$R$ is a commutative ring and $$NP = \{a\in R|a^{n}=0\}$$
Why there is closure to substraction ?
Let $a,b\in NP| a^{n}=0, b^{m}=0$, therefore I want to prove that $(a-b)^{nm}=0$. $$$$ Using the binomeal theorem: $(a-b)^{nm}=\left(^n_0\right)a^{nm}+\left(^n_1\right)a^{nm-1}(-b)+...+\left(^n_n\right)b^{nm}$, so from here the first and the last ae in nilpotent, but is there a reason I can say the whole sum is in nilpotent ?
I do want to say that for any of the sum elements: $$\left(^n_i\right)a^{nm-i}(-b)^{i}=\left(^n_i\right)a^{n^{m}}(a^{-i})(-b)^{i}$$ and then because of the absorbtion (which is easier to show). But what bothers me is that writing $a^{-i}$ is meanless because nobody said that it's a division ring (and it must be so this algebraic expression exist and will be meaningful, isn't it?).
The collection $\mathfrak{R}$ of all nilpotent elements in a commutative ring $A$ is indeed an ideal, called the nilradical of $A$.
To prove this, consider $a, b \in \mathfrak{R}$. That is, $a^n = b^m = 0$. Then expanding $(a + b)^{m+n - 1}$ by bionomial theorem, you get a sum whoose terms are $a^p b^q$ where $p + q = m + n - 1$. Clearly one cannot have both $p < m$ and $q < n$, as otherwise $p + q$ would be smaller than $m + n - 1$. Thus, each term vanishes and $(a + b)^{m+n - 1} = 0$. Hence $a + b$ is also a nilpotent element.
It is clear that if $a \in \mathfrak{R}$, $c \in A$ an arbitrary element, then $ac$ is also nilpotent, as $a^n = 0$ implies $(ac)^n = a^n c^n = 0$.
Now specialize to the case $a = x, b = (-1)y$, and you get closure under subtraction.
As a side-note, the nilradical $\mathfrak{R}$ of $A$ is also the intersection of all primes ideals in $A$.