Let $\frak{R}$ $(A)$ denote the nilradical of the commutative ring $A$ i.e. the ideal containing all nilpotent elements of $A$, which is also the intersection of all prime ideals of $A$.
For $A$ commutative ring with $1$, then,
$\frak{R}$$(A[x])=\bigcap\limits_{P \text{ prime}\\ \text{ in }A[x]}P\subset \bigcap\limits_{p \text{ prime}\\ \text{ in }A}p[x]$
since every prime ideal $p$ of $A$ yields a prime ideal $p[x]$ in $A[x]$ and the intersection of the bigger family of sets is contained in that of the smaller one.
$\bigcap\limits_{p \text{ prime}\\ \text{ in }A}p[x]=\big(\bigcap\limits_{p \text{ prime}\\ \text{ in }A}p\big)[x]=:\frak{R}$$(A)[x]$
since any $f:=\sum a_ix^i$ in the former set is such that $a_i\in p$ for all prime $p$ in $A$ and as such each of its co-efficients
$a_i\in \big(\bigcap\limits_{p \text{ prime}\\ \text{ in }A}p\big)$.
I have just begun self-studying Commutative Algebra from Atiyah-Macdonald and this is in aid of proof of '$\implies$' of exercise 1.2.ii (f is nilpotent $\iff$ all of its co-efficients are nilpotent). I have a different proof that involves using the reduction ring homomorphism and using that a domain doesn't contain non-zero nilpotents. But I wanted to try to figure out if the argument I have above also works? Any help will be greatly appreciated!!
