Invertibility of elements in $A[x]$ with coefficients in the Jacobson radical

Question

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?

Answer 1

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 $M = A + \mathfrak{R}((x) + \cdots + (x^n))$!

Answer 2

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.