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Let $A$ be a commutative ring. I'm trying to prove that in $A[x]$, the Jacobson radical $\mathcal{J}$ is a subset of the nilradical $\mathcal{P}$. This is an exercise from Atiyah & MacDonald

Let $a_0+a_1x+a_2x^2+\dots +a_nx^n\in\mathcal{J}$. Then $$1-(a_0+a_1x+a_2x^2+\dots+ a_nx^n)y,$$ is a unit for all $y\in A[x]$. Putting $y=1$, we get that $$(1-a_0)-a_1x-\dots-a_nx^n,$$ is a unit. This is only possible if $(1-a_0)$ is a unit and $a_i$ are nilpotent for $i\geq 1$.

If I can somehow prove that

$$(1-a_0)\text{ is a unit} \implies a_0\text{ is nilpotent}$$ then I'll be done. This is because if all coefficients $a_0,a_1,\dots, a_n$ are nilpotent, then the polynomial $a_0+a_1x+a_2x^2+\dots +a_nx^n$ too will be nilpotent. However, I'm having problems proving this.

Is my assertion even true? A preliminary investigation shows that it may be false, but I'm not sure. If it is, any helpful hints as to how to prove it would be great.

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    $\begingroup$ This is false; take for example $a_0=2$ where $A$ is a ring that is not of characteristic $2$. $\endgroup$
    – Servaes
    Commented Jun 13, 2014 at 11:08

1 Answer 1

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HINT: In stead of putting $y=1$, try putting $y=x$.

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  • $\begingroup$ Oh my god this is brilliant! Did you come up with it just like that? I've been working on this problem since last night! $\endgroup$
    – freebird
    Commented Jun 13, 2014 at 11:27
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    $\begingroup$ I recall doing the exact same exercise two years ago ;) $\endgroup$
    – Servaes
    Commented Jun 13, 2014 at 11:29
  • $\begingroup$ @freebird Same as this answer yesterday. $\endgroup$
    – Bill Dubuque
    Commented Jun 13, 2014 at 13:41

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