The Jacobson radical of polynomial ring

Question

Let $R$ be a commutative ring with $1$ and $R[x]$ be the polynomial ring over $R$. We know that the Jacobson radical $J(R[x] )$ of $R[x]$ is the intersection of all maximal ideals of $R[x]$. Now let $M$ be a fixed (but arbitrary) maximal ideal of $R[x]$. How can we show that $J(R[x])=\cap_{M\not=N\in \mathrm{Max}(R[x])}N$, where $\mathrm{Max}(R[x])$ is the set of all maximal ideals of $R[x]$?

Answer 1

What you want to show is the following:

$f\not \in M \implies f\not \in N $ for some maximal ideal $N\neq M$. $(*)$

Let $M'=R\cap M$, $R'=R／M'$ and $f'=[f]\in R'$. Then any maximal ideal in $R'[x]$ not containing $f'$ correspond to a maximal ideal in $R[x]$ not containing $f$. So if we can prove $(*)$ for the ring $R'$, then $(*)$ is also true for $R$. Hence we may assume $R$ is an integral domain, and $R\cap M=0$.

Note that we already have a contradiction if $R$ is a field, this follows from that $R[x]$ has infinitely many prime elements, and only a finite number of them can divide $f$. So we may assume $R$ is not a field, in particular, we assume $R$ is infinite.

Let $K=R[x]/M$, regard $R$ as a subring of $K$. Suppose $M$ is the kernel of the surjective map $R[x]\to K$ sending $x$ to $\alpha \in K$. For any $r\in R$, let $M_r$ be kernel of the surjective map $R[x] \to K$ sending $x$ to $\alpha+r$ (and fix $R$). Since $R$ is infinite, we may can $r$ so that $\alpha+r$ is not a conjugate of $\alpha$, and $f(\alpha+r)\neq 0$. Then $M_r\neq M$ is another maximal ideal not containing $f$.