I am studying some algebra during my spare time. In particular I am learning about Noetherian rings. A friend sent me the following excersise, and I am not able to solve it.
Suppose that a ring $A$ is generated by a subring $R$ and a nilpotent element $n$ such that $R+nR=R+Rn$. (Dis)prove:
$R$ left noetherian implies that $A$ is left noetherian.
I believe that the statement above is correct. However I failed in trying to prove that $A$ satisfies the ascending chain condition on ideals, or in proving that every ideal of $A$ is finitely generated. I also thought about using induction as the statement is trivial if $n=0$ but I do not see really how to progress from there on. Moreover I thought about writing $A$ as a ring isomorphic to a quotient of $R[X]$ and then using the Hilbert basis theorem, but I am not convinced.
Would somebody be so kind to shed light on this problem?
Thanks in advance!
Since $A$ is generated by $R$ and $n$, every element of $A$ can be written as a sum of products of elements of $R$ and $n$. Moreover, we can use $R+nR=R+Rn$ to ensure that in those products, the $n$'s always come at the end, after all the elements of $R$: if we have an expression of the form $nr$ for $r\in R$, we can rewrite it as $s+tn$ for some $s,t\in R$, and by iterating this we can move all the $n$'s to the right of all our products.
This means that we can express every element of $A$ as a sum $r_0+r_1n+r_2n^2+\dots r_mn^m$ for some $r_0,\dots,r_m\in R$. Moreover, since $n$ is nilpotent, we can always have $m<k$ where $k$ is such that $n^k=0$. This means exactly that $A$ is generated as a left $R$-module by the finitely many elements $1,n,\dots,n^{k-1}$. Since $R$ is left Noetherian, this means $A$ is Noetherian as a left $R$-module. But any left ideal in $A$ is in particular an $R$-submodule of $A$, so this implies $A$ is left Noetherian.