Ring of germs and induced isomorphism

Question

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.

Answer 1

For the first question: the map is injective simply because the kernel is the set of germs of analytic functions all of whose derivatives at $0$ are zero, and there is clearly only one such germ. To show that the map is surjective on the completion it is enough to show that it is surjective modulo powers of the maximal ideals. Can you do that?

For the second question: There is a famous theorem of Borel that says that every formal series in $\mathbb R[[T]]$ is the Taylor series of a $C^\infty$ function. It follows easily from this that in the $C^\infty$ case the map is not surjective.

Answer 2

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.

Answer 3

Half answer: Let $f$ be analytic, and suppose $f$ is carried by the above map to $0$. Then all the derivatives of $f$ at $0$ vanish, and by definition of analytic function, it means $f\equiv0$.

The claim fails to hold when considering $C^\infty$ functions, since there are smooth functions which are not the zero function, but satisfy $f^{(n)}(0)=0$, for all $n$.