I have a unital commutative ring $A$. And we know that an element $a \in A$ is said to be nilpotent if $a^n=0$ for some $n \in \mathbb{R}$. Prove the following assertions:
a) $A$ does not contain nonzero nilpotent elements if and only if zero is the unique element in $A$ whose square is zero.
Here, what've done is:
$\Rightarrow$: Trivial. We know $ \nexists\ 0\neq a \in A $ such that $a^n=0$ fore some $n \in \mathbb{N}$, in special for n=2.
$\Leftarrow$: $0$ is the unique element in $A$ whose square is $0$. Then $ \exists\ 0\neq a \in A $ such that $a^n=0$ fore some $n \in \mathbb{N}$ ?
We know that if $a \neq 0$, $a\in A$, then $a^2\neq 0$. Suppose $a^n=0$ with $a\neq 0$
Notice that $n \in \mathbb{N}$, so because of the algorithm division we can write $n$ as $n=n_1\cdot 2+r$ where $r<2$ then, $r=1$ or $r=0$. In both cases we can write
$0= a^n=a^{2n_1+r}=a^{2n_1}a^r=(a^{n_1})^2a^r$
where $a^r\neq 0$ (because if $r=1$, then $a^r=a$, if $r=0$ then $a^0=1_A$ so $a^{n_1}=0$ and we can write $n_1$ as: $n_1=n_2\cdot2+r_1$ where $r_1<2$ then, $r_1=1$ or $r_1=0$ and continue as before.
Whe now that it will have a finite number of steps because $n_k<n_{k-1}<\ldots<n_2<n_1<n$ for some $k$ that:
$n_k=2\cdot 1+r_k$ with $r_k<2$
Then $0= a^n_k=a^{2\cdot1+r_k}=a^{2}a^{r_k}$ and we know $a^{r_k}\neq 0$ so $a^2=0$ CONTRADICTION with the hipothesis. So $a=0$ if $a^n=0$, $a\in A$
QED.
It is well proved? I don't know that if I do the "continue as before" is ok with the proof.
Thank you.
Best Answer
I think the nontrivial part can be more or less as you did but with the following fixing: suppose $\;0\neq a\in A\;$ is such that $\;a^{2n+1}=0\;$ and this is the lowest possible exponent (the nilpotency degree or something), but then
$$\left(a^{n+1}\right)^2=a^{2n+1}a=0\cdot a=0\implies a^{n+1}=0$$
and we get contradiction to nilpotency degree.