# Understanding Cayley-Hamilton Theorem for Modules

I'm trying to understand the Cayley-Hamilton Theorem for modules, here is what I'm having trouble with.

PROPOSITION (Cayley-Hamilton) Suppose $$I \subseteq A$$ is an ideal and $$\phi:M→M$$ is an A-module homomorphism, where $$M$$ is an $$A$$-module generated by $$\{x_1,...,x_n\}$$. And suppose $$\phi(M) \subseteq IM$$. Then $$\phi$$ satisfies an identity $$\phi^n+a_1 \phi^{n−1}+···+a_n = 0 \in End(M)$$,with $$a_j \in I^j$$.

The proof goes as follows.

PROOF. Write $$\phi x_i= \sum_j a_{ij} x_j$$, and form the $$n \times n$$ matrix $$(\phi Id_n−(a_{ij}))$$ over the ring $$A[\phi]$$. This matrix acts on $$M^n$$, with $$(\phi Id_n−(a_{ij}))(x_1,...,x_n)^t= (0)^t$$. There is the adjugate matrix $$(\phi Id_n−(a_{ij}))^*$$ with $$(\phi Id_n−(a_{ij}))^*(\phi Id_n−(a_{ij}))=det(\phi Id_n−(a_{ij}))Id_n$$. It follows that $$det(\phi Id_n−(a_{ij}))$$ annihilates all $$x_j$$, hence is the 0 endomorphism of M. Expanding the determinant shows the coefficient of $$\phi^{n-j}$$ belongs to $$I^j$$.

The part I'm having trouble with is the following:

-From the fact that $$(\phi Id_n - (a_{ij}))(\vec{x})^t = (\vec{0})^t$$ has a non-zero solution $$(x_1 \ldots x_n)^t$$, can we not conclude $$det(\phi Id_n−(a_{ij}))$$ is the zero homomorphism?

-How does it follow from $$(\phi Id_n−(a_{ij}))^*(\phi Id_n−(a_{ij}))=det(\phi Id_n−(a_{ij}))Id_n$$ that $$det(\phi Id_n−(a_{ij}))$$ annihlates all $$x_i$$?

-Why is the assumption $$\phi(M) \subseteq IM$$ neccessary? (The notes I'm reading from sets $$\phi \colon M \to IM$$, but I believe there's no difference if we set $$\phi \colon M \to M$$ and $$\phi(M) \subseteq IM$$.

1/ For your third question: Yes, $$\phi \colon M \to IM$$ is the same as $$\phi \colon M \to M$$ and $$\phi(M) \subseteq IM$$. The reason why we need this assumption is because we want $$a_{ij} \in I$$ so that the coefficients of $$\phi^{n-j}$$ belongs to $$I^j$$ (in the determinant the coefficient of $$\phi^{n-j}$$ is a product of $$j$$ elements $$a_{ij}$$).
2/ For your second question: since $$(\phi Id_n−(a_{ij}))(x_1,...,x_n)^t= (0)^t$$ if we apply $$(\phi Id_n−(a_{ij}))^*$$ to the left we will get $$[det(\phi Id_n−(a_{ij}))Id_n](x_1,...,x_n)^t = [(\phi Id_n−(a_{ij}))^*(\phi Id_n−(a_{ij}))](x_1,...,x_n)^t = (0)^t$$ Note that $$[det(\phi Id_n−(a_{ij}))Id_n](x_1,...,x_n)^t = det(\phi Id_n−(a_{ij})) \left( Id_n(x_1,...,x_n)^t \right) = det(\phi Id_n−(a_{ij}))(x_1,...,x_n)^t$$ Therefore each entry of the last vector must be $$0$$, i.e, $$det(\phi Id_n−(a_{ij}))x_j =0$$ for all $$j$$.