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

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?

• What is $x_n$? Did you mean $(x) + \dots + (x^n) + \dots$ ? Commented May 4, 2020 at 14:50
• @lisyarus Yes, thank you! Commented May 4, 2020 at 14:50
• 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? Commented May 4, 2020 at 15:02
• @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 $\mathfrak RN' = 0$. Commented May 4, 2020 at 15:47
• @BenSteffan $x \notin M$ but $x \in N'$. Commented May 4, 2020 at 15:59

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.

• 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. Commented May 4, 2020 at 15:35
• @BenSteffan where do you see this other version in print? Commented May 4, 2020 at 16:04
• This is from our lecture notes, which unfortunately aren't public. Commented May 4, 2020 at 16:11
• 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 $N=\{0\}$. Commented May 4, 2020 at 17:14
• 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. Commented May 4, 2020 at 18:04