Nilpotency of finite ideal Suppose we have a commutative local ring $R$ with unit. I'm curious about whether the following statements are correct:
1- every proper finite ideal is nilpotent.
2-every proper finitely generated ideal is nilpotent.
3- If maximal ideal be finite then it's nilpotetnt.
4- If maximal ideal be finitely generated then it's nilpotetnt.

For $1$:
Suppose $I$ is a proper finite ideal and $|I|=n$. Take $x\in I$, $x\neq 0$ and think about $\{x,x^2,\ldots,x^n\}\subseteq I$. Suppose none of these powers of $x$ in this set are zero: then  $x^k=x^m\neq 0$ for some $m> k$, but then $(x^{m-k}-1)x^k=0$, but the left hand factor must be a unit. See the end?
For $2$ look at $(x)$ in $\Bbb R[[x]]$.
For $3$ look at $1$.
For $4$ look at $2$.

Update: Just so there is no mistake, "the end" is not "right away." Where I left off, one should be able to conclude that $x^n=0 $ for all $x\in I$, and then conclude that $I$ is nilpotent by a pigeonhole-principle argument.
