# Condition so that coefficients nilpotent implies power series nilpotent for non-noetherian ring $R$ with char$(R) = 0$

For a noetherian ring $R$ the following holds:

$$\sum\limits_{i=0}^{\infty}a_iX^i \mbox{ nilpotent } \iff a_i \mbox{ nilpotent } \forall i, \mbox{ where } a_i \in R.$$ If $R$ is non-noetherian $\implies$ still holds, but $\impliedby$ does not. However, under certain conditions $\impliedby$ does hold. For example, if char$(R) = p >0$, then the condition $$\exists N \in \mathbb{N}: \forall i: a_i^N=0$$ is a necessary and sufficient condition for $\impliedby$ to hold.

However, what about the case char$(R) = 0$? Is there a similar necessary and sufficient condition that can give us the desired implication?

• Have you already read "Zero divisors and nilpotent elements in power series rings" (D. E. Fields) and "On nilpotent power series with nilpotent coefficients" (T. K. Kwak, Y. Lee)? – Martin Brandenburg Oct 20 '13 at 12:47
• @MartinBrandenburg I had already read the first one, which confirmed the things I mentioned in my post but didn't answer the question. I've just read the second one, but again it only confirms my post (unless I misread something). – user49719 Oct 20 '13 at 15:32
• So you already know that if $\langle a_0,a_1,\dotsc \rangle$ is nilpotent, then $f$ is nilpotent? This is part of Theorem 2 in Fields paper. Probably you want a weaker assumption? – Martin Brandenburg Oct 20 '13 at 17:15
• Sorry, I should've specified that I was looking for a necessary and sufficient condition - I'm looking to describe "what the nilpotent elements of R[[X]] look like", so to speak. Theorem 2 gives a sufficient but not necessary condition, and I believe Corollary 1 does too (where the sufficient but not necessary part is that $A_f$ has to be finitely generated). – user49719 Oct 21 '13 at 6:16