# The ring of power series in z with a positive radius of convergence

Let $A$ be the ring of power series in $z$ with a positive radius of convergence, where all coefficients are complex numbers.

1. What is the name of $A$? Does it have a special name?
2. I want to show that if $f(z)\in A$ has nonzero constant, then $f(z)$ is invertible in $A$. How can I prove this? I know that any formal power series is invertible if the constant is not zero. But how can I show that the radius of convergence of $1/f(z)$ is positive?
3. How can I know this result from complex analysis? I studied complex analysis for a semester, but I don't remember whether I learned this or not. I thought I have heard some similar thing, but I can't find out it in the textbook(Silverman)
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I thought I might as well make an earlier comment into an answer:

1. I'd call it the ring of germs of holomorphic/analytic functions at $0$.

2. Look at the explicit inverse formal power series and recall the formula for the radius of convergence.

3. Every such power series represents a holomorphic function in some neighborhood of $0$. Now use that $1/f$ is holomorphic wherever $f$ is nonzero and the identification of holomorphic functions with functions that are locally developable as power series.

You asked about proving that the formal inverse of a power series with positive radius of convergence is again a power series with positive radius of convergence "without using complex analysis". This can be done with a nice trick that goes back to Abel, Hurwitz and Weierstrass, using comparison with geometric series only.

Step 1. A formal power series $f(z) = \sum_{n=0}^{\infty} a_n z^n$ has positive radius of convergence $\rho \gt 0$ if and only if either one of the following three conditions holds:

1. We have $\rho^{-1} = \limsup\limits_{n \to \infty}\; |a_n|^{1/n} \lt \infty$.
2. There are $q \gt 0$ and $C \geq 0$ such that $|a_n| q^n \leq C$ for all $n \in \mathbb{N}$.
3. There is $r \gt 0$ such that $|a_n| \leq r^n$ for all $n \geq 1$.

This should be clear from the proof of the formula for the radius of convergence.

Step 2. Let $f(z)= 1 + \sum_{n=1}^\infty a_n z^n$ have positive radius of convergence. Then the inverse formal power series $g(z) = 1 + \sum_{n=1}^{\infty} b_n z^n$ with $b_0 = 1$ and $b_n = -\sum_{i=1}^{n} a_i b_{n-i}$ for $n \geq 1$ has positive radius of convergence as well.

By step 1 there is $r \gt 0$ such that $|a_n| \leq r^n$ for all $n \geq 1$. Note that $|b_1| = |a_1| \leq \frac{1}{2} \cdot 2 r$ and by induction $$|b_n| \leq \frac{1}{2} (2r)^n.$$ Indeed, $$|b_n| \leq \sum_{i=1}^{n-1} |a_i|\,|b_{n-i}| + |a_{n}| \leq \frac{1}{2} \sum_{i=1}^{n-1} r^i (2r)^{n-i} + r^n \leq \frac{1}{2}(2r)^n.$$ Putting $q = 1/(2r)$ we get $|b_n| q^n \leq 1/2$, and we conclude by step 1 again.

Step 3. The element $f(z) = \sum_{n=0}^\infty a_n z^n$ of $A$ is a unit if and only if $a_{0} \neq 0$.

Reduce to step 2.

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If you need more details, I'll be happy to add them, but I suggest that you try it first on your own. – t.b. Jun 8 '11 at 11:36
For answer2, if $f(z)=\sum a_n z^n$, the inverse of it is $g(z)= \sum b_n z^n$ where $b_0 = a_{0}^{-1}, b_n = -a_{0}^{-1} \sum_{i=1}^{n} a_i b_{n-i}$. But how can I know that $\limsup b_n$>0? – Gobi Jun 9 '11 at 0:54
Hint: if $f$ is a formal power series with positive radius of convergence, then $f$ defines a holomorphic function in some disk of radius $\delta>0$ centered at the origin. Moreover, since $f(0)\neq 0$, the continuity of $f$ implies that we can choose $\delta>0$ sufficiently small such that $f$ has no zeros in this disc. The function $\frac{1}{f}$ is now holomorphic in this disc. Hence $\frac{1}{f}$ has a formal power series representation in this disc. – Amitesh Datta Jun 9 '11 at 0:59
@Amitesh Yes, I understand it now after reviewing some of complex analysis. (I specially missed that $1/f$ is holomorphic if $f$ has no zero, just by the quotient rule $(1/f)'=-f'/f^2$.) But is there any proof with the explicit inverse formal power series, not using the complex analysis background? – Gobi Jun 9 '11 at 4:18
@gobi: see edit. – t.b. Jun 9 '11 at 16:03

Theo's (excellent) answer is sufficient; however, let me also add the following exercises which I think are useful if you wish to develop a deeper understanding of the ring $A$:

Definition

A ring $R$ is said to be a local ring if it has a unique maximal ideal $m$. In other words, the ring $R$ is local if every proper ideal of $R$ is contained in $m$.

