# The Jacobson radical of a polynomial ring is contained in the nilradical.

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.

• This is false; take for example $a_0=2$ where $A$ is a ring that is not of characteristic $2$. – Servaes Jun 13 '14 at 11:08

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