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.

Solution 1:

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.

Solution 2:

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.