In this article, I want to discuss four versions of Nakayama’s Lemma, ranging from quite algebraic to quite geometric. I’ll prove that the first version is true, and then show that all versions are logically equivalent.
As usual, all rings are commutative with unit.
Nakayama’s Lemma (I). Let be an -module of finite type and let be an ideal of such that . Then there exists an element such that and (mod ). (Equivalently, there exists an element that acts on as the identity, i.e. for all .)
Proof. Write for the generators for . The equation implies that for each of those, there’s an equation where the are elements of . By writing for the square matrix and for the vector consisting of the generators, we can conveniently rewrite all of these equations at once in a single matrix equation: Here is the identity matrix of the appropriate size. Recall that for any square matrix has an adjugate matrix whose entry at is the determinant of the submatrix obtained by removing the -th row and the -th column, times . In other words, the adjugate matrix is the transpose of the cofactor matrix. Laplace expansion says that we always have the relation In our case, therefore, we have From this we see the element anihilates all of the generators . Hence anihilates the whole of , that is, . Notice that the expansion of the determinant gives a polynomial with constant term , and all other monomials products of elements in . Therefore, when passing to the quotient , all that’s left of is .
Given such an , one can choose which is an element of and which verifies for all . On the other hand, given such an , one can choose , so that mod and for all .
Notice this proof is constructive: not only does it assert the existence of the element , but it gives us a receipe to actually construct it. Let’s see a concrete example. Take the gaussian integers mod , seen as a -module. This is finitely generated by the elements . For the ideal, choose . For any element of , we can write it as since mod . Therefore, . In fact, and , so our matrix above looks like , which is nine times the identity matrix. Therefore the element we’re after is , which is . And indeed, while mod .
Nakayama’s Lemma (II). Let be an -module of finite type and let be an ideal of such that . If is contained in the Jacobson radical of , then .
Proof that (I) (II). Recall that the Jacobson radical is the intersection of all maximal ideals in the ring. In this context, notice that any element mod is invertible in . To see this, suppose it is not. Then the principal ideal is contained in some maximal ideal , and so . Therefore mod , which is a contradiction with . Now Nakayama’s Lemma (I) guarantees the existence of an element (mod ) such that . We’ve just shown that must be invertible, whence .
Nakayama’s Lemma (III). Let be an -module of finite type, let be a submodule, and let be an ideal of which is contained in the Jacobson radical of . If is surjective, then .
Proof that (II) (III). Notice that if and only if . The reason I’m pointing this out is because that looks like the conclusion of (II); the quotient of a module of finite type by any submodule is also of finite type (hint: being of finite type means there’s a surjection for some integer ), so we reduced to problem to showing the equality of -modules . Let be any element of . From the surjectivity hypothesis, there exists an element such that . In particular, we have so there exists an integer and elements and such that Passing to the quotient , we find . Since was arbitrary, this shows any element of is in .
Nakayama’s Lemma (IV). Let be an -module of finite type, with being a local ring, so is a finite-dimensional vector space over . Let be elements of such that their images in form a basis. Then these elements form a minimal set of generators for .
Proof that (III) (IV). Let be the submodule of generated by the ’s. From (III), we simply have to show that is a surjective morphism. But this is obvious!
I’ll come back later and add more explanations and examples.