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There was a question on showing that every subgroup of a cyclic group is cyclic which might help (although largely focused on the finite case): – Jonas Meyer Jan 22 '12 at 17:55
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HINT: Use the division algorithm and a proof by contradiction.

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$\textbf{Hint-}$Every ideal is an additive subgroup and every subgroup of cyclic group is cyclic. I hope this will help.

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Let $I \subset \mathbb{Z}$ be an ideal; we want to show that $I$ is generated by only one element of $I$, that is, it is a principal. If $I=0$, then $I= \langle 0 \rangle$. Let us assume then that $I \ne 0$, so there is a smallest positive number on $I$. We call this number $n$ and claim that $I = \langle n \rangle $. Let us pick $m \in I$ and divide it by $n$ with remainder. The division theorem (a proof of Euclid's the division theorem is found in the link below: states that there should be a quotient $q$ and a remainder $r$ such that $m = q n + r$, with $0 \le r < n$. We proof that $r=0$. We have that $r=m - q n \in I$, and since we assumed that $n$ is the smallest positive in $I$, then $r=0$. This means that every number on $I$ is of the form $q n$, and the set $\mathbb{Z}$ is PID.

We need to be careful here. Assume that we have the following ideal: \begin{equation} I_{23} = \langle \{2, 3 \} \rangle = \{ 2 n + 3m : n, m \in \mathbb{Z} \}. \end{equation} That is the ideal consists of all the multiples of $2$ together with the multiples of $3$, and their linear combinations. Then by choosing $n=0$ we get all multiples of $3$, and by choosing $m=0$ we get all multiples of $2$. We could say that $2$ is the smallest positive integer in the ideal $I_{23}$, but then that $3$ being in $I_{23}$ can not be obtained from $2$ with an integer multiplication, so the proof above is wrong. In fact that is not the case. We let you to prove that $I_{23}=\mathbb{Z}$, and that in general $\langle \{ p, q \} \rangle$ is equal to $\langle \{ o \} \rangle$, where $o=\text{gcd}(p,q)$. More generally, for ideals in $\mathbb{Z}$, $\langle \{ p_1, \cdots, p_k \} \rangle = \langle \{ o \} \rangle$ with $o=\text{gcd}(p_1, \cdots, p_k)$.

The same argument used here can be applied to polynomials in one variable $\mathbb{K}[x]$, over the field $\mathbb{K}$. For polynomials in two or more variables the property breaks.

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protected by user26857 Jun 25 '15 at 12:08


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