Exercise 1: If $R$ is local with unique maximal ideal $m$, prove that every element of $R$ not in $m$ is a unit in $R$.

Exercise 2: Prove that if the set of non-units in a commutative ring $A$ is an ideal in $A$, then $A$ is a local ring.

Exercise 3: Prove that the ring $A$ of your question is a local ring. Moreover, prove that the unique maximal ideal of $A$ is a principal ideal. (Hint: the set of all elements of $A$ with constant term $0$ is an ideal; prove that it is the unique maximal ideal of $A$. Theo's answer and exercise 2 are relevant.)

Exercise 4: Prove that every proper non-zero ideal of $A$ (of your question) is a power of $m$, the unique maximal ideal of $A$. (Recall that if $I$ is an ideal in a ring $A$, the power $I^n$ ($n\geq 1$) is defined to be the ideal generated by the set of all products consisting of $n$ terms where each term is an element of $I$.) In particular, we say that $A$ is a discrete valuation ring (DVR is a common abbreviation).

Exercise 5: Let $A$ be an arbitrary local ring with unique maximal ideal $m$. Is it true that every proper non-zero ideal of $A$ is a power of $m$? If $A$ is a Noetherian ring, is it true that the intersection of all powers of $m$ is the zero ideal? (This question is difficult.)

Exercise 6: Let $x\in \mathbb{R}$ and consider the set of all ordered pairs $(f,U)$ where $U$ is open and $f:U\to \mathbb{R}$ is continuous. We define an equivalence relation on this set by setting $(f,U)\equiv (g,V)$ if $f|W=g|W$ for some open subset $W\subseteq U\cap V$. Let us define a ring structure on the set of all equivalence classes $A$ by the rules: $(f,U)+(g,V)=(f+g,U\cap V)$ and $(f,U)\times (g,V)=(f\times g,U\cap V)$. Prove that these operations are well-defined and prove that the ring $A$ in question is a local ring. (Hint: The set of all equivalence classes corresponding to representatives of the form $(f,U)$ such that $f(x)=0$ is the unique maximal ideal of $A$.)

Exercise 7: Is the result of the above exercise true if the word "continuous" is deleted throughout? Is the result of the above exercise true if the words "continuous" and "open" are deleted throughout? Is the result of the above exercise true if the word "continuous" is replaced by either "differentiable", "smooth" or "analytic"?

Exercise 8: Is every proper non-zero ideal in the ring of exercise 5 a power of the unique maximal ideal of this ring?

I hope these exercises are useful! (They are all important since they provide good practice with the notion of a local ring; local rings are ubiquituous in commutative algebra.)

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Wow, that's what I call a nice and informative answer! Certainly useful since the question is tagged commutative algebra by the OP. By the way: From your answers I've seen so far, I wouldn't have guessed that your age is what it is, that's quite impressive. Thanks for the nice words in the first sentence! – t.b. Jun 8 '11 at 12:08
@gobi Of course, if you are stuck with any of the above exercises, please feel very free to comment and I will help you. – Amitesh Datta Jun 8 '11 at 12:11
Dear Theo, thank you very much for the compliments! – Amitesh Datta Jun 8 '11 at 12:15
Thanks for really good exercises. And I can't believe your age, too. I think I can solve exercises 1~4. 1~2 is elementary, and there's also a solution in CrazyProject‌​. 3~4 is simliar to Atiyah's Ex6.4, from which I wrote this question. But other exercises need time to solve. – Gobi Jun 9 '11 at 4:47
@gobi I should emphasize that exercise 5 is not easy (in fact, it is a theorem known as the "Krull intersection theorem") but I think that as one progresses in one's study of commutative algebra (or, indeed, any branch of mathematics), one should start proving difficult theorems on one's own. This is why I gave the exercise. However, if you are truly stuck, it might be better to focus mainly on the other exercises. – Amitesh Datta Jun 9 '11 at 6:50