# $p$-adic Ring extensions vs. “ordinary” Ring extensions

I read about inverse limits in this post, and found the example by Arturo Magidin quite interesting (his "approximate" solution of $x^2 = -1$ in $\mathbb Z$).

By his construction we get a Ring which is an extension of $\mathbb Z$, i.e. the ring of $p$-adic (here $5$-adic) integers where this equation has a solution. On the other side it is well know that $i$ solves this equation, so $\mathbb Z[i] \equiv \mathbb Z[x] / (x^2+1)$ is also a Ring which is an extension of $\mathbb Z$ such that this equation is solvable. Could something be said about the relation on these different ring extensions?

EDIT: Correction from comment $x^2 - 1 \to x^2 + 1$.

• ITYM $x^2 + 1$ in that quotient... – Ben Millwood Jul 14 '13 at 22:57

I’ll assume that you do understand the relation between $\mathbb Z$ and $\mathbb Z[i]$ on the one hand and between $\mathbb Z$ and $\mathbb Z_5$ on the other hand. I take it that you’re asking about what the relation between $\mathbb Z[i]$ and $\mathbb Z_5$ might be. Let’s call $I$ the solution that @Arturo told you about, of the equation $X^2+1=0$ in $\mathbb Z_5$. Maybe it was the solution that is congruent to $2$ modulo $5$ there. So the question becomes what the relation between $i\in\mathbb Z[i]$ and $I\in\mathbb Z_5$ might be. The two of them satisfy the same polynomial relation over $\mathbb Z$, after all. The answer is simple enough, if you know the language of ring (homo)morphisms. There is a morphism, and only one, $\varphi_1\colon\mathbb Z[i]\to\mathbb Z_5$, for which $\varphi_1(i)=I\,$; and similarly there is a unique morphism $\varphi_2$ with same domain and target, for which $\varphi_2(i)=-I$.
In other words, you can embed $\mathbb Z[i]$ into $\mathbb Z_5$ in precisely two ways, depending on where you choose to send $i$.