# A non-nilpotent formal power series with nilpotent coefficients

Does anyone have an example of a formal power series $$p=a_0+a_1x+ a_2x^2 + \cdots \in R[[x]]$$ ($R$ is a commutative ring) all of whose coefficients $a_i$ are nilpotent in $R$ such that $p$ is not nilpotent in $R[[x]]$?

I know that if $p$ is nilpotent in $R[[x]]$ then all $a_i$'s are necessarily nilpotent in $R$, but I can't figure out a simple example that shows that the opposite is not true in general. Any help is appreciated.

• Note that the converse is true if $R$ is Noetherian. This is an exercise somewhere in Atiyah–Macdonald. It seems to me that if you figure out a proof for that, then it's easy to come up with a counterexample for the general case. – Dylan Moreland Aug 28 '12 at 17:02
• A silly noncommutative example is to let $R$ be a 2×2 matrix ring, and $p=E_{12}+E_{21}x$. Then $p^2 = x$, so $p$ is not nilpotent, even though its coefficients are. – Jack Schmidt Aug 28 '12 at 17:16
• @DylanMoreland. I did proof in one direction like this: $p^n=0$ for some $n$ implies $a_0^n=0$, that is, $a_0$ is also nilpotent. Then $p-a_0$ is nilpotent, what, by induction, implies $a_1, a_2, \ldots$ nilpotent. I couldn't see how this could be reversed, and also how to show a counterexample. Now, by the answers, I got it. – fmoura2005 Aug 29 '12 at 14:01
• @fmoura2005 Ah, good. I wasn't implying that this would give the result immediately, but the idea that the orders of nilpotency will have to grow is a good one, and I think Georges's example follows naturally from that. Proving that this works is another matter, of course. – Dylan Moreland Aug 30 '12 at 17:36

## 2 Answers

Consider an integer $N\geq 2$, the polynomial ring in infinitely many variables $\mathbb Q[T_1,T_2,T_3,\ldots, T_n,...]$ and its quotient $$R=\mathbb Q[T_1,T_2,T_3,\ldots, T_n,...]/\langle T_1^N,T_2^N,T_3^N,\ldots, T_n^N,...\rangle=\mathbb Q[t_1,t_2,t_3,\ldots, t_n,...]$$ The formal power series series $p(x)=t_1x+t_2x^2+t_3x^3+\ldots+t_nx^n+\ldots \in R[[x]]$ clearly has all its coefficients nilpotent but is nevertheless not nilpotent: this is not trivial but proved in this article by Fields (Proc.AMS,Vol. 27, Number 3, March 1971).

However, he proves that if $R$ is a ring of characteristic $p\gt 0$, then a power series $f(X)=\sum a_ix^i\in R[[x]]$ all of whose coefficients $a_i$ are nilpotent is itself nilpotent iff the orders of nilpotence of the $a_i$'s are bounded : all $a_i^N=0$ for some integer $N$.
Hence if you replace $\mathbb Q$ by $\mathbb F_p$ in the above example, the resulting formal series $p(x)$ is nilpotent.

Fields's article is interesting throughout and extremely elementary: the level is that of the first few chapters of Atiyah-Macdonald's Commutative Algebra.

• Thanks Georges for the paper. I've seen the proof. That's what I needed. – fmoura2005 Aug 29 '12 at 13:43

For an example, take $$R=\mathbb{Q}[t,t^{1/2},t^{1/3},\ldots]/(t)$$ Form the power series $p(x)=\sum_{n\geq 1} a_nx^n$, with $a_n=t^{1/n}$. I leave it to you to check that $p(x)$ is not nilpotent.

• Dear Pink Elephant, you are welcome to let the checking of non-nilpotence to whomever you like, but that is the only difficult point: it is very easy to come up with non-nilpotent power series $\sum a_nx^n$ where $a_n$ is nilpotent of order $n$ as long as you don't have to write down the proof of non-nilpotence of the series rigorously. – Georges Elencwajg Aug 28 '12 at 17:08
• Perhaps it would be better to take $R=\mathbb{F}_2[t,t^{1/2},\ldots]/(t)$. Then for all non-negative integers $k$, we can directly compute $p(x)^{2^k}$ and see that it's non-zero. – Julian Rosen Aug 28 '12 at 17:36
• Dear Pink, yes the version with $\mathbb F_2$ replacing $\mathbb Q$ is guaranteed correct. The version with $\mathbb Q$ as in your answer is probably correct too but the proof seems a bit messy to write down . – Georges Elencwajg Aug 28 '12 at 19:54
• @julianrosen, how can we compute $p(x)^{2^{k}}$? Is some kind of Freshman's Dream for formal power series involved? Thank you. – Srinivasa Granujan Feb 25 '15 at 21:49
• @Empty Over $\mathbb{F}_2$, for each positive integer $k$, $p(x)^{2^k}=\sum_{n\geq 1} t^{2^k/n} x^{2^k n}$, which is not $0$ because the coefficient of $x^{2^k n}$ is not $0$ once $n>2^k$. – Julian Rosen Dec 11 '17 at 5:02