Answer to Question #222101 in Abstract Algebra for Komal

Question

Prove that subring of Noetherian Ring need not be noetherian.

Answer 1

Let K be a field and consider $S= K[x_1,x_2,...]$ be the polynomial ring in infinitely many variables over K. Let R denote the field of fractions of S. R is Noetherian being a field but S is not, because there exists a non terminating strictly increasing chain of ideals of S

$\langle x_1\rangle \sub \langle x_1,x_2\rangle \sub \langle x_1, x_2, x_3\rangle \sub ... \langle x_1, x_2, ... ,x_n\rangle \sub ...$