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Let $k=\mathbb{C}$ or $\mathbb{R}$, and let k{x} denote the ring of power series with appropriate coefficients that are convergent around 0. Check that k{x} is a local ring.

I have a similar question: https://math.stackexchange.com/questions/765780/ring-of-formal-power-series-is-local-ring

Could someone give an example of what k{x} is for $\mathbb{C}$ and $\mathbb{R}$? I do not really understand what k{x} would even be.

For the proof, I know that a ring is local iff it has a unique maximal ideal. Would I want to use this to prove the statement?

Community
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math1234567
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  • There is no need to post the same question twice. Clearly once you answer the other question you will have answered this one! – fretty Apr 23 '14 at 11:03
  • I do not understand how it is the same question. I don't see how to relate "the coefficients that are convergent around 0" to "coefficients in k". – math1234567 Apr 23 '14 at 14:05
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    OK, I retracted my close vote. But the two questions are very similar --- in particular, the maximal ideal in both cases will be more or less the same thing. –  Apr 23 '14 at 14:58

2 Answers2

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The ring (actually $\mathbb C$-algebra) $\mathbb C\{x\}$ is the subring of $\mathbb C[[x]]$ consisting of those power series $\sum a_n x^n$ which represent the development at zero of an analytic function defined on some neighbourhood of zero. Notice that the neighbourhood in question may vary from power series to power series.

While the ring of formal power series $k[[x]]$ may be defined for an arbitrary field (or even ring) $k$, the ring $\mathbb C\{x\}$ depends for its definition on the topology of $\mathbb C$.
To simplify things outrageously, $\mathbb C\{x\}$ is in the realm of analysis and $\mathbb C[[x]]$ in that of algebra.
The great analyst Hadamard has given a wonderful criterion for a power series $\sum a_n x^n \in \mathbb C[[x]]$ to belong to $\mathbb C\{x\}$: $$\sum a_n x^n \in \mathbb C\{x\} \subset \mathbb C[[x]] \iff \limsup (|a_n|^{1/n} )\lt \infty $$

And now, back to your actual question!
The ring $\mathbb C\{x\}$ has a maximal ideal $\mathfrak m$ consisting in the power series $\sum a_n x^n$ with $a_0=0$.
That ring is local because every converging power series $\sum a_n x^n \notin \mathfrak m$, that is with $a_0\neq 0$, is invertible, so that $\mathfrak m$ is the only maximal ideal in the ring .
Indeed $\sum a_n x^n \notin \mathfrak m$ is invertible in $\mathbb C[[x]]$ [by the (virtual) answer to your preceding question :-)] and its inverse in $\mathbb C[[x]]$ can be shown to actually be in $\mathbb C\{x\}$ too.

(Everything I wrote above is still true if you replace $\mathbb C$ by $\mathbb R$)

(Non-) example
Hadamard's theorem trivially shows that the formal power series $\sum n^nx^n \in \mathbb C[[x]]$ is not a convergent power series: $$\sum n^nx^n \notin \mathbb C\{x\}$$

Georges Elencwajg
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  • This is a nice answer. If $f: = \sum a_n x^n \not \in \mathfrak{m}$, how do we show that $f^{-1} \in \mathbb{C} { x }$? The fact that $f \not \in \mathfrak{m}$ implies that $f$ is invertible in $\mathbb{C}[[x]]$ is a straightforward exercise in Atiyah--MacDonald, but we (or at least I) do not obtain a closed form a expression for the inverse. Can you point out what you have in mind for this? Thanks – AmorFati Jul 12 '18 at 22:02
  • Are we actually using the fact that $f \in \mathbb{C} {x }$ instead of $\mathbb{C}[[x]]$ anywhere ? In fact the exact same proof works for that case too right ? – Lelouch Apr 17 '20 at 10:44
  • @Lelouch: Yes, the proof works for $\mathbb C[[x]]$ and is even much simpler because we don't have to prove convergence of the inverse of a series not in $\mathfrak m$. This was implicit in the sentence starting with "Indeed..." – Georges Elencwajg Apr 17 '20 at 16:30
  • @AmorFati I guess the easiest way to see this is to use that $f$ is analytic and nonzero in a small neighbourhood around zero. Hence $1/f$ is also analytic in this neighbourhood. – user920957 Feb 13 '23 at 13:10
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If you know some theory of analytic functions, you know that a power series $$ \sum_{n\ge0}a_nx^n $$ with coefficients in $\mathbb{C}$ which converges in a neighborhood of zero converges in an open circle (which might be the entire plane); on the boundary of the circle convergence is not guaranteed. Let's call $r$ the maximal open circle where the series converges and let's set $$ f(x)=\sum_{n\ge0}a_nx^n $$ for $x$ in the (interior of) this circle. Then $f$ is an analytic function.

The series converges (uniformly) in every closed circle of radius $s$ with $0<s<r$. Since this set is compact, the function $f$ has only a finite number of zeros in it, because an analytic function whose set of zeros has an accumulation point is identically zero. Therefore we can choose $s>0$ such that the only possible zero of $f$ in the open circle with radius $s$ is at $0$. This happens, of course, if and only if $a_0=0$.

If $a_0\ne0$, then $g(x)=1/f(x)$ is an analytic function in the circle with radius $f$, because it has a (complex) derivative at each point. On the other hand, if $a_0=0$, the function $f$ can't have a multiplicative inverse in any neighborhood of $0$. Note that an analytic function in a circle around $0$ is the sum of its Taylor development at $0$, so $g$ is the sum of a power series.

Thus the non invertible power series are those with $a_0=0$ and this is clearly an ideal.

For $\mathbb{R}\{x\}$ there's nothing to prove, because a power series with coefficients in $\mathbb{R}$ (converging in some neighborhood of $0$) defines a complex analytic function in a circle and the same reasoning applies: the multiplicative inverse in $\mathbb{C}\{x\}$ is still $1/f(x)$, which assumes real values for real $x$; so the coefficients of its Taylor series at $0$ are real.

egreg
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