elements in the completion of a ring and binomial theorem My question is the elements in the completion of a ring. As far as I know, if $A$ is a ring and $\widehat{A}$ is a completion w.r.t. $I$-adic topology where $I\subset A$ is an ideal, then if $0\neq a\in I$ then $(1+a)^{-1}\in\widehat{A}$ expressed as $1-a+a^2-a^3+\cdots$. Also, $\sqrt{a+1}\in\widehat{A}$ expressed as $1+\frac{1}{2}a-\frac{1}{8}a^2+\frac{1}{16}a^3-\cdots$. I think all these forms came from the binomial theorem. Is that correct? How can I just say those $(1+a)^{-1}$ and $\sqrt{a+1}$ are equal to those infinite sum in $\widehat{A}$?
 A: This is one example of a kind of infinite "permanence of identities", which says that identities which are true for integer polynomials are automatically true in every ring. I have a blog post where I go over this in the polynomial case, and I've been meaning to write a follow-up about power series for a while now. When I get around to it, I'll update this answer with a link.
As for the idea, we know that the binomial theorem is true for rational power series. Then since (continuous) homomorphisms preserve identities, we can transfer our knowledge of things like the binomial theorem to any ring we like.
As a concrete example, we know that $\sqrt{1 + x} = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \ldots$ in $\mathbb{Q} [ \! [ x ] \! ]$. We don't know how to interpret $\sqrt{\cdot}$ in general rings, though, so we need to come up with a slightly massaged version of this fact if we want to transfer it. It doesn't take long to land on
$$
\left ( 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \ldots \right )^2 = 1+x
$$
This is now a simple equation in $\mathbb{Q} [ \! [ x ] \! ]$, and if $\hat{A}$ is a ring where $a^n \to 0$ and the coefficients exist, there's an obvious continuous homomorphism
$\mathbb{Q} [ \! [ x ] \! ] \to \hat{A}$ which sends $x$ to $a$. Since homomorphisms preserve equations (including infinite equations, by continuity) this tells us
$$
\left ( 1 + \frac{1}{2}a - \frac{1}{8}a^2 + \ldots \right )^2 = 1+a
$$
which gives us a way of talking about "$\sqrt{1+a}$" inside of $\hat{A}$. Life is too short to write out the series all the time, so in practice we often (abusively) write $\sqrt{1+a}$ and manipulate it in the expected way, safe in the knowledge that we're really abbreviating some power series which happens to solve the defining equation for $\sqrt{1+a}$.
Notice whenever we have a power series solution to some polynomial equation, we can transfer that power series solution to the setting of general complete rings (provided all the coefficients exist). This is extremely useful in practice, but the reason that it works is often left implicit.

I hope this helps ^_^
