M is a module such that , then M does not admit a maximal submodule, since if there is a maximal submodule , and so , a contradiction. Since a nonzero finitely generated module admits a maximal submodule, in particular, one has: If and M is finitely generated, then A maximal ideal is a prime ideal and so one has where the intersection on the left is called the nilradical of R. As it turns out, is also the set of nilpotent elements of R. If R is an Artinian ring, then is nilpotent and . (Proof: first note the DCC implies for some n. If (DCC) is an ideal properly minimal over the latter, then . That is, , a contradiction.) Extension and contraction of an ideal Let A and B be two commutative rings, and let f : A → B be a ring homomorphism. If is an ideal in A, then need not be an ideal in B (e.g. take f to be the inclusion of the ring of integers Z into the field of rationals Q). The extension of in B is defined to be the ideal in B generated by . Explicitly, If is an ideal of B, then is always an ideal of A, called the contraction of to A. Assuming f : A → B is a ring homomorphism, is an ideal in A, is an ideal in B, then: is prime in B is prime in A. It is false, in general, that being prime (or maximal) in A implies that is prime (or maximal) in B. Many classic examples of this stem from algebraic number theory. For example, embedding . In , the element 2 factors as where (one can show) neither of are units in B. So is not prime in B (and therefore not maximal, as well). Indeed, shows that , , and therefore . On the other hand, if f is surjective and then: and . is a prime ideal in A is a prime ideal in B. is a maximal ideal in A is a maximal ideal in B. Remark: Let K be a field extension of L, and let B and A be the rings of integers of K and L, respectively. Then B is an integral extension of A, and we let f be the inclusion map from A to B. The behaviour of a