Cayley-Hamilton theorem
General case
Traditionally, endomorphisms (over a vector space, for example) were viewed as matrices. Thus the natural analogy is as follows: given an endomorphism, there exists a polynomial…
Theorem. (Cayley-Hamilton). Let $R$ be a ring, $I\subset R$ an ideal, and $M$ an $R$- module that can be generated by $n$ elements. Let $\phi$ be an endomorphism of $M$. If $$\phi(M)\subset IM,$$ then there is a monic polynomial $$p(x)=x^n+p_1x^{n-1}+\cdots+p_n$$ with $p_j\in I^j$ for each $j$, such that $p(\phi)=0$ as an endomorphism of $M$.
Proof. Let $m_1,\dots,m_n$ be a finite set of generators of $M$. Write each $\phi(m_i)$ in terms of $m_j$ using coefficients in $I$: $$\phi(m_i)=\sum a_{ij}m_j,\quad\text{with }a_{ij}\in I.$$ (We allow some $a_{ij}$ to take the value of zero as necessary, so long as there are precisely $n$ coefficients.) We can thus regard $M$ as an $R[x]$-module by letting $x$ represent the action of $\phi$. This is because, in particular, multiplying something in $M$ by something in $R[x]$ amounts to multiplying it by values in $R$ and applying and endomorphism to it, so it lands back in $M$.
Let $A$ be the $n\times n$ matrix with entried $a_{ij}$. Let $\textbf{1}$ be the $n\times n$ identity matrix. Let $m$ be the column vector whose entries are the $m_j$. Then $$(x\textbf{1}-A)\cdot m=0.$$ This is because we can take the $m_j$ to be represented by the matrix $\textbf{1}$, and $x$, which is the application of $\phi$, is precisely what sends this to $A$.
Multiplying both sides by the matrix of cofactors of $x\textbf{1}-A$, we get $$[\text{det}(x\textbf{1}-A)]\textbf{1}\cdot m=0,$$ which means $\text{det}(x\textbf{1}-A)m_i=0$ for all $i$, so $$[\text{det}(x\textbf{1}-A)]\textbf{1}\cdot M=0.$$
So clearly if we let $p(x)=\text{det}(x\textbf{1}-A)$ then $p(\phi)=0$.
Corollaries
- If $M$ is a finitely generated $R$-module and $I$ is an ideal of $R$ such that $IM=M$, then there is an element $r\in I$ that acts as the identity on $M$; that is, such that $(1-r)M=0$.
- Nakayama’s lemma
Proofs
1
Let $\phi$ be the identity. By the Cayley-Hamilton theorem, there is a polynomial $p$ such that $p(1)M=0$. This means $$ (1+p_1+\dots+p_n)M=0, $$ with $p_j$ in $I^j\subset I$. Then just take $r=-(p_1+\cdots+p_n)$.