Prove that every ideal of $ \mathbb Z_{1000}$ is principal. I just want to check I'm doing this right:
Let $1000$ be the smallest element in the ideal $I$.
Then, we have that $m \in I$ such that $m = q \cdot 1000 + r$ for $0 \leq r \leq 1000$.
We get that $m - q \cdot 1000 \in I$, such that  $r \equiv 0$. 
Therefore, $I = \langle 1000 \rangle$.
Is this valid? Did I missed something? 
 A: Let $I$ be an ideal of $\mathbb{Z}_{1000}.$ Since $\mathbb{Z}_{1000}$ is finite, there is a finite set of generators $a_1,\ldots,a_n$ of $I.$ Let $d$ be the greatest common divisor of the $a_i.$ Then show that $I=\langle d\rangle.$
Español:
Sea $I$ un ideal de $\mathbb{Z}_{1000}.$ Como $\mathbb{Z}_{1000}$ es finito, $I$ tiene un número finito de generadores $a_1,\ldots,a_n.$ Sea $d$ el máximo común divisor de los $a_i.$ Entonces (se debe probar que) $I=\langle d\rangle.$
A: The ideals of $\mathbb{Z}_{1000}$ correspond to the ideals of $\mathbb{Z}$ that contain $1000\mathbb{Z}$. These ideals are all principal.
It is a general fact that $R$ and $S$ are commutative rings and $\phi: R \to S$ is a ring homomorphism and $R$ is a principal ideal ring, then so is $S'=\phi(R)$. Indeed, every ideal of $S'$ is of the form $\phi(I)$ for $I$ an ideal of $R$. If $I=(a)=aR$, then $\phi(I)=\phi(a)\phi(R)=\phi(a)S'=(\phi(a))$.
A: A somewhat more abstact and general approach, based upon the following
Proposition:  Let $R$ be a PIR; then every homomorphic image of $R$ is also a PIR.  End of Proposition.
Here, PIR stands for Principle Ideal Ring; that is, a commutative, unital ring in which every ideal is principle, that is, of the form $\langle d \rangle = Rd$, where $d \in R$.  
Proof of Proposition:  Let $\phi:R \to S$ be a homomorphism such that $S = \phi(R)$; that is, $\phi$ is surjective.  Let $J \subset S$ be an ideal.
It is both well-known, and easy to see, that $I = \phi^{-1}(J) \subset R$ is an ideal as well.  Indeed, if $a, b \in I$, then $\phi(a), \phi(b) \in J$, whence
$\phi(a - b) = \phi(a) - \phi(b) \in J, \tag{1}$
whence
$a - b \in \phi^{-1}(J) = I; \tag{2}$
and if $i \in I = \phi^{-1}(J)$ and $r \in R$, then
$\phi(ri) = \phi(r)\phi(i) \in J, \tag{3}$
since $\phi(i) \in J$, an ideal in $S$, whence
$ri \in I; \tag{4}$
(2) and (4) show that $I = \phi^{-1}(J) \subset R$ is an ideal.  Since $R$ is a PIR, we have
$I = \langle d \rangle = Rd \tag{5}$
for some $d \in R$; then for any $j \in J$ there is $i \in I$ such that $\phi(i) = j$; but such $i = rd$ for some $r \in R$; so
$j = \phi(i) = \phi(rd) = \phi(r) \phi(d), \tag{6}$
which shows every $j \in J$ is of the form $s\phi(d)$ for some $s = \phi(r) \in S$; thus
$J \subset \langle \phi(d) \rangle; \tag{7}$
furthermore, since $\phi:R \to S$ is surjective, every $s \in S$ satisfies $s = \phi(r)$ for some $r \in R$, whence
$s\phi(d) = \phi(r)\phi(d) = \phi(rd) \in J, \tag{8}$
since $rd \in I$.  This shows that
$\langle \phi(d) \rangle = S\phi(d) \subset J; \tag{9}$
combining (7) and (9) yields
$J = \langle \phi(d) \rangle = S\phi(d); \tag{10}$
that is, the ideal $J$ is principle in $S$; since this argument applies to any ideal $J \subset S$, we see that $S$ is a PIR.  End:  Proof of Proposition.
Since for any $n \in \Bbb N$ the natural homomorphism
$\theta: \Bbb Z \to \Bbb Z_n \tag{11}$
given by
$\theta(i) = i + \langle n \rangle \tag{12}$
is surjective, it follows that $\Bbb Z_n$ is PIR; in partiular, $Z_{1000}$ is a PIR; every ideal in $\Bbb Z_{1000}$ is principal.
