a finite combination for some . Each can be further decomposed as for some . Therefore, we have . Since is a (two-sided) ideal in , we have for every , and thus this becomes for some , . Putting and applying distributivity, we obtain . Choose some . If the right ideal were proper, then it would be contained in a maximal right ideal and both and would belong to , leading to a contradiction (note that by the definition of the Jacobson radical). Thus and has a right inverse in . We have . Therefore, . Thus is a linear combination of the elements from . This contradicts the minimality of and establishes the result. Graded version There is also a graded version of Nakayama's lemma. Let R be a ring that is graded by the ordered semigroup of non-negative integers, and let denote the ideal generated by positively graded elements. Then if M is a graded module over R for which for i sufficiently negative (in particular, if M is finitely generated and R does not contain elements of negative degree) such that , then . Of particular importance is the case that R is a polynomial ring with the standard grading, and M is a finitely generated module. The proof is much easier than in the ungraded case: taking i to be the least integer such that , we see that does not appear in , so either , or such an i does not exist, i.e., . See also Module theory Serre–Swan theorem Notes References . . . . . . . Links How to understand Nakayama's Lemma and its Corollaries Theorems in ring theory Algebraic geometry Commutative algebra Lemmas in algebra