Ring of germs and induced isomorphism Let $A$ be the ring of germs of real analytic functions in $0\in\mathbb R$. Let $x\in A$ be the identity map on $\mathbb R$. How can I show that the map $f\mapsto \sum_n(f^{(n)}(0)/n!)T^n$ from $A$ to $\mathbb R[[T]]$ is injective, and induces an isomorphism from the $xA$-adic completition of $A$ onto $\mathbb R[[T]]$? And why this is false if we replace $A$ by $C^\infty$-functions in $0$? Qing Liu: Algebraic Geometry and Arithmetic Curves problem 1.3.3.

The injectivity is trivial : if a germ maps to zero, then it corresponds to a function analytic in a neighbourhood of $0 $ that has all it's derivatives vanishing at zero, which implies by analyticity (expend the function at $0$ !) that the function is zero, and same for the germ. To see that it induces and isomorphic from the $(x)$-adic completion to $\mathbf R[[T]]$, first see that if you note $\varphi$ this map, you have $\varphi((x^n)) \subseteq (T^n)$ for each $n\in\mathbf{N}$, which implies by definition of the $(x)$-adic completion $B$ fo $A$ as projective limit that $\varphi$ extends to $\psi : B \to \mathbf R[[T]]$. This latter is injective by construction as $\cap_{n>0} (x^n) = \{0\}$. For the surjectivity, as $B = \varprojlim A / (x^n)$ and $\mathbf R[[T]] = \varprojlim \mathbf R[[T]] / (T^n)$ it is enough to see that the local maps $A / (x^n) \to \mathbf R[[T]] / (T^n)$ are surjective, but this is trivial.
This is false for $C^\infty$ functions because you have flat functions having all their successive derivatives vanishing at $0$ and which are not equal to zero. Think of $x\mapsto e^{-1/x}$ extended by zero on $\mathbf{R}_{<0}$, for instance. Its germs will map to $0$ by your map because this function (simple exercise) has all derivatives vanishing at $0$, but this function is non-zero.
