Every ideal of Z is a principal ideal

Dependencies:

1. Principal ideal
2. Integer Division Theorem

Let $I$ be an ideal of $\mathbb{Z}$. Then $I$ is a principal ideal, i.e. $I$ is of the form $n\mathbb{Z}$ for some $n \in \mathbb{Z}$.

Proof

Let $n$ be the smallest positive element of $I$. Let $a \in I$.

By the integer division theorem, $a = qn + r$, where $0 \le r \lt n$. $r = a - qn \in I$ by absorption and commutativity. Since $r \in I$, $r < n$ and $n$ is the smallest positive element of $I$, $r = 0$.

Therefore, every element of $I$ is a multiple of $n$. Also, all multiples of $n$ are in $I$ by the absorption property.

Therefore, $I = n\mathbb{Z}$.

Dependency for: None

Info:

• Depth: 6
• Number of transitive dependencies: 11

Transitive dependencies:

1. Group
2. Ring
3. Ideal
4. Identity of a group is unique
5. Subgroup
6. Inverse of a group element is unique
7. Conditions for a subset to be a subgroup
8. Condition for a subset to be a subgroup
9. Conditions for a subset of a ring to be a subring
10. Principal ideal
11. Integer Division Theorem