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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?

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  • $\begingroup$ Don't you mean $I$ an ideal of $\mathbb{Z}$ (Not $\mathbb{Z}_{1000}$)? $\endgroup$ – positrón0802 Jul 26 '17 at 21:03
  • $\begingroup$ Well, the exercise says exactly that.. not sure what you mean. It's in spanish, "Demuestre que todo ideal de$ \mathbb Z_{1000}$es principal". $\endgroup$ – user2371916 Jul 26 '17 at 21:06
  • $\begingroup$ En ese caso la solución no está bien. Ésta solución sirve si estuviéramos trabajando en $\mathbb{Z}$: se toma el menor entero positivo en $I$ y se demuestra que ese entero genera $I.$ En el caso de $\mathbb{Z}_{1000}$ el mismo argumento no sirve. $\endgroup$ – positrón0802 Jul 26 '17 at 21:12
  • $\begingroup$ I see.. damn. Thank you, anyways. I'll keep trying other approaches then. $\endgroup$ – user2371916 Jul 26 '17 at 21:16
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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.

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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) $I=\langle d\rangle.$

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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 $\phi(I)=\phi(a)\phi(R)=\phi(a)S'=(\phi(a))$.

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  • $\begingroup$ is this enough to answer my question? I keep reading that every ideal of Z is principal, but I still don't see it clearly. $\endgroup$ – user2371916 Jul 26 '17 at 22:51
  • $\begingroup$ @chrisoname any ideal in $\Bbb Z$ is generated by the least positive integer in that ideal. This is really just the division algorithm. $\endgroup$ – PVAL-inactive Jul 26 '17 at 23:47

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