My question concerns some steps of a proof. Let's start with some definitions.
Definition 1. Let $R$ a ring, an element $x\in R$ is said to be nilpotent if exists $n\in\mathbb{N}\setminus\{0\}$ such that $x^n=0$.
Definition 2. An ideal $I$ of a ring $R$ is said to be nil ideal if each element $x\in I$ is nilpotent.
Definition 3. An ideal $I$ of a ring $R$ is said nilpotent if $I^n=\{0\}$ for same $n\in\mathbb{N}\setminus\{0\}$.
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Note: If $f\colon R\to R'$ is a morfism of rings and $a\in R$ is nilpotent, then $f(a)$ is nilpotent.
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Proposition. Let $f\colon R\to R'$ a morfism of rings. It seems that:
$1.$ If $R$ is an nil-ring (nilpotent ring), then each subring $S\subseteq R$ is an nil-ring (nilpotent-ring). Moreover, too $f(R)$ is a nil-ring (nilpotent-ring).
$2.$ If $R$ contains an ideal $I$ such that $I$ and $R/I$ are both nil (nilpotent), then $R$ is a nil (nilpotent) ring. $$$$
proof. $1.$ If $R$ is an nil-ring, then for all $x\in R$, $\exists n\in\mathbb{N}\setminus\{0\}$ such that $x^n=0$. Let $S\subseteq R$ a subring. Let $y\in S$, then, in particular, $y\in R$, therefore $\exists m\in\mathbb{N}\setminus\{0\}$ such that $y^m=0$. Then $S$ is a nil- ring. Moreover, for the note $f(x)^n=0_{R'}$, then $f(R)$ is a nil-ideal.
Questioin 1. How can I proceed in the event that $R$ is nilpotent?
$2.$ If $I$ and $R/I$ are nil rings and $a\in R$. Then, exists same positive integer $n$ for which $$(a+I)^n=a^n+I=I,$$ this means that $a^n\in I$. But, for hypotesis, $I$ is a nil-ideal, then exists $m$ such that $(a^n)^m=a^{nm}=0$, for some $m\in\mathbb{Z}_+$. Then $a$ is nilpoten as a member of $R$, in consequence $R$ is a nil ring.
Question 2. How can I proceed in the event that $I$ and $R/I$ are both nilpotent?