# Show that every ideal of the ring $\mathbb Z$ is principal

Let $\mathbb Z$ be the ring of integers. The question asks to show that every ideal of $\mathbb Z$ is principal. I beg someone to help me because it is a new concept to me.

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(I assume you mean $\mathbb{Z}$ the ring of integers, not only "a ring"...) An ideal is, in particular, a subgroup. Do you know what the subgroups of $\mathbb{Z}$ are? –  lentic catachresis Jan 22 '12 at 16:32
"Let Z be a ring" is wrong. The question is about the particular ring whose proper name is $\mathbb Z$, namely the ring of ordinary integers under ordinary addition and multiplication. –  Henning Makholm Jan 22 '12 at 16:32
Well one thing is if you know that $\mathbb{Z}$ is an euclidean domain then it is automatically a PID. –  user38268 Jan 22 '12 at 17:50
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): math.stackexchange.com/q/6998 –  Jonas Meyer Jan 22 '12 at 17:55

Let $I$ be an ideal of $\mathbb Z$.

Goal: $\mathbb{Z}$ is a principal ideal domain (or PID)

If $I={0}$ then $0$ generates $I$. And we are done.

Suppose $I\neq {0}$, and let $a$ be the smallest element in $I$.

Claim: $a$ generates $I$ i.e $(a)=I$

To prove my claim, clearly $a\subset I$ Since $(a)$= {$ar :r \in \mathbb Z$}, $ar\in I$

Let $b \in I$ if $b=0$ then $b=a0 \in (a)$.

If $b\neq 0$, we may assume $b>0$, by euclidean algorithm

$$b=aq+r$$ also $0\le r<a$ of course $q,r \in \mathbb Z$.

Now $r=b-aq \in I$ since $b,a \in I$. this implies $r=0$ since $r<a$ and $a$ is the smallest element in $I$.

So, $b=aq \in I$. Thus, $(a)=I$. meaning that $a$ generates I.

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"we may assume $b>0$": There is no need to assume $b>0$, or even to treat $b=0$ as a separate case. –  Jonas Meyer Jan 22 '12 at 18:12
Remark that in $\mathbb Z$ it is a bit simpler to use (repeated) subtraction, versus division (with remainder) - see my answer. This corresponds to the subtractive (vs. divisive) form of the Euclidean algorithm. –  Bill Dubuque Jan 22 '12 at 18:13
@Mwanginde vote up my answer if have you accepted it. –  Hassan Muhammad Jan 25 '12 at 17:17
@Hassan The proof has a (minor) error in that it implicitly assumes $a$ is positive. Instead, it should say: $I\ne 0$ implies $I$ has a positive element. Let $a$ be the least positive element in $I$. –  Bill Dubuque Feb 28 '12 at 21:58
@Jack Maney: Thanks for the edit. –  Hassan Muhammad Mar 1 '12 at 7:17

HINT $\$ In $\rm\:\mathbb Z\:,\:$ descent via the Division (Euclidean) algorithm has especially simple form, viz.

LEMMA $\ \$ If a nonempty set of positive integers $\rm\: M\:$ satisfies $\rm\ n > m\ \in\ M \ \Rightarrow\ \: n-m\ \in\ M$
then every element of $\rm\:M\:$ is a multiple of the least element $\rm\:m_{\:1} \in M\:.$

Proof $\ \:$ If not there is a least nonmultiple $\rm\:n\in M\:,$ contra $\rm\:n-m_{\:1} \in M\:$ is a nonmultiple of $\rm\:m_{\:1}.$

REMARK $\$ Note that the lemma depends only on the fact that $\rm M$ is discrete and closed under subtraction, so it applies much more generally, e.g. to $\:\mathbb Z$-modules $\subset \mathbb Q\:.\$ The study of these "fractional ideals" essentially go back to Euclid, who studied the application of the Euclidean algorithm to "line segments" to determine their "greatest common measure". This leads quite naturally to the study of the continued fraction expansion of a real number.

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Suppose that $I$ is an ideal in $\mathbb{Z}$. If $I=(0)$, it’s certainly principal, so assume that it contains a non-zero element. Since $I$ is a subgroup of $\mathbb{Z}$, if it contains a non-zero element, it must contain a positive element. Let $m$ be the smallest positive member of $I$. Show that $I=(m)$, the set of multiples of $m$.

HINT: Use the division algorithm and a proof by contradiction.

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If $I$ is an ideal of $\mathbb{Z}$, then, considering only the addition operator, $(I,+)$ is a subgroup of $(\mathbb{Z},+)$ (ie $I$ is an additive subgroup of $\mathbb{Z}$, when viewed as a group under addition). Since $(\mathbb{Z},+)$ is cyclic, it follows that $(I,+)$ is cyclic (if you aren't convinced that every subgroup of a cyclic group is cyclic, you should sit down and prove it).

Therefore, $(I,+)$ as a group is generated by some $n\in \mathbb{Z}$. So, for every $x\in I$, there exists $m\in \mathbb{Z}$ such that $$x=\underbrace{n+n+\cdots+n}_{m\textrm{ times}}$$ if $m>0$ or $$x=\underbrace{-n-n-\cdots-n}_{\vert m \vert \textrm{ times}}$$ if $m<0$ (and of course $x=0$ if $m=0$). Thus, $I=\{nm\,\vert\,m\in \mathbb{Z}\}$, and now you only need show that, as an ideal, $I$ is generated by $n$.

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