Nilpotent elements in ring homomorphism

Question

Let $f:R\rightarrow S$ be a ring homomorphism, if $x\in R$ is a nilpotent element, then $f(x)$ is also nilpotent with a nilpotent degree smaller/equal to the nilpotent degree of. That statement is easy to proof and I dont have a problem with it. My question is that: Is the nilpotent degree of $f(x)$ divides the nilpotent degree of $x$?

Answer 1

No, not necessarily. Take $k[x]/(x^3)\to k[y]/(y^2)$ defined by sending $x\mapsto y$. Then $x^3=0$, while $y^2=0$ but $2\nmid 3$.