In Commutative Algebra we have the following standard facts which I am going to state in a slightly different form than usually found in textbooks. Namely, let $A$ be a commutative unital ring of characteristic 0 and let $f(x) \in A[x]$, then:

  1. $f(x) \in U(A[x])$ $\Leftrightarrow$ $f(0) \in U(A)$ and $f'(x) \in \operatorname{nil}(A[x])$;
  2. $f(x) \in \operatorname{nil}(A[x])$ $\Rightarrow$ $f'(x) \in \operatorname{nil}(A[x])$;

It is a natural question to ask whether something similar holds if we replace $A$ by a noncommutative unital ring $R$ and consider $R[x]$ ($x$ is assumed to commute with $R$). Of course, this would be much less straightforward because the nilpotent elements of a noncommutative ring in general do not form an ideal. The only result I know of along these lines is the following

Lemma: Let $R$ be a (noncommutative) $\mathbb{N}_0$-graded ring and let $r \in R$ be an element of positive degree. Then $1+r$ is a unit iff $r$ is nilpotent.

But I have not been able to trace this result throughout the literature. It is stated in Bass, Connell, and Wright's paper on the Jacobian Problem, but without proof from what I can see, likely because it is not too difficult to prove on one's own. Nevertheless, I would like to have a precise reference for this, because I would like to use it, but would rather avoid including a proof of what appears to be a standard result in noncommutative graded algebra.

Question 1: Do you know of a reference where the above lemma is stated with a proof?

And more importantly:

Question 2: Are there generalizations of the above lemma, i.e. is anything known about units of the form $1 + r_1 + \cdots + r_n$ for $r_1,\dots,r_n \in R$ homogeneous elements of respective degrees $1 \leq d_1 < \dots < d_n$?

I fear this might be too broad of a question or one that is likely impossible to answer in full generality, but I would be happy with some starting places. The motivating example is $U(R[x])$ for $R$ of characteristic 0, for which I would like to find out about analogous properties to the aforementioned commutative algebra facts.

  • $\begingroup$ Your "standard facts " are not correct. In $\mathbb{F}_p[x]$, $1+x^p$ satisfies the hypotheses of 1. but is not invertible, and $g(x)=x^p$ is not nilpotent though $g'(x)=0$. $\endgroup$
    – abx
    Commented Dec 16, 2022 at 7:34
  • $\begingroup$ However, you can easily fix (1) either by replacing $f'$ by $f-f(0)$, or by making it a one-sided implication "$\Rightarrow$" like in (2). $\endgroup$ Commented Dec 16, 2022 at 9:23
  • $\begingroup$ @abx: My apologies, I implicitly meant characteristic 0. I will fix that immediately. $\endgroup$
    – M.G.
    Commented Dec 16, 2022 at 12:48
  • 2
    $\begingroup$ In your lemma you don't need $r$ to have positive degree. Rather, you need the grade-$0$ component of $r$ to be zero. For question 2, you might look at some of the work Agata Smoktunowicz did on graded nil rings. Really weird things can happen. $\endgroup$ Commented Dec 16, 2022 at 16:08
  • $\begingroup$ @PaceNielsen: Thanks for the reply! I basically copied the lemma from the BCW's paper. If I understand you correctly, you are saying that the non-trivial direction of the equivalence in the lemma also holds for non-homogeneous elements $r$ as long as all the components are of positive degree? $\endgroup$
    – M.G.
    Commented Dec 16, 2022 at 16:14

1 Answer 1


The lemma for homogeneous $r$ is easy, so that there is no needs for special reference.

Proof: Let $(1+r)(1+y_1+\dots +y_n)=1$ in an associative graded ring, where $\deg r = d>0$, $\deg y_i=i$, $y_i$ are homogeneous. Then $y_i+ry_{i-d}=0$ for all $i$ (we assume here $y_0 =1$ and $y_t=0$ for $t$ not in $[0,n]$), so, the only nonzero $y_i$ with $i>0$ are $y_d = -r$, $y_{2d } =r^2$, etc. Moreover, for the last nonzero $y_{md} = \pm r^d$ we get $0 = r y_{md} = \pm r^{m+1}$.

  • 1
    $\begingroup$ Dear @Dmitri Piontkovski, thanks, but, as stated in my original post, I am aware that the proof is not difficult. I was looking for an actual physical reference with the proof stated rather than the proof itself. I guess, you are right that whether I include a proof or not, it's a non-issue either way. $\endgroup$
    – M.G.
    Commented Jan 19, 2023 at 23:45

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