I have a question which is as follows:
Let $R$ be a unitary, commutative, local, noetherian ring with J principal.
Prove that if $J$ is not nilpotent then R is an integral domain and that $0$ and $J$ are the only prime ideals
I will use the following theorem:
Let R be a unitary commutative local noetherian ring with J principle. Then every non-zero ideal is a power of $J$
So suppose that $R$ was not an ID then we would have $x,y\neq 0$ such that $xy=0$. But this gives that $ann(x)\neq 0$ and so $ann(x)=J^k$ for some $k$
We also have that $xR=J^s$ for some $s$. But then as $ann(x)xR=0$ this gives that $J^kJ^s=J^{k+s}=0$ contradicting the non-nilpotency of $J$
Proving that $0$ and $J$ are the only prime ideals of $R$
Suppose we have another prime ideal, say $P$ then we have that $P=J^m$. Now as $JJ^{m-1}\subset J^{m}$ this implies that (w.lo.g) that $J^{m-1}\subset J^{m}$ and hence $J^{m}=J^{m-1}$.
This then gives that any descending chain has DCC and and so this is an artinian ID and hence a field which gives as contradiction as $P$ was supposed to be an ideal with $P\neq R$
Is the above correct?
Thanks for any help
Things look ok, but there is a direct proof of the second fact you might consider.
Starting with $P=J^m$, you could immediately conclude "since $P$ is prime, $J\subseteq P$. Since we already had $P\subseteq J$, we now have $J=P$.
Remember that for commutative rings, the "ideal-wise" versions of the "element-wise" definitions are actually equivalent. By that I mean $P$ is prime iff $AB\subseteq P$ implies $A$ or $B$ is a subset of $P$, and as a consequence of this, $I^k\subseteq P$ implies $I\subseteq P$.