# Showing that $\overline{\mathbb{Z}}/n\overline{\mathbb{Z}}$ is infinite

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.

• Well, these are distinct, because $x_i^{(2^j)}=0$ in the quotient iff $j\ge i$ (for big enough $i$, at least). – Berci Oct 20 '19 at 23:18
• @Berci could you explain it a little further? This isn't clear to me – Gabriel Oct 21 '19 at 6:07

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.

• 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? – Gabriel Oct 22 '19 at 14:33
• @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). – AmirHosein Sadeghimanesh Oct 22 '19 at 18:49

$$a=b$$ 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.