I know that principal ideals are ideals which are generated by a single element, but I have no clue how to show that $\mathbb Z: I= \langle 14, 21 \rangle$ is a principal ideal.
Could anyone please give me a rough idea or a hint in the right direction.
$21-14=7\in I$, implies that $7\mathbb{Z}\subset I$, on the other hand $14a+21b=7(2a+3b)$ implies that $I\subset 7\mathbb{Z}$ we deduce that $I=7\mathbb{Z}$.
Let $a \in \mathbb Z_+$ be the smallest element of $Z_{+}$ belonging to $\langle 14,21\rangle$.
Then, note that $a$ must generate $\langle 14,21 \rangle$.
(why? this is a standard argument : if $a$ did not, then there is some $b$ not a multiple of $a$ which is inside the ideal , and you can show that the remainder when $b$ is divided by $a$ also belongs in the ideal, which would be smaller than $a$ but not zero, contradicting the choice of $a$. This is how one shows that every ideal of $\mathbb Z$ is principal).
Show that $7 \in \langle 14,21\rangle$.
Show that if $a \in \langle 14,21 \rangle$ then $a$ is a multiple of $7$. Therefore, $7$ is the smallest positive element in the ideal, and we are done.
