A ring $R$ is said to have a bounded index (of nilpotency) if there is a positive integer $n$ such that $x^n=0$ for every nilpotent $x∈R$.
Can anyone give me an example of a ring $R$ which has a bounded index, but $R[x]$ has no bounded index? Same question for $M_n(R)$ instead of $R[x]$.
Just a particular case showing that the property transfers from $R$ to $R[x]$.
$n=2$ (where $n$ denotes the bounded index of $R$):
A polynomial $f\in R[x]$ is nilpotent iff its coefficients are nilpotent. We prove that in such case $f^2=0$. Write $f=a_0+a_1x+\cdots+a_mx^m$ with $a_i\in R$ nilpotent. The coefficients of $f^2$ are of the form $\sum_{i+j=k}a_ia_j$. If into the product $a_ia_j$ we have $i=j$, then $a_ia_j=0$. Otherwise recall that $a_i^2=a_j^2=(a_i+a_j)^2=0$, so $2a_ia_j=0$. (Here I've used that the sum of two nilpotent elements is also nilpotent.) But in our sum appears $a_ia_j$ and also $a_ja_i$, that is, $2a_ia_j$, and we are done.
For the general case we can prove that $f^t=0$ where $t=1+n(n-1)/2$.