See also determinant.
Let be a commutative ring, and let be an matrix with entries in . Then there exists an matrix with entries in such that .
We may as well take to be the polynomial ring , since we are then free to interpret the indeterminates however we like along a ring map . Let denote the corresponding generic matrix.
Guided by Cramer’s rule (see determinant), put
the being columns of and , the column vector with in the row and ‘s elsewhere, appearing as the column. If we pretend is invertible, then we know by Cramer’s rule. We claim this holds for general .
Indeed, we can interpret this as a polynomial equation in and check it there. As an equation between polynomial functions on the space of matrices , it holds on the dense subset . Therefore, by continuity, it holds on all of . But a polynomial function equation with coefficients in implies the corresponding polynomial identity, and the proof is complete.
(Cayley-Hamilton) Let be a finitely generated free module over a commutative ring , and let be an -module map. Let be the characteristic polynomial of , and let be the unique -algebra map sending to . Then is the zero map .
Via , regard as an -module, and with regard to some -basis of , represent by a matrix . Now consider as an matrix with entries in . By definition of the module structure, this matrix , seen as acting on , annihilates the length column vector whose row entry is .
By the previous lemma, there is such that is times the identity matrix. It follows that
i.e., for each . Since the form an -basis, the -scalar annihilates the -module , as was to be shown.
The Cayley-Hamilton theorem easily generalizes to finitely generated -modules (not necessarily free) as follows. Let be a module endomorphism, and suppose is an epimorphism. Since is projective, the map can be lifted through to a map . Let be the characteristic polynomial of .
Write . We already know . From , it follows that for any . Hence . Since is epic, follows.
We give an interesting and perhaps surprising consequence of the Cayley-Hamilton theorem below, after establishing a lemma close in spirit to Nakayama's lemma.
Suppose is a finitely generated -module, and is a module map such that for some ideal of . Then there is a polynomial , with all , such that .
For some finite , we have a surjective map . Tensoring with , we obtain a surjective map , fitting in a commutative diagram
By projectivity of , we can lift to a map making the diagram commute. Let be the -module map , regarded as a matrix. Then the characteristic polynomial of satisfies the conclusion, by the Cayley-Hamilton theorem and Proposition 1.
Let be a finitely generated module over a commutative ring , and let be a surjective module map. Then is an isomorphism.
Regard as a finitely generated -module via . Since is assumed surjective, we have for the ideal of . Now take as in the preceding lemma, a module map over the ring . By the lemma, we see that where , in other words the -scalar
as an operator on . Write for polynomials . Now we may rewrite the previous displayed equation as
for all , which translates into saying that , i.e., that is a retraction of . Since is epic, we now see is an isomorphism.
The proof of the Cayley-Hamilton theorem follows the treatment in
The proof of Proposition 2 on surjective endomorphisms of finitely generated modules was extracted from