How many units are there in $\mathbb{Z}[i]/ (13^{2021})$?

Question

Clearly, there are $(13^{2021})^2$ elements in $\mathbb{Z}[i]/(13^{2021})$ and I wish to determine how many of these are units.

There are $\phi(13^{2021})=13^{2021}-13^{2020}=13^{2020}\cdot 12$ invertible elements if we consider $\mathbb{Z}/13^{2021}\mathbb{Z}$. Because me may place these elements with or without the imaginary component, there are at least:

$$|U|\geq 2\cdot\phi(13^{2021})=24\cdot13^{2020}$$

Next, I tried to study when an element writen as $u_1+u_2 i$ is a unit if $u_1$ and $u_2$ are invertible modulo $13^{2021}$. In this case, we need that the following equations to be satisfied for certain $z_1, z_2\in \mathbb{Z}$:

$$ \begin{cases} u_1z_1-u_2z_2 \equiv 1 \mod13^{2021} \\ u_1z_2+u_2z_1\equiv 0 \mod13^{2021} \end{cases}$$

Equivalently this means we need:

$$z_2\equiv -u_2z_1 u_1^{-1} \quad \quad \text{and}\quad \quad z_1 u_1^{-1}(u_1^2+u_2^2)\equiv 1$$

There will be $z_1$ and $z_2 $ satisfying this if and only if $u_1^2+u_2^2$ is invertible $\mod 13^{2021}$. Therefore we have $2+3i+(13^{2021})$ non invertible as $(2^2+3^2,13)=13$ but $2+5i+(13^{2021})$ invertible as $(2^2+5^2,13)=1$. This means:

$$|U|> 2\cdot\phi(13^{2021})=24\cdot13^{2020}$$

But this line of reasoning makes it very hard to count exactly how many units are in $\mathbb{Z}[i]/(13^{2021})$. Perhaps it would be nice to establish an isomorphism to a ring we have a better handle of, but it seems not very clear how to do that.

EDIT: Here is Servae's answer in more detail, but I still couldn't fill all of the details without computational aid:

Lemma: $$[\mathbb{Z}[i]/(13^{2021})]^*=\{a+bi+(13^{2021}) | a+bi+(13)\in [\mathbb{Z}[i]/(13)]^* \:\text{and}\: a,b\in [1,...13] \}+13\mathbb{Z}[i]/(13^{2021})$$ Proof: Indeed, if $a+bi+(13^{2021})\in [\mathbb{Z}[i]/(13^{2021})]^*$, we have:

$$13^{2021}|(a+bi)z-1\Rightarrow 13|(a+bi)z-1\Rightarrow a+bi+(13) \in [\mathbb{Z}[i]/(13)]^*$$

Reciprocally, if we have $a+bi+13x+(13^{2021})$ on the right handside, this is such that $(a+bi)z-1=13k$, we have $(a+bi+13x)z-1=13\hat{k}$ and taking the 2021-th power:

$$(a+bi+13x)Z-1^{2021}=13^{2021}\hat{k}^{2021}\Rightarrow a+bi+13x +(13^{2021})\in [\mathbb{Z}[i]/(13^{2021})]^* \quad \quad \square$$

$13\mathbb{Z}[i]/(13^{2021})$ has $(13^{2020})^2$ elements.

I was only able to count the number of units in $[\mathbb{Z}[i]/13]^*$ numerically. If anyone has a clever argument, I would like to see it. Here is the matlab code if anyone is interested:

    close all
    clear all
    clc

    Ziquotient13=[];
    k=1;
    for l=1:13
        for m=1:13
            Ziquotient13(k)=l+i*m;
            k=k+1;
       end
   end

   units=[];
   k=1;
   for l1=1:length(Ziquotient13)
       for l2=1:length(Ziquotient13)
           num=Ziquotient13(l1)*Ziquotient13(l2);
           if mod(real(num),13)==1 & mod(imag(num),13)==0
               units(k)=Ziquotient13(l1);
               k=k+1;
           end
       end
   end

   display(length(units))

With this code, $[\mathbb{Z}[i]/13]^*$ has $144$ elements. Hence, there are in total $144\cdot (13^{2020})^2$ units in $\mathbb{Z}[i]/(13^{2021})$

Answer 1

Here is a sketch of an approach that you can make precise yourself: The natural map $$\varphi:\ \Bbb{Z}[i]/(13^n)\ \longrightarrow\ \Bbb{Z}[i]/(13),$$ is a ring homomorphism, so it maps units to units. Its kernel is precisely $13\Bbb{Z}[i]/(13^n)$ so the units are contained in the set $$(\Bbb{Z}[i]/(13))^{\times}+13\Bbb{Z}[i]/(13^n).$$ On the other hand $13\in\Bbb{Z}[i]/(13^n)$ is nilpotent, so if $u\in\Bbb{Z}[i]/(13^n)$ is a unit then so is $u+13x$ for any $x\in\Bbb{Z}[i]/(13^n)$. This shows that $$(\Bbb{Z}[i]/(13^n))^{\times}=(\Bbb{Z}[i]/(13))^{\times}+13\Bbb{Z}[i]/(13^n).$$ Then it remains to determine $(\Bbb{Z}[i]/(13))^{\times}$, which should not be too difficult.