which this does not hold?
In a PID, every ideal is principal. That includes every ideal generated by a pair of elements. So such an ideal $(a, b)$ must always be generated by a single element $c$, which therefore must be the greatest common divisor $\gcd(a, b)$ (up to a unit). This is one definition of $\gcd(a, b)$, in fact. In $\mathbb{Z}$ you get the $\gcd$ in the usual sense. The fact that this works is more or less Bezout's identity: for any integers $a, b$ there exist integers $x, y$ such that
$$ax + by = \gcd(a, b).$$
For example, $(2, 4) = (2)$.