Polynomial ring $R[X] $ has zero Jacobson radical, for $R$ a domain

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.

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.