a generalized version of the Cayley–Hamilton theorem, and the proof proceeds along the same lines. On the generators xi of M, one has a relation of the form where aij ∈ I. Thus The required result follows by multiplying by the adjugate of the matrix (φδij − aij) and invoking Cramer's rule. One finds then det(φδij − aij) = 0, so the required polynomial is To prove Nakayama's lemma from the Cayley–Hamilton theorem, assume that IM = M and take φ to be the identity on M. Then define a polynomial p(x) as above. Then has the required property. Noncommutative case A version of the lemma holds for right modules over non-commutative unital rings R. The resulting theorem is sometimes known as the Jacobson–Azumaya theorem. Let J(R) be the Jacobson radical of R. If U is a right module over a ring, R, and I is a right ideal in R, then define U·I to be the set of all (finite) sums of elements of the form u·i, where · is simply the action of R on U. Necessarily, U·I is a submodule of U. If V is a maximal submodule of U, then U/V is simple. So U·J(R) is necessarily a subset of V, by the definition of J(R) and the fact that U/V is simple. Thus, if U contains at least one (proper) maximal submodule, U·J(R) is a proper submodule of U. However, this need not hold for arbitrary modules U over R, for U need not contain any maximal submodules. Naturally, if U is a Noetherian module, this holds. If R is Noetherian, and U is finitely generated, then U is a Noetherian module over R, and the conclusion is satisfied. Somewhat remarkable is that the weaker assumption, namely that U is finitely generated as an R-module (and no finiteness assumption on R), is sufficient to guarantee the conclusion. This is essentially the statement of Nakayama's lemma. Precisely, one has: Nakayama's lemma: Let U be a finitely generated right module over a (unital) ring R. If U is a non-zero module, then U·J(R) is a proper submodule of U. Proof Let be a finite subset of , minimal with respect to the property that it generates . Since is non-zero, this set is nonempty. Denote every element of by for . Since generates ,. Suppose , to obtain a contradiction. Then every element can be expressed as