Exercise: Show that every ideal $I$ of $\mathbb{Z}$ is principal.
Attempt: Since $I$ is principal, it can be generated by one element. Also, my tutor said that if $I \subset \mathbb{Z}$ is an ideal and we consider any $a \in I-\{ 0 \}$ with $|a|$ minimal, we need to show that $a$ generates $I$ (which I have trouble understanding). Thank you.
Are you familiar with the Division Theorem, which says that given any integers $c$ and $d$, $d\ne0$, there exist (unique) integers $q$ and $r$ with $c=dq+r$ and $0\le r\lt|d|$? If so, take $c$ to be an arbitrary element of the ideal $I$, take $d$ to be the smallest (in the sense of absolute value) nonzero element of $I$, and think about what happens to $r$.
What your tutor means is that if $I \subseteq \mathbb{Z}$ is an ideal, then if $a$ is the smallest positive element in $I$, $I = (a)$. The inclusion $(a) \subseteq I$ is clear. The other inclusion can be shown via the Euclidean algorithm. The solution is below.
Let $x \in I$. We may assume WLOG that $x$ is positive. By the Euclidean algorithm, there exists integers $q$ and $0 \le r < a$ such that $x = aq + r$. Then $r = x - aq \in I$. Since we chose $a$ to be the smallest positive element of $I$, $r = 0$, so $x = aq \in (a)$.
