This seems like a pretty simplistic question, but I can't find a solid, non-ambiguous answer to it. The question I'm given:
Is $I$ a principal ideal of $R$? Given: $R=\mathbb{Z}$ and $I=\left\langle 6,10\right\rangle_{R}$
$I$ is clearly an ideal, however for it to be principal, there must be a $a \epsilon R$ such that $I=\left\langle a\right\rangle = \left\{ ra|r,a\epsilon R\right\}$
Now, since any element of the ideal has the form $6r+10s$ where $r,s \epsilon R$ clearly the ideal could be generated by {2}, since all the elements in the ideal will be even.
My question: is the ideal principal if it contains all elements generated by $a$? Or do all elements of a principal ideal need to be able to be generated by a single element - in which case the ideal would form a subset of the generated group. In our example, all elements of the ideal area generated by $2$, but not all elements generated by $2$ are in the ideal. Most sources that I've found don't particularly specify, as they seem to assume something that I'm missing (apparently).
Thanks!
In your example, there is equality: $\langle 6, 10 \rangle = \langle 2 \rangle$. Using the Euclidean algorithm, we know we can find $s, r \in \mathbb{Z}$ such that $6r + 10s = 2$. From here, you can just multiply the entire thing through by any $x \in \mathbb{Z}$:
$$6xr + 10xs = 2x$$
By definition, for any (two-sided) principal ideal:
$$\langle x_1, x_2, ..., x_n \rangle = \langle y \rangle = \{r \cdot y:r \in R\} = \{y \cdot r:r \in R\}$$
for some $y \in R$. In other words, this will indeed be all elements generated by $y$.
Side note: $\mathbb{Z}$ is a principal ideal domain - i.e. all ideals are principal. A useful fact is that any ring endowed with some form of the Euclidean algorithm as a means of finding GCD's and applying Bezout's identity will be a principal ideal domain.