91Ó°ÊÓ

Assume that R is a Noetherian ring, which means that every ascending chain of ideals in R becomes stationary. Let I be an arbitrary ideal in R. We want to show that I has a finite generating set. To begin, we will create a chain of ideals. Let \(x_1\) be any element of I. Thus, we have the ideal \(R \cdot x_1\) generated by \(x_1\). If this generated ideal is equal to I, then we are done. If not, there must be an element \(x_2 \in I\) such that \(x_2 \notin R \cdot x_1\). Now, consider the ideal generated by \(x_1\) and \(x_2\), denoted by \(R \cdot x_1 + R \cdot x_2\). We can repeat this process, each time adding a new element to the generated ideal if it is not equal to I. If this process stops after a finite number of steps, then we have found a finite generating set for I. If it does not stop, we would have an infinite ascending chain of ideals: \(R \cdot x_1 \subset R \cdot x_1 + R \cdot x_2 \subset \ldots \) This is a contradiction, since R is assumed to be Noetherian and every ascending chain of ideals must become stationary. Thus, every ideal in R must have a finite generating set.

Now, assume that every ideal in R has a finite generating set. We want to show that R is Noetherian. Suppose there is an ascending chain of ideals in R that does not become stationary: \(a_1 \subset a_2 \subset a_3 \subset \ldots \) Let J be the union of all these ideals, i.e., \(J = \bigcup_{i=1}^{\infty} a_i\). By assumption, J must have a finite generating set, denoted as \(\{y_1, y_2, \ldots, y_n\}\). Now, for each \(y_i\), there must be some index \(k_i\) such that \(y_i \in a_{k_i}\). Let \(m = \max\{k_1, k_2, \ldots, k_n\}\). Since the chain is ascending, all of the generators \(y_i\) are in \(a_m\), and since they generate J, it follows that \(J \subseteq a_m\). On the other hand, it is clear that \(a_m \subseteq J\), since J is the union of all the ideals in the chain. Therefore, \(a_m = J\). Now, for any index \(l > m\), it must be the case that \(a_l \subseteq J\). But, since the chain is ascending, we also have \(J = a_m \subseteq a_l\). Thus, \(J = a_l\) for every \(l > m\), meaning the chain becomes stationary: \(a_1 \subset a_2 \subset \ldots \subset a_m = a_{m+1} = a_{m+2} = \ldots \) Therefore, R is Noetherian. In conclusion, we have shown that a ring R is Noetherian if and only if every ideal in R has a finite generating set.