Show that $\mathbb{Z}[\sqrt{-5}]_3 \simeq \mathbb{Z}_3[\sqrt{-5}]$ I have a somewhat tedious question.  Is $\mathbb{Z}[\sqrt{-5}]_3 \simeq \mathbb{Z}_3[\sqrt{-5}]$ ? It might help to describe what these are:


*

*$\mathbb{Z}[\sqrt{-5}]_3$ is the $3$-adic completion of the ring $\mathbb{Z}[\sqrt{-5}]$.

*$\mathbb{Z}_3[\sqrt{-5}]$ is the extention of the $3$-adic numbers $\mathbb{Z}_3$ with the element $\sqrt{-5}$.  We could even write it as $\mathbb{Z}_3[x]/(x^2 + 5)$. 
Are these two constructs the same?  So we could write these as inverse limits of a kind:
$$ \mathbb{Z}_3[\sqrt{-5}] = \big(\lim_{\leftarrow} \mathbb{Z}/3^k\mathbb{Z} \big) [\sqrt{-5}] $$
Can we have $x^2 + 5 = 0$ in $\mathbb{Z}/3\mathbb{Z}$ ?  In fact $x^2 + 5 = (x+1)(x+2)$ in $\mathbb{Z}/3\mathbb{Z}$ so that $\sqrt{-5} \in \mathbb{Z}_3$.   The ring can still exist it just has zero divisors.
In the other case, we can always adjoin $\sqrt{-5}$ to $\mathbb{Z}$.  And we can take this ring modulo powers of $3$.  And I wonder what the properties of this ring are:
$$ \mathbb{Z}[\sqrt{-5}]_3 = \lim_{\leftarrow} \mathbb{Z}[\sqrt{-5}] /3^k \mathbb{Z}[\sqrt{-5}] $$
The answer is looking like a no.
 A: Here’s one approach:
Let’s look first at $\bigl(\Bbb Z[\sqrt{-5}\,]\bigr)_3$. This is, as you have said, the $(3)$-adic completion of the ring $R=\Bbb Z[\sqrt{-5}\,]$.
But what is $(3)$ as an ideal of $R$? There are many ways of seeing that $3$ splits in this ring, but I think it’s easiest to just notice that $(3,\sqrt{-5}-1)(3,\sqrt{-5}+1)=(3)$, as ideals of $R$, by direct computation. Call these two maximal ideals $\mathfrak p_{3}$ and $\bar{\mathfrak p}_3$.
Aha. This is like asking for the $(10)$-adic completion of $\Bbb Z$: $10$ is the product of two rational primes, and in the same way, $(3)$ is the product of two primes of $R$.
But it’s “well known” that the ten-adic integers form a ring isomorphic to $\Bbb Z_2\oplus\Bbb Z_5$, with coordinatewise addition and multiplication, so a ring with zero-divisors. In the same way, $R_3\cong R_{\mathfrak p_{3}}\oplus R_{\bar{\mathfrak p}_3}$. I think this accords with your intuition here. (It does need to be verified that both factors are isomorphic to $\Bbb Z_3$.)
Now, what about $\Bbb Z_3[\sqrt{-5}\,]$? The notation is just a bit unclear. Do you mean the smallest ring containing $\Bbb Z_3$ and $\sqrt{-5}$, as I in fact understand this notation? In that case, $\Bbb Z_3[\sqrt{-5}\,]=\Bbb Z_3$, since $\Bbb Z_3$ already has a square root of $-5$. In justification of my understanding, what is $\Bbb R[\sqrt5\,]$? The way we teach Galois Theory at least, this is simply $\Bbb R$. I think it should be the same thing with $\Bbb Z_3[\sqrt{-5}\,]$.
But if what you mean $\Bbb Z_3[X]/(X^2+5)$? In this case, since $X^2+5=(X+\sqrt{-5}\,)(X-\sqrt{-5}\,)$, a good $\Bbb Z_3$-factorization, then by Chinese Remainder, the ring is $\Bbb Z_3\oplus\Bbb Z_3$, the same as the three-adic completion of $\Bbb Z[\sqrt{-5}\,]$. I do prefer the other interpretation, though, making the two rings of your title quite different.
