While solving an exercise about invertibility of elements in a polynomial ring, I came up with the following "proof" that a polynomial is invertible if its zeroth coefficient is invertible and all higher coefficients are in the Jacobson radical:

Let $A \neq 0$ be a commutative ring with unit, $\mathfrak{N}$ its nilradical and $\mathfrak{R}$ its Jacobson radical, and consider the $A$-module $M = A + \mathfrak{R}((x) + \cdots + (x^n)) \subset A[x]$. By the Nakayama lemma, since $(x) + \cdots + (x^n)$ is finite, we obtain that $M = A$; in particular, the invertible elements of $M$ are exactly those of $A$.

This is certainly wrong, since it is well known that we in fact need a stronger condition: the higher coefficients must not only lie in $\mathfrak{R}$, but in $\mathfrak{N}$! I have, however, been unable so far to spot my mistake. Where am I going wrong?

  • $\begingroup$ What is $x_n$? Did you mean $(x) + \dots + (x^n) + \dots$ ? $\endgroup$
    – lisyarus
    May 4, 2020 at 14:50
  • $\begingroup$ @lisyarus Yes, thank you! $\endgroup$ May 4, 2020 at 14:50
  • $\begingroup$ What well-known condition are you talking about that says the coefficients must be in $\mathfrak N$, just so everyone is on the same page? $\endgroup$
    – rschwieb
    May 4, 2020 at 15:02
  • 2
    $\begingroup$ @BenSteffan That only works if $N'$ is a submodule of $M$. Otherwise, you could take $N = 0$ and $M = \mathfrak RN'$, which would imply that $\mathfrak RN' = 0$. $\endgroup$
    – jb78685
    May 4, 2020 at 15:47
  • 1
    $\begingroup$ @BenSteffan $x \notin M$ but $x \in N'$. $\endgroup$
    – jb78685
    May 4, 2020 at 15:59

2 Answers 2


It does not have to be the case that $\mathfrak{R}((x) + \cdots + (x^n))$ is finite, so Nakayama is not applicable here in general.

There is a second, much more serious error, however: for the argument to work, one would need that $M = A + \mathfrak{R} \cdot \mathfrak{R}((x) + \cdots + (x^n))$, but that does not follow from $M = A + \mathfrak{R}((x) + \cdots + (x^n))$!


consider the $A$-module $M = A + \mathfrak{R}((X) + \cdots + (X^n)) \subset A[X]$.

I will suppose you mean $AX^i$ rather than $(X)^i$, which would be a much larger subset of $A[X]$, and certainly not finitely generated over $A$. Even then with this interpretation the product with the radical is not obviously finitely generated.

But I think you have a bigger problem. What you want to consider is $M'= \sum_{i=1}^n AX^i$, but the problem is that Nakayama's Lemma says:

Let $N$ be a submodule of $M$ such that $M=N+\mathfrak R M$.

The way you've picked $N$, it is not a submodule of the thing multiplied by $\mathfrak R$. You've got something of the form $M=A+\mathfrak R M'$ where $M'\neq M$, so Nakayama doesn't seem to apply.

EDIT: nope, never mind, there is a version of Nakayama's Lemma that I hadn't seen. Right now it is item #4 here.

  • $\begingroup$ To the best of my knowledge, that is not correct: Nakayama already tells you that if you have $M = N + \mathfrak{R}N'$, and $\mathfrak{R}N'$ is finite, then $M = N$. No further requirements on $N, M$ or $N'$ are posed. $\endgroup$ May 4, 2020 at 15:35
  • $\begingroup$ @BenSteffan where do you see this other version in print? $\endgroup$
    – rschwieb
    May 4, 2020 at 16:04
  • $\begingroup$ This is from our lecture notes, which unfortunately aren't public. $\endgroup$ May 4, 2020 at 16:11
  • $\begingroup$ But the claim in your first comment is absurd because of the following example: let $R$ be a Noetherian ring with nonzero Jacobson radical, and set $N=\{0\}$ and $N'=R$. Your conclusion is that $M=\{0\}+J(R)$ is the same as $N=\{0\}$. $\endgroup$
    – rschwieb
    May 4, 2020 at 17:14
  • $\begingroup$ Sorry, I did forget to write that $N, N' \subset M$ need to be submodules, which should rule that case out. I'm not entirely sure whether that allows you to salvage the argument from your answer. $\endgroup$ May 4, 2020 at 18:04

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