every integral domain has no non zero nilpotent

Question

So this is what I've done so far. Let $R$ be an arbitrary integral domain, and suppose to the contrary it has a non zero nilpotent say $x$; then $x^n=0$ for some natural number $n$, but this implies $x \cdot x^{n-1}=0$, and since $x$ is not equal to zero and $R$ is an integral domain, we must have $x^{n-1}=0$, but again we can write $x^{n-1}$ as $x \cdot x^{n-2}$, and so repeating inductively eventually we get $x^2=0$, and so $x=0$, contradicting how we chose $x$; hence $R$ has no non-zero nilpotent. but since $R$ was an arbitrary integral domain, we conclude that any integral domain has no non zero nilpotent. Just wanted to check whether the proof seemed alright.

Answer 1

Of course, our OP lard's proof is more or less conceptually correct--lard has the right idea-- although a little more care might be taken to handle cases like $xx^{n - 1} = 0$ with $n =1$, since $x \ne 0$ implies $1 = x^0 = 0$, a contradiction indicating $n = 1$ might require special attention.

One might also argue that, if $x \in R$ is nilpotent,

$\exists k \in \Bbb N, \; x^k = 0; \tag 1$

let $m \in \Bbb N$ be the minimum amongst all such $k$; if

$m = 1, \tag 2$

then

$x = x^1 = 0, \tag 3$

and we are done; if

$m \ge 2, \tag 4$

we may write

$xx^{m - 1} = x^m = 0; \tag 5$

but the minimality of $m$ implies

$x^{m - 1} \ne 0; \tag 6$

since $R$ is an integral domain, it has no non-zero zero divisors, and thus we conclude from (5), (6) that

$x = 0. \tag 7$