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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.

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    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.2012-08-28
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    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.2012-08-28
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    @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.2012-08-29
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    @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.2012-08-30

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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.

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    Thanks Georges for the paper. I've seen the proof. That's what I needed.2012-08-29
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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.

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    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.2012-08-28
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    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.2012-08-28
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    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 .2012-08-28
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    @julianrosen, how can we compute $p(x)^{2^{k}}$? Is some kind of Freshman's Dream for formal power series involved? Thank you.2015-02-25
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    @Gauloises Yes, it's not hard to check that the Freshman's Dream $(a+b)^2=a^2+b^2$, which holds in characteristic 2, extends to formal power series: $(\sum a_n x^n)^2=\sum a_n^2x^{2n}$2015-02-25
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    @JulianRosen Can you please tell me how you can say that $p(x)$ is not nilpotent in $\Bbb F_2[t,t^{1/2},...]/(t)$ ?2017-12-11
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    @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 $n>2^k$.2017-12-11
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    Got it. Thanks. !2017-12-11