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Let $\overline{\mathbb{Z}}$ be the ring of algebraic integers. Since $n\overline{\mathbb{Z}}$ is an ideal of $\overline{\mathbb{Z}}$, we can form the quotient ring $\overline{\mathbb{Z}}/n\overline{\mathbb{Z}}$. I want to prove that it is infinite.

The exercise has a hint to consider $x_i=n^{1/{2^i}}$, for all $i\geq 1$. However it is not clear to me why would someone think about these numbers nor why they are distinct in the quotient ring.

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  • $\begingroup$ Well, these are distinct, because $x_i^{(2^j)}=0$ in the quotient iff $j\ge i$ (for big enough $i$, at least). $\endgroup$ – Berci Oct 20 '19 at 23:18
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    $\begingroup$ @Berci could you explain it a little further? This isn't clear to me $\endgroup$ – Gabriel Oct 21 '19 at 6:07
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I don't know about the idea behind the hint (it may be something interesting), but I would choose a simpler set, namely $$S=\{\sqrt{p}\mid p\text{ is a prime natural number}\}.$$ First of all each of elements of $S$ is inside the integral closure of $\mathbb{Z}$ in $\mathbb{C}$ which is also called the ring of algebraic integers, because they are satisfying a monic polynomial with integer coefficients, $f_p(x)=x^2-p\in\mathbb{Z}[x]$. Now I show that for each pair of distinct prime numbers $p$ and $q$, we have $\bar{p}$ and $\bar{q}$ are distinct classes in the ring $\frac{\bar{\mathbb{Z}}}{n\bar{\mathbb{Z}}}$ when $n$ is an integer not equal to $\pm 1$ or 0. To show that two classes are different, you should show that the difference of their representative elements doesn't belong to the ideal. If $\sqrt{p}-\sqrt{q}\in n\bar{\mathbb{Z}}$, then there should exist an algebraic integer $a$ such that $na=\sqrt{p}-\sqrt{q}$. Now the trick is to play with the minimal polynomials of the two sides. First let's play with the right side. $$\begin{array}{lll} x=\sqrt{p}-\sqrt{q} & \Longleftrightarrow & x^2=p+q-2\sqrt{pq}\\ & \Longleftrightarrow & x^2-p-q=-2\sqrt{pq}\\ & \Longleftrightarrow & x^4-2(p+q)x^2+(p-q)^2=0 \end{array}$$ Note that $x^4-2(p+q)x^2+(p-q)^2\in\mathbb{Z}[x]$ and it has exactly four complex roots (which are also reals), $\pm(\sqrt{p}+\sqrt{q})$ and $\pm(\sqrt{p}-\sqrt{q})$.

Now going for the left side. If the minimal polynomial of $a$ is $f(x)$ (which is in $\mathbb{Z}[x]$ and is monic), then for an integer $n$, we have that $na$ satisfies $n^{deg(f)}f(\frac{1}{n}x)$ and this new polynomial is in $\mathbb{Z}[x]$ and monic (check for yourself why, we use $n\neq 0$ here). Besides because $na=\sqrt{p}-\sqrt{q}$, so $na$ should satisfy $x^4-2(p+q)x^2+(p-q)^2=0$ and this implies that the minimal polynomial of $na$ should divide this polynomial. $$n^{deg(f)}f(\frac{1}{n}x)\mid x^4-2(p+q)x^2+(p-q)^2.$$ This tells us that degree of the left polynomial in above should be $1$, $2$, $3$ or $4$. Note that degree of the left polynomial is the same as degree of $f$. If $deg(f)=1$, then we get a contradiction with the roots of the right side polynomial are not integers. If $deg(f)=2$, then we get contradiction with $\sqrt{pq}$ is not an integer. If $deg(f)=3$, then the right side polynomial has to have a linear facor over $\mathbb{Z}$ which again makes a same contradiction with that none of its roots are integer. If $deg(f)=4$, then the right side polynomial should be an integer multiple of the polynomial in the left, however both are monic, thus they should be equal. This implies that $na$ should be one of the four roots of the right side polynomial. Calling the right side polynomial by $g(x)$ we can compute $f(x)$ (the minimal polynomial of $a$); $$f(x)=\frac{1}{n^4}g(nx)=x^4+\frac{p+q}{n^2}x^2+\frac{(p-q)^2}{n^4}$$ Which is in $\mathbb{Z}[x]$ only for $n=\pm 1$. This causes contradiction by we assumed $n\neq \pm 1$. (Note that if $n^2$ divides both $p+q$ and $p-q$, then $n^2$ should divides $2p$ and $2q$ as well. This implies $n$ should divides $p$ and $q$ and so their g.c.d which is 1).

Therefore there is no such an algebraic integer $a$, and we proved for $n\in\mathbb{Z}-\{-1,0,1\}$, the elements of $S$ make distinct classes in the residual ring $\frac{\bar{\mathbb{Z}}}{n\bar{\mathbb{Z}}}$ and since $S$ is an infinite set (because prime numbers are infinite), the main claim of your question is proved.

There are three cases we didn't covered, namely $n=-1$, $n=0$ and $n=1$. For $n=0$ we have $n\bar{\mathbb{Z}}=\{0\}$ and so the residual ring is the algebraic integer ring itself which is clearly infinite (for example contains $\mathbb{Z}$ itself). For $n=\pm 1$, the ideal $n\bar{\mathbb{Z}}$ is the whole ring, and thus the resudual ring is a singleton $\{0\}$ with obvious structure. This case clearly is not infinite, so in your question's face has to be excluded.

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    $\begingroup$ Wait a little... I agree that the minimal polynomial of $na$ has to divide $g$. But why does $n^{deg(f)}f(x/n)$ divide it? $\endgroup$ – Gabriel Oct 22 '19 at 14:33
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    $\begingroup$ @Gabriel It is the minimal polynomial of $na$. It is already monic and $na$ satisfies it, the only further thing you need to show is that no polynomial of a lower degree can be satisfies by $na$. To do so use contradiction, if there is a lower degree polynomial that $na$ satisfies it, then you can make a polynomial of lower degree than $deg(f)$ that $a$ satisfies it (Look at one of the lines where we wrote f using g). $\endgroup$ – AmirHosein Sadeghimanesh Oct 22 '19 at 18:49
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$a=b$ in $\overline{\mathbb{Z}}/n\overline{\mathbb{Z}}$ iff $\frac{a-b}{n}$ is an algebraic integer,

Let $c=\frac{n^{2^{-i}}-n^{2^{-j}}}{n} $ with $j >i$

since $n^{1-2^{-i}}$ is an algebraic integer, if $c$ was an algebraic integer then $cn^{1-2^{-i}}=1-n^{2^{-j}-2^{-i}}$ thus $n^{2^{-j}-2^{-i}}$ and $(n^{2^{-j}-2^{-i}})^{2^j} = n^{1-2^{j-i}}$ would be algebraic integers, a contradiction because it is immediate a non-integer rational number cannot be a root of a monic integer polynomial.

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