TITLE: Simple minded details about why a particular map is étale
QUESTION [5 upvotes]: In the category of (commutative, unital) $\mathbb{C}$-algebras,
let ${f : \mathbb{C}[t, t^{-1}]  \rightarrow   \mathbb{C}[t, t^{-1}, x]/(x^2 -t)}$ be the obvious map sending $t$ in the domain to $t$ in the codomain.
Let $N$ be a nilpotent ideal in an algebra $C$ and let ${q : C \rightarrow C/N}$ be the resulting quotient. Let ${g : \mathbb{C}[t, t^{-1}] \rightarrow C}$ and ${h : \mathbb{C}[t, t^{-1}, x]/(x^2 -t) \rightarrow C/N}$ be such that ${h f = q g}$ (in other words, such that  `the square commutes'). 
Why is it that  there exists a unique ${k : \mathbb{C}[t, t^{-1}, x]/(x^2 -t) \rightarrow C}$ such that ${k f = g}$ and ${q k = h}$ (in other words, a unique $k$ making the inner triangles of the square commute)?
(Please, no appeal to schemes, references to EGA etc. I just want to see a direct proof using only the simplest commutative-algebraic facts. Thanks in advance.)

REPLY [3 votes]: Let me give a bare-hands answer.
Consider the Taylor expansion for $(1+u)^{1/2}$:
$$ s(u) = \sum_{n\geq 0} \binom{1/2}{n} u^n.$$
This power series converges in a small disk around $1$ in $\mathbb C$, so at the level of formal power series with coefficients in $\mathbb Z[1/2]$ as well we have $s(u)^2 = 1+ u$. Thus, for any nilpotent element $v$ of a ring where $2$ is a unit, we have $s(v)^2 = 1+v$.
Now suppose we are in your situation. We are given $g(t) \in C$ a unit and some $y \in C$ such that $y^2 - g(t) \in N$. We wish to find a square root of $g(t)$ in $C$. But $ g(t)/y^2 - 1  \in N$, so 
$$s( (g(t)/y^2 - 1))^2 = 1 + (g(t)/y^2 - 1) = g(t)/y^2$$
and thus
$$v = y s(g(t)/y^2 - 1)$$
is a square root of $g(t)$ in $C$. Uniqueness follows since if $v$ and $w$ are square roots, then $w^2/y^2 - 1 = g(t)/y^2 - 1 \in N$, so
$w/y = s(1 + ((w/y)^2 - 1)) = s(1 + ((v/y)^2 - 1)) = v/y$
as we have an equality
$s(1 + (u^2 -1 )) = u$ 
of formal functions.

This method seems mysterious on the face of it, relying on the Taylor expansion of the square root. I suspect it is more general.