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.

  • $\begingroup$ 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. $\endgroup$ – Dylan Moreland Aug 28 '12 at 17:02
  • 1
    $\begingroup$ 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. $\endgroup$ – Jack Schmidt Aug 28 '12 at 17:16
  • $\begingroup$ @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. $\endgroup$ – fmoura2005 Aug 29 '12 at 14:01
  • 1
    $\begingroup$ @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. $\endgroup$ – Dylan Moreland Aug 30 '12 at 17:36

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.

  • $\begingroup$ Thanks Georges for the paper. I've seen the proof. That's what I needed. $\endgroup$ – 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.

  • 3
    $\begingroup$ 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. $\endgroup$ – Georges Elencwajg Aug 28 '12 at 17:08
  • 6
    $\begingroup$ 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. $\endgroup$ – Julian Rosen Aug 28 '12 at 17:36
  • $\begingroup$ 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 . $\endgroup$ – Georges Elencwajg Aug 28 '12 at 19:54
  • $\begingroup$ @julianrosen, how can we compute $p(x)^{2^{k}}$? Is some kind of Freshman's Dream for formal power series involved? Thank you. $\endgroup$ – Srinivasa Granujan Feb 25 '15 at 21:49
  • 2
    $\begingroup$ @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 $n>2^k$. $\endgroup$ – Julian Rosen Dec 11 '17 at 5:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.