Fix a field of zero characteristic, $k$, e.g. $\Bbb{R}$ or $\Bbb{C}$. Suppose $k$ is normed (and complete for its norm). Consider the ring extensions: $k[x_1,..,x_n]\subset \ k<x_1,..,x_n> \ \subset k\{x_1,..,x_n\}$.

The algebraic power series are in some sense "controlled" by the ring polynomials. (e.g. because they satisfy polynomial equations). Is there some way "to control" the analytic power series by the algebraic ones?

e.g. given a morphism of two "large" rings, $R\stackrel{\phi}{\to}S$, each containing subrings of algebraic/analytic power series, suppose I know how $\phi$ acts on polynomials (and thus on the algebraic power series). Is there any way to determine how $\phi$ acts on the analytic power series? (Here $\phi$ is just an algebraic morphism, without any assumption of continuity in the classical topology, coming from the topology on $k$.)

In the extreme case: suppose an automorphism $\phi\circlearrowright R$ is identity on the subring of algebraic power series. Is $\phi$ identity on the analytic power series? At least, does $\phi$ send analytic to analytic?

(Here $R$ can be non-Noetherian, e.g. $C^\infty(\Bbb{R}^1,0)$.)


1 Answer 1


I don't think that analytic power series can be controlled by algebraic power series in an algebraic way, because they are often transcendental over the former.

Suppose an automorphism $\phi$ of $R$ is identity on the subring of algebraic power series. Is $\phi$ identity on the analytic power series?

The answer is no.

For a counter-example let $k$ be any field of characteristic zero and let $K \subseteq k((X))$ be the field of elements that are algebraic over $k(X)$. By the accepted answer of What's an example of a transcendental power series?, the exponental function $\text{exp}$ is transcendental over $k(X)$ and by transitivity of algebraic extensions, $\text{exp}$ is also transcendental over $K$. Extend the identity on $K$ to the automorphism $$\phi: K(\text{exp}) \to K(\text{exp}),\,\,\text{exp}\mapsto \text{exp} + X$$ (the inverse sends $\text{exp}\mapsto \text{exp} - X)$.

Let $R$ be an algebraic closure of $k((X))$. Then $\phi$ extends to an automorphism of $R$ (cf. https://en.wikipedia.org/wiki/Transcendence_degree, "Applications"). This $\phi$ is the identity on algebraic power series but not on analytic power series.

At least, does $\phi$ send analytic to analytic ?

I think the answer is no, but I have no example. To produce an example one could look for a non-analytic power series $f$ such that $\text{exp}, f$ are algebraically independent over $K$. Then modify the $\phi$ above by mapping $\text{exp} \mapsto f,\,\, f \mapsto \text{exp}$.

Added Apr 11, 2017: The answer to the second question is also no:

Let $k = \mathbb{C}$. As indicated above, it suffices to find a non-analytic power series $f$, that is algebraically independet over $K(\text{exp})$.

By the Newton-Puiseux Theorem (see http://www.emis.de/journals/UIAM/actamath/PDF/38-279-282.pdf), the field $P_a$ of analytic Puiseux series over $\mathbb{C}$ is algebraically closed. In particular, each non-analytic power series is transcendental over $P_a \supseteq K(\text{exp})$. So, we can take, for example, $f := \sum_{n\ge 1} n^n X^n$.

  • $\begingroup$ I meant mostly the local rings, and your example is not local, right? If the local ring has no flat functions, i.e. $\mathfrak{m}^\infty=\{0\}$, then the power series can be controlled by $\mathfrak{m}$-adic topology. But for the ring $C^\infty(\Bbb{R}^1,0)$ the situation seems to be tricky. $\endgroup$ Commented Apr 3, 2017 at 19:34
  • $\begingroup$ In my example $R$ is a field and hence local. BTW: What's the meaning of the $0$ in $C^\infty(\mathbb{R},0)$ ? $\endgroup$ Commented Apr 4, 2017 at 7:58
  • $\begingroup$ Well, I meant a local ring with the non-trivial maximal ideal. (0 denotes the zero ideal) $\endgroup$ Commented Apr 4, 2017 at 8:06
  • $\begingroup$ So $C^\infty(\mathbb{R},0)$ is the (one-element) set of functions $\mathbb{R}\to \{0\}$ ? $\endgroup$ Commented Apr 4, 2017 at 8:15
  • $\begingroup$ Sorry for the confusion, $C^{\infty}(\Bbb{R}^1,0)$ is the ring of germs of infinitely differentiable functions (defined on some neighborhoods of $0\in \Bbb{R}^1$) $\endgroup$ Commented Apr 4, 2017 at 9:03

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