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This should be a pretty easy problem, but I'm a dummy so I'm stuck. Here's the statement:

Let $R$ be a $\mathbb Z$-graded ring, and $M$ a graded $R$-module, and let $x \in R_k$ for some non-zero integer $k$. Then $u = 1-x$ is not a zero divisor. Show that $u$ is a unit if and only if $x$ is nilpotent.

Now I know that a similar question has been asked here many times before, so let me say that I know how to show $u$ is not a zero divisor, and I can show that if $x$ is nilpotent, $u$ is a unit. This is easy and has been done on this site a million times. My struggle is in the converse, that is to say, if $u$ is a unit, then I want to prove that $x$ is nilpotent.

Apologies if this has also already been done on this site, but I can't seem to find the question on hand.

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    I suppose you don't need a graded module to state the question. – 2017-02-03

1 Answers 1

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Assume that $$(1-x)y=1$$ and let $y=\sum y_i$ be a sum of homogeneous elements $y_i$ then we have $$\sum y_i-xy_i=1$$ Now we see that $$y_0=1$$ and that $$y_{i+k}=xy_i$$

Since the sum is finite, $x$ is nilpotent.

Note that if $-l$ is the smallest negative index where $y_{-l}$ is non zero then (assuming, $k$ positive by symmetry) we have $$y_{-l}+xy_{-l}=0$$ and since these have different degrees $y_{-l}=0$.

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    This seems to only work if $R_n=0$ for negative $n$, no? – 2017-02-03
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    Yeah, I guess I was implicitly assuming that, force of habit. – 2017-02-03
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    No wait it follows. – 2017-02-03
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    @ReneSchipperus is the argument that the sum has to be telescoping? I'm not sure I see why that's necessary, although I can see that it would be sufficient. – 2017-02-03
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    @ReneSchipperus nice catch! – 2017-02-03
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    The argument is $y_0=1$ so $y_k=x$ and $y_{2k}=x^2$ etc. Now for large values all $y_n=0$ so a power of $x$ is zero. – 2017-02-03
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    @ReneSchipperus Ah I see now. Thanks. I was really close, I just forgot the fundamental property of the grading - $R_iR_j \subset R_{i+j}$! It was in fact, an easy question. – 2017-02-03