# 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 $(x) + \dots + (x^n) + \dots$ ? – lisyarus May 4 at 14:50
• @lisyarus Yes, thank you! – Ben Steffan May 4 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? – rschwieb May 4 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 $\mathfrak RN' = 0$. – jb78685 May 4 at 15:47
• @BenSteffan $x \notin M$ but $x \in N'$. – jb78685 May 4 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. – Ben Steffan May 4 at 15:35
• @BenSteffan where do you see this other version in print? – rschwieb May 4 at 16:04
• This is from our lecture notes, which unfortunately aren't public. – Ben Steffan May 4 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 $N=\{0\}$. – rschwieb May 4 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. – Ben Steffan May 4 at 18:04