Usbrth

I'm interested in understanding why $(mathbbZ[i]/8mathbbZ[i])^*cong mathbbZ/4mathbbZtimesmathbbZ/4mathbbZtimesmathbbZ/2mathbbZ$ (I have confirmed this by looking at element orders). This is of course the group of units in the ring $mathcalO_K/8mathcalO_K$ for $K=mathbbQ(i)$.

The literature on problems like this seems pretty scarce; the only real article I've found is here.

Is there a simple way of understanding the specific example above without resorting to looking at individual element orders? I see that $(mathbbZ[i]/8mathbbZ[i])^*cong ((mathbbZ/8mathbbZ)[i])^*$, but it's not clear how to proceed from there.

Here is a direct approach, not appealing to local fields (*). In order to answer also to @Shimrod, let me start with a general imaginary quadratic field $K$ in which the prime $2$ is totally ramified (this is the meaning of the condition "$2$ divides the discriminant"), i.e. $(2)$ is the square of a prime ideal of $mathfrak O_K=mathfrak O$, say $(2)=P^2$, so that $mathfrak O/2^k=mathfrak O/P^2k$ for all $kge 1$. The natural surjective ring homomorphism $mathfrak O/P^2kto mathfrak O/P^2$ for $kge 2$ induces a homomorphism of groups $f_k :(mathfrak O/P^2k)^*to (mathfrak O/P^2)^*$, and we aim to determine its kernel and cokernel. Note that $(mathfrak O/P^2k)^*$ contains the image of the group of units $mathfrak O^*$ under the natural surjection $mathfrak O to mathfrak O/P^2k$, in particular it contains the image of $mu_K$, the group of roots of unity of $K$. In the sequel, for simplification, we'll always beforehand divide $mu_K$ out, but without changing the notation $mathfrak O$. It will be also convenient to write $[a]_k$ for the class of $a$ mod $P^2k$.

Now add the hypothesis that $mathfrak O$ is a PID (which is the case here with $mathbf Z[i]$), and denote $P=(pi)$. Then $[a]_1$ is invertible iff $pi nmid a$, iff $[a]_k$ is invertible too (just apply Bezout's thm. in $mathfrak O$). It follows that $f_k$ is surjective. Besides it is straightforward that Ker$f_k= (1+(pi^2)/1+(pi^2k),times)$, and the map $x to x-1$ induces an isomorphism $(1+(pi^2)/1+(pi^2k),times)cong ((pi^2)/(pi^2k), +)$ (just check that $(xy-1)-(x-1)-(y-1)$ $=(x-1)(y-1))$. The latter group is the cyclic group $((2)/(2^k), +)$, and its order can be computed by introducing the descending filtration $..(2^n)> (2^n+1)>...$ and its successive quotients are $(2^n)/(2^n+1)cong$ $ (mathbf F_2,+)$, where $mathbf F_2$ is the residual field at $(pi)$. It follows immediately that Ker$f_kcong mathbf Z/2^k-1$. It remains to compute the order of $(mathfrak O/P^2)^*$ following the same steps: the map $(mathfrak O/P^2)^* to (mathfrak O/P)^*$ is surjective, but $(mathfrak O/P)^*=mathbf F_2^*=(1)$, so $(mathfrak O/P^2)^*$ has the same order $2$ as the kernel $(pi)/(pi^2)$, and finally (recall our initial convention) $(mathfrak O/P^2k)^*cong$ Im $mu_Ktimes mathbf Z/2^k-1 times mathbf Z/2$. In your present case, $k=3$, Im $mu_Kcong (mathbf Z/4,+)$ and we recover the announced result.

(*) @user655377. I don't recover the formula given in your comments. Putting apart possible calculation errors, I contend that the passage from local to global is not so "easy": you introduce the completion of $K$ at $pi$ (not at $2$, which is not a prime of $K$), say $K_pi$, and its group of units, say $U$, which is non canonically $cong W times U_1$, where $W$ is the group of roots of unity of odd order in $K_pi$ and $U_1$ the group of principal units. . Then you compute (I think) $U_1$ mod $2^k$-th powers using the $mathbf Z_2$-structure of $U_1$. But there is a priori no direct relation between $U$ mod $2^k$-th powers and $(mathfrak O/2^k)^*$.