Show that if \(\phi\) is an endomorphism of a free \(k\)-module \(E\) of finite rank, then \(p_\phi(\phi) = 0\).
Hint: If \(A\) is the matrix of \(\phi\) and \(B = x I_n - A\) then \[ B^a B = |B| I_n = p_\phi I_n \in M_n(k[x]) .\] If \(E\) is a \(k[x]\)-module with structure induced by \(\phi\), and \(\psi\) is the \(k[x]\)-module endomorphism \(E\to E\) with matrix given by \(B\), then \[ \psi(u) = x u -\phi(u) = \phi(u) - \phi(u) = 0 && \forall u\in E .\]
Solution
Let \(E\) be free of rank \(n\), choose a basis, and let \(A\in M_n(k)\) be the matrix of \(\phi\). Write \[ p_\phi(x)=\det(xI_n-A). \] We prove that \(p_\phi(A)=0\), which is equivalent to \(p_\phi(\phi)=0\).
1. In \(M_n(k[x])\), the adjugate identity gives\[ \operatorname{adj}(xI_n-A)(xI_n-A)=p_\phi(x)I_n. \]
For every square matrix \(M\) over a commutative ring, \[ \operatorname{adj}(M)M=\det(M)I_n. \] Apply this to \(M=xI_n-A\) over the polynomial ring \(k[x]\).
2. Give \(E\) the structure of a \(k[x]\)-module by letting \(x\) act as\(\phi\), let \(e_1,\ldots,e_n\) be the chosen basis with\(\phi(e_j)=\sum_i a_{ij}e_i\), and let \(M_n(k[x])\) act on \(E^n\) bymatrix multiplication on columns. With \(B=xI_n-A\) and\(\mathbf e=(e_1,\ldots,e_n)^t\in E^n\), one has \(B^t\mathbf e=0\).
The \(j\)th entry of \(B^t\mathbf e\) is \[ \sum_{i=1}^n(\delta_{ij}x-a_{ij})e_i=xe_j-\sum_{i=1}^na_{ij}e_i =\phi(e_j)-\phi(e_j)=0. \]
3. Multiplication by \(p_\phi(x)\) on this \(k[x]\)-module is the zero map.
Since \(E\) is a module over the commutative ring \(k[x]\), the action of \(M_n(k[x])\) on \(E^n\) satisfies \((MN)\mathbf v=M(N\mathbf v)\). Applying 1 to \(B^t\), whose determinant is also \(p_\phi(x)\), and using 2, \[ p_\phi(x)\mathbf e=\operatorname{adj}(B^t)B^t\mathbf e =\operatorname{adj}(B^t)\,0=0. \] Thus \(p_\phi(x)e_i=0\) for every basis vector \(e_i\), and since the \(e_i\) generate \(E\), \(p_\phi(x)u=0\) for every \(u\in E\).
4. Therefore \(p_\phi(\phi)=0\).
By definition of the \(k[x]\)-module structure, multiplication by a polynomial \[ a_0+a_1x+\cdots+a_mx^m \] acts as \[ a_0I+a_1\phi+\cdots+a_m\phi^m. \] Hence the zero action of \(p_\phi(x)\) from 3 is exactly the identity \[ p_\phi(\phi)=0. \]