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?

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 04 '20 at 15:02
@rschwieb "$f = a_0 + a_1 x + \cdots + a_n x^n \in A[x]$ is invertible iff $a_0$ is invertible in $A$ and $a_1, \ldots, a_n$ are nilpotent", see for instance this question. — Ben Steffan, May 04 '20 at 15:06
How does the Nakayama lemma imply $M=A$? For any $0 \neq a \in \mathfrak R$, $ax \in M$ but $ax \notin A$. — jb78685, May 04 '20 at 15:17
@jb78685 The Nakayama lemma implies that if $M = N + \mathfrak{R}N'$ for $A$-modules $M, N, N'$ such that $\mathfrak{R}N'$ is finite, then we already have $M = N$. — Ben Steffan, May 04 '20 at 15:31
@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 04 '20 at 15:47

Ben Steffan · Accepted Answer · 2020-05-04T18:08:43.847

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

edited May 04 '20 at 18:08

answered May 04 '20 at 15:45

Ben Steffan

2,672

rschwieb · Answer 2 · 2020-05-04T18:22:42.733

1

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.

edited May 04 '20 at 18:22

answered May 04 '20 at 15:18

rschwieb

153,510

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 04 '20 at 15:35
@BenSteffan where do you see this other version in print? – rschwieb May 04 '20 at 16:04
This is from our lecture notes, which unfortunately aren't public. – Ben Steffan May 04 '20 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 04 '20 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 04 '20 at 18:04
@BenSteffan I'm not sure how you're reading it but that final thing is really only the third thing I mentioned as being a problem. I dismissed that you were actually using $(X^i)$ because I thought it was just a mistake of notation, and then I did mention that the product you chose is not obviously finitely generated... – rschwieb May 04 '20 at 18:11
@BenSteffan OK, y'know I have found it here: https://stacks.math.columbia.edu/tag/07RC. I've never seen that formulation before, but I see how one refines the proof to do it. I'll retract that part of my answer, then. Thanks for leading me to this conclusion. – rschwieb May 04 '20 at 18:20

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

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