Polynomial ring $R[X] $ has zero Jacobson radical, for $R$ a domain Let $R$ be a commutative domain. 
Prove that the Jacobson radical of $R[X]$, i.e. the intersection of all maximal ideals, is the zero ideal.
Thank you.
 A: Exercise 1.4 of Atiyah - Macdonald tells you that in any polynomial ring $R[x]$, the Jacobson radical and nilradical are equal. For your problem let us throw in the additional hypothesis that $R$ is an integral domain. Then the nilradical of $R[x]$ is zero because $R[x]$ is an integral domain and hence the Jacobson radical is zero.
A: Hint $\rm\ f\in J(R[x])\Rightarrow f\:$ in all max $\rm M$ $\Rightarrow\:\overbrace{\rm 1\! +\! x\:\!f\:\text{ in no max}\ \rm M}^{\small\textstyle \rm 1\!+\!x\:\!f,f\in M\Rightarrow 1\in M}\Rightarrow \rm 1\!+\!x\:\!f\,$ unit $\rm \Rightarrow f = 0$
Remark $ $ More generally this shows that $\rm R[x]$ has Jacobson radical equal to its nilradical: the above shows $\rm\ f\in J(R[x])\Rightarrow 1+x\:\!f\,$ a unit, so by here all coef's of $\rm\,f\,$ are in $\rm\sqrt R$ so $\rm\,f\in \sqrt{R[x]}$
Perhaps the following is of interest, from my post giving a constructive generalization of Euclid's proof of infinitely many primes (for any ring with fewer units than elements).
Theorem $ $ TFAE in ring $\rm\:R\:$ with units $\rm\:U,\:$ ideal $\rm\:J,\:$ and Jacobson radical $\rm\:Jac(R)$
$\rm(1)\quad  J \subseteq Jac(R),\quad $ i.e.  $\rm\:J\:$  lies in every max ideal $\rm\:M\:$ of $\rm\:R$
$\rm(2)\quad  1+J \subseteq U,\quad\  $  i.e. $\rm\: 1 + j\:$  is a unit for every $\rm\: j \in J$
$\rm(3)\quad  I\neq 1\ \Rightarrow\  I+J \neq 1,\qquad $  i.e.  proper  ideals survive in $\rm\:R/J$
$\rm(4)\quad M\:$ max $\rm\:\Rightarrow  M+J \ne 1,\quad $  i.e. $ $ max $ $ ideals $ $ survive $ $ in $\rm\:R/J$
Proof $\: $ (sketch) $\ $  With  $\rm\:i \in I,\ j \in J,\:$ and max ideal $\rm\:M,$
$\rm(1\Rightarrow 2)\quad  j \in all\ M\ \Rightarrow\ 1+j \in no\ M\ \Rightarrow\ 1+j\:$ unit.
$\rm(2\Rightarrow 3)\quad i+j = 1\ \Rightarrow\ 1-j = i\:$ unit $\rm\:\Rightarrow  I = 1\:.$
$\rm(3\Rightarrow 4)\ \:$  Let $\rm\:I = M\:$ max.
$\rm(4\Rightarrow 1)\quad  M+J \ne 1 \Rightarrow\ J \subseteq M\:$  by  $\rm\:M\:$ max.
