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Suppose $R$ is a commutative unital ring with identity $1$ such that the equation $nx = 1$ has a unique solution for each integer $n \ge 1$, and let $\xi$ be a nilpotent element of $R$ with nilpotency index $v$. Then fix a positive integer $k$ and set $a = \sum_{i=0}^{v-1} \binom{1/k}{i} \xi^i$. Is it true that $a^k = 1 + \xi$?

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This is quite interesting as it shows that every unipotent element of a commutative ring has roots of arbitrary order. I haven't found this in the literature. The following answer is quite long and consists of three parts, as it evolved through an attempt to make the proof as natural as possible. Those interested only in the proof can skip the second part and parts of the first one.

Of course we would like to apply the binomial series and write $a=a+\xi^v = (\xi+1)^{1/k}$, but the exponent here does not make sense in an arbitrary ring. However, with a pinch of abstraction, we can make this idea work:

It is enough to consider the universal commutative $\mathbb{Q}$-algebra containing a nilpotent element of nilpotency index $v$. It is given by $\mathbb{Q}[\xi]/(\xi^v)$. The claim holds for every $v$ if and only if it holds in the projective limit ${\varprojlim}_v ~ \mathbb{Q}[\xi]/(\xi^v)$. It is isomorphic to the ring of formal power series $\mathbb{Q}[[\xi]]$ in one variable. Here $a$ becomes the power series $\sum_{i=0}^{\infty} \binom{1/k}{i} \xi^i$, and it suffices to check that $a^k=1+\xi$ as formal power series, i.e. that the coefficients coincide. But these are the same coefficients as in the well-known equality of analytic power series $(\sum_{i=0}^{\infty} \binom{1/k}{i} z^i)^k = ((1+z)^{1/k})^k=1+z$ for $z \in \mathbb{C}$, $|z|<1$. This proves the claim.

Somehow this is magic, but it works!


Here is an attempt to make proof above a bit more explicit. We calculate

$$a^k = \sum_{i_1,\dotsc,i_k \geq 0} \binom{1/k}{i_1} \cdot \dotsc \cdot \binom{1/k}{i_k} \cdot \xi^{i_1+\dotsc + i_k}=\sum_{p=0}^{v-1} \lambda_p \cdot \xi^p,$$ where $$\lambda_p := \sum_{i_1+\dotsc+i_k=p} \binom{1/k}{i_1} \cdot \dotsc \cdot \binom{1/k}{i_k} \in \mathbb{Q}.$$ Clearly $\lambda_0=1$ and $\lambda_1 = \sum_{i=0}^{k} \binom{1/k}{1}=k \cdot \frac{1}{k}=1$. We have to show $\lambda_p=0$ for all $p>1$. This has nothing to do with $R$ anymore, it is just an equality of rational numbers. As above one can prove this looking at the coefficients of the binomial series. But a direct proof should also be possible. For example,

$$\lambda_2=\sum_{i=1}^{k} \binom{1/k}{2} + \sum_{1 \leq i<j \leq k} \binom{1/k}{1} \cdot \binom{1/k}{1}=k \cdot \binom{1/k}{2} + \binom{k}{2} \cdot \frac{1}{k^2}=\frac{ \frac{1}{k}-1}{2}+\frac{1- \frac{1}{k}}{2}=0.$$


More generally, let $R$ be a commutative $\mathbb{Q}$-algebra, $r \in R$, and $f \in R[[x]]$ a formal power series with $f \equiv 1 \bmod x$. Then define the power $$f^{r} := \sum_{n=0}^{\infty} \binom{r}{n} \cdot (f-1)^n.$$ The binomial theorem implies that the power agrees with the usual power when $r \in \mathbb{N}$. Recall that the binomial coefficient is defined recursively by $\binom{r}{0}=1$ and $\binom{r}{n+1}=\frac{r-n}{n+1} \cdot \binom{r}{n}$.

Proposition. For $r,s \in R$ we have $f^{r+s}=f^r \cdot f^s$.

Proof. It suffices to prove that $\binom{r+s}{n}= \sum_{p+q=n} \binom{r}{p} \cdot \binom{s}{q}$. Both sides are polynomials in $r,s$. Therefore it suffices to prove this in the universal example $R=\mathbb{Q}[x,y]$ with $r=x$, $s=y$. The polynomials on both sides agree when evaluated at $\mathbb{N} \times \mathbb{N}$ by Vandermonde's identity. This implies that they are equal. For a more direct proof, see here. QED

From this Proposition it follows by induction that $f^{r_1+\dotsc+r_k}=f^{r_1} \cdot \dotsc \cdot f^{r_k}$, in particular that $f = (f^{1/k})^k$ for $k \in \mathbb{N}$. For $f=1+x$ this is exactly what we wanted. QED

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Thanks for your answer. While I agree that the natural setting to deal with the question is the ring of power series in one variable $x$ over the rationals, and that the formal sum $f(x)$ defined by $\sum_{i=0}^\infty \binom{1/k}{i} x^i$ satisfies the identity $f(x)^k = 1 + x$ (taken in $\mathbb Q[x]$), I don't understand how this can be translated into a genuine identity between elements of $R$ (the unital ring in the OP) by replacing $x$ with a nilpotent $\xi \in R$. It may be obvious, but I'm not very practical with the analysis of convergence of formal power series, so please let me ask. –  idle questioner Mar 19 '13 at 11:44
    
Btw, I think that proving $f(x)^k = 1 + x$ in $\mathbb Q[x]$ does already require all sort of algebraic manipulations needed to prove that $a^k = 1 + \xi$ directly in $R$. Right? –  idle questioner Mar 19 '13 at 11:57
    
I have added several explanations and alternative proofs. –  Martin Brandenburg Mar 19 '13 at 19:24
    
Thank you, Martin, this is finally very clear. Btw, I wrote $\mathbb Q[x]$ in the above comments, but it's just a reiterated typo. –  idle questioner Mar 19 '13 at 19:53

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