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 $\phi(M) \subseteq IM$.

Thanks for your help!

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1 Answer 1


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 $j$.

3/ I'm not fully understand your first question since it was proved in the proof (by using the above).


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