When is $rad(L)[x_1,\ldots]$ radical in $Ker(\varphi_\ast)$?

Question

Suppose we have a local ring $L$ (not necessarily commutative) such that $L/rad(L)$ is a division algebra (here $rad(L)$ is the Jacobson radical of $L$). We clearly have the canonical surjection $\varphi:L\to L/rad(L)$ from which we may induce the surjection $\varphi_\ast:L[x_1,\ldots,x_n]\to(L/rad(L))[x_1,\ldots,x_n]$ and where $Ker(\varphi_\ast)=rad(L)[x_1,\ldots,x_n]$. Now, my question is this: is $Ker(\varphi_\ast)$ radical in $L[x_1,\ldots,x_n]$ in general? If not, are there conditions upon $rad(L)$ that can ensure $Ker(\varphi_\ast)$ is radical?

Answer 1

Consider $L=\mathbb{Z}_{(2)}$ (the set of rational numbers which in reduced form have denominators coprime to $2$). This is a local ring with ${\rm rad}(L)=2L$ and $L/2L\cong \mathbb{F}_2$. However $2L[x_1,\ldots,x_n]$ is not contained in the Jacobson radical of $L[x_1,\ldots, x_n]$. For instance, $1+2x_1$ is not a unit (by a simple degree argument).

A sufficient condition for ${\rm rad}(L[x_1,\ldots, x_n])\supseteq {\rm rad}(L)[x_1,\ldots,x_n]$ is that ${\rm rad}(L)$ is nilpotent. It is necessary that ${\rm rad}(L)$ be nil.

Proof: (Sufficient): If ${\rm rad}(L)$ is nilpotent, then so is ${\rm rad}(L)[x_1,\ldots, x_n]$, hence is contained in the Jacobson radical.

(Necessity): Suppose ${\rm rad}(L)[x_1,\ldots, x_n]\subseteq {\rm rad}(L[x_1,\ldots, x_n])$. Fix $a\in {\rm rad}(L)$. Then $1-ax_1$ is a unit in $L[x_1,\ldots, x_n]$. In the overring $L[[x_1,\ldots, x_n]]$, we know that $1-ax_1$ has an inverse $1+ax_1+a^2x_1^2+\cdots$. But since $1-ax_1$ has a polynomial inverse, and inverses are unique, we must have that $1+ax_1+a^2x_1^2+\cdots$ is a polynomial. In other words, $a^n=0$ for some $n\geq 0$. Thus every element of ${\rm rad}(L)$ is nilpotent, hence it is a nil ideal.