Let $(A, +, \cdot )$ be a ring such that $A/<x>$ is finite, for every $0 \neq x \in A$. Show that $A$ is noetherian.

Question

I tried to show that every ideal $I$ of $A$ is finite. Let $0 \neq x \in I$. By hypothesis there is $x_1, x_2, ..., x_k \in A$ such that $$A/<x> = <x_1+<x>,..., x_k+<x>>$$ I was trying to show that $I = <x_1, x_2, ..., x_k, x>$, but I was not able to prove the $\supseteq$ inclusion. There is an easier way to show this? Any help would be great.

Answer 1

Let $I_1\subset I_2\subset\cdots$ be an ascending chain of ideals. Assume wlog $I_1=(x)$ for some $x\neq0$ (why?). Take the quotient modules on the chain modulo $I_1=(x)$ and obtain

$$0=(x)/(x)\subset I_2/(x)\subset I_3/(x)\subset\cdots\subset A/(x)$$

Next, you would want to show this chain cannot strictly increase indefinitely. This is the where the hypothesis comes into play. And recall that there is a 1-1 correspondence between ideals of $A$ and ideals of $A/(x)$ containing $(x)$. It follows that the original chain is stationary.