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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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    $\begingroup$ (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? $\endgroup$ – Bruno Stonek Jan 22 '12 at 16:32
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    $\begingroup$ "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. $\endgroup$ – hmakholm left over Monica Jan 22 '12 at 16:32
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    $\begingroup$ Well one thing is if you know that $\mathbb{Z}$ is an euclidean domain then it is automatically a PID. $\endgroup$ – user38268 Jan 22 '12 at 17:50
  • $\begingroup$ 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 $\endgroup$ – Jonas Meyer Jan 22 '12 at 17:55
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Goal: show that $\mathbb{Z}$ is a principal ideal domain (or PID).

Let $I$ be an ideal of $\mathbb Z$. If $I={0}$ then $0$ generates $I$. And we are done.

Suppose $I\neq {0}$, and let $a$ be the smallest positive 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 $b>0$, and by the euclidean algorithm we have

$$b=aq+r.$$ Moreover, $0\le r<a$, and 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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    $\begingroup$ "we may assume $b>0$": There is no need to assume $b>0$, or even to treat $b=0$ as a separate case. $\endgroup$ – Jonas Meyer Jan 22 '12 at 18:12
  • $\begingroup$ 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. $\endgroup$ – Bill Dubuque Jan 22 '12 at 18:13
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    $\begingroup$ @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 $I$. $\endgroup$ – Bill Dubuque Feb 28 '12 at 21:58
  • $\begingroup$ @Jack Maney: Thanks for the edit. $\endgroup$ – Hassan Muhammad Mar 1 '12 at 7:17
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    $\begingroup$ @HassanMuhammad "let $a$ be the smallest element in $I$" without a further explanation it is not clear that in $I$, a subset of $\mathbb Z$, a smallest element exists. $\endgroup$ – Sergey Zykov Dec 24 '15 at 16:43
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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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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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$\textbf{Hint-}$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: http://www.math.fsu.edu/~pkirby/mad2104/SlideShow/s5_1.pdf 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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  • $\begingroup$ Explanation with an example is really helpful. $\endgroup$ – Sejwal Mar 18 '16 at 6:28
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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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Let $d$ be the smallest integer in ideal $I \subseteq \mathbb{Z}$. Assume for contradiction $\exists a \in I$ so that $a$ is not divisible by $d$. Then we can represent $a$ as $a = qd + b$, for $q \in \mathbb{Z}$ and $ 0 <b < d$. As $d$ is the smallest integer in $I$ and $d > b$, we know that $b \notin I$. As I is an ideal, we know that $qd \in I$. Therefore we have \begin{align} a = qd + b \in I \\ qd \in I \\ b \notin I\\ \end{align} Because I is an ideal, we know that $I$ is a group under addition, which implies $-qd \in I$, and $a - qd = b \in I$, which contradicts the assumption that $b \notin I$.

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