If $S=R/Nil(R)$ is a quotient ring then $Nil(S)=\{0_s\}$

Question

Let $R$ be a commutative ring (why do we need this here?), $Nil(R)\trianglelefteq R$ is the subring of all nilpotent elements of $R$. Let $S=R/Nil(R)$. Prove that $Nil(S)=\{0_S\}$.

My approach:

The cosets of $S$ are of the form $r+Nil(R), \forall r\in R$. If we pick an element $r_1 \in R\backslash Nil(R)$ then $r_1+Nil(R)\in R/Nil(R)$. We can now observe that $(r_1+Nil(R))^m\ne 0$, for any integer $m>k$, where $k$ satisfies $n^k=0$. We can thus conclude that $Nil(S)=\emptyset$. But I'm probably wrong because what needs to be proved is that $Nil(S) = \{0_S\}$. But what is $0_S$ after all? I thought it should be $Nil(S)$, or is $Nil(S)=1_S$?

Would appreciate some clarification.

Answer 1

Since $R$ is commutative, then $(r + Nil(R))^m = r^m + Nil(R)$. Now if $(r + Nil(R))^m = 0$, then $r^m \in Nil(R)$ thus there exists an integer $n$ such that $r^{mn} = 0$, this means that $r \in Nil(R)$, then $r + Nil(R) = 0_S$.

Answer 2

$0_S$ is the coset $Nil(R) \in R/Nil(R)$. Your issue is that you started with $r_1 \in R \setminus Nil(R)$, rather than arbitrary $r_1$.

Suppose $[r] \in S$ is the coset containing $r \in R$. If $[r] \in Nil(S)$, then $[r]^n = [r^n] = 0_S$. That means $r^n \in Nil(R)$, and subsequently $r \in Nil(R)$, so $[r] = 0_S$.

Answer 3

Let be $I=Nil(R)$, we will prove that $Nil(R/I)=0$ :

Let be $a+I\in S$ and $(a+I)^n=0\Rightarrow a^n+I=0\Rightarrow a^n \in I$ , then exists $m$ where $(a^n)^m=a^{nm}=0 $, then $a$ inpotent and $a\in I$ ,so $a+I=I$ and $Nil(R/I)=0$