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?



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


