# Questions about the Cayley-Hamilton theorem for modules

Having recently learned the proof of CH for vector spaces from Hoffman&Kunze (I've known the statement of the theorem for a while now, but have never really bothered with the proof), I am now trying to wrap my head around the proof for finitely-generated modules, following Eisenbud's in his commutative algebra book and have some questions about this more general version.

Let $$R$$ be a ring, $$I\subset R$$ an ideal, and $$M$$ an $$R$$-module that can be generated by $$n$$ elements. Let $$\varphi$$ be an endomorphism of $$M$$. If $$\varphi(M)\subset IM$$ 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 $$p(\varphi)=0$$ as an endomorphism of $$M$$.

Firstly, why is it necessary to require that $$\varphi(M)\subset IM$$? I understand why this isn't required in the version for vector spaces, as the only ideals of a field are $$(0)$$ and $$(1)$$, so the condition is always true, but why is this required in the general case? As far as I can tell this condition isn't used in the proof itself. Secondly, why is $$p_j\in I^j$$ true? I can see intuitively that due to the inclusion condition the "matrix" (if we're allowed to talk about such things) of $$\varphi$$ will have entries in $$I$$ (is this why we require that $$\varphi(M)\subset IM$$, so that we can say where the coefficients of the matrix of $$\varphi$$ lie?), so the coefficients of the characteristic polynomial must be in $$I$$ as well, but how can we show formally that each is a $$j$$-fold power of $$I$$? Next, about the proof itself. The proof for modules uses the same idea as the one for vector spaces, wherein you view the vector space/module as a module over the ring of polynomials in $$\varphi$$ and then you use the defining property of the adjugate matrix, and I think I more or less understand it, except for the final part.

View the equation $$\varphi(m_i)=\sum_j a_{ij}m_j$$ (where $$m_j$$ are the generators of $$M$$ and $$a_{ij}\in I$$) as the equation $$(\varphi\mathbf{1}-A)\cdot m=0$$, multiply it by the adjugate to get $$[\det(\varphi\mathbf{1}-A)]\mathbf{1}\cdot m=0$$, therefore $$\det(\varphi\mathbf{1}-A)\cdot m_j=0$$ for each generator $$m_j$$, from which follows that $$[\det(\varphi\mathbf{1}-A)]\cdot M=0$$, i.e. $$\det(\varphi\mathbf{1}-A)$$ is the $$0$$ endomorphism.

I think I understand how this works for vector spaces (it still somehow feels tautological, as if we're cheating somehow by just plugging in $$\varphi$$), but I don't understand the following bit: in the case of vector spaces, from the fact that $$\det(\varphi\mathbf{1}-A)\cdot m_j=0$$ for the generators $$m_j$$ we can definitively confirm that $$\det(\varphi\mathbf{1}-A)=0$$ (because $$a\cdot x=0,x\neq0\Rightarrow a=0$$ in vector spaces), but how can we conclude the same for modules? Isn't there such a thing as torsion elements, i.e. $$a\cdot x=0$$ such that neither $$a$$ nor $$x$$ are $$0$$?

• For the second question: if a linear map is zero over a set of generators, it must be the zero map. Does this answer that part? Mar 20, 2021 at 11:56
• For the first: as you said, the condition $\varphi(M)\subseteq IM$ implies that the entries of a matrix associated to $\varphi$ (notice that I said "a", not "the") are in $I$. In order to prove that the coefficients $p_{i}$ lie in $I^{i}$, I'd exploit the fact that $p=\operatorname{det}(\varphi 1-A)$ is a candidate, and try to expand that determinant using the Laplace formula. Either that or using induction on the numer of generators of $M$. Mar 20, 2021 at 12:05

1. Yes, we require $$\varphi(M) \subseteq IM$$ to guarantee that $$p_j \in I^j$$ (this is particularly useful for commutative algebra when $$I$$ is prime). Note that $$I = R$$ simply ignores where each $$p_i$$ lives; the condition $$M \subseteq IM$$ tells you that you can pick $$a_{ij} \in I$$, and thus their $$k$$-products belong to $$I^k$$.
The implicit usage of $$\varphi(M) \subseteq IM$$ goes as follows: since $$M$$ is finitely generated (say, $$\{m_i\}$$ is a generating set), then $$IM$$, the submodule generated by elements of the form $$rx$$ where $$r \in I,\ x \in M$$, is simply $$IM = \left\{\sum\limits_i r_i m_i\colon\ r_i \in I\right\}$$. To see that, note that RHS is an $$R$$-module and any element of the form $$rx$$ belongs to it (so the generated module, $$IM$$, does too). The reverse inclusion is obvious. So $$\varphi(m_i) \in IM$$ and the coefficients belong to $$I$$.
1. We can conclude that $$p_j \in I^j$$ because the determinant is an homogeneous polynomial of degree $$n$$ in $$n^2$$ variables. If we group together the terms of degree $$n-j$$ in $$\varphi$$, we are left with sums of $$j$$ products of elements from $$I$$, i.e. the coefficient lies in $$I^j$$.
2. The important and last part here is a bit tricky. What we do is consider the "generating vector" $$m = \begin{pmatrix} m_1 \\ \vdots \\ m_n\end{pmatrix}$$. By construction of such a matrix $$A$$, we have $$(\varphi\mathbf 1-A)\cdot m = \begin{pmatrix} 0 \\ \vdots \\ 0 \end{pmatrix}$$ - but it wouldn't make any sense to say that this matrix $$\varphi\mathbf 1 - A$$ has coefficients in $$R$$. It doesn't! It has coefficients in $$R[\varphi] \subseteq \mathrm{End}(M)$$! So what we do here is, actually, consider this ring acting over $$M$$, with $$\varphi\cdot m = \varphi(m)$$. So now we can in fact use Cramer's rule to get that $$\det(\varphi\mathbf 1 - A)\cdot m = \vec{\mathbf 0}$$. But that means, by linearity, that the operator $$\det(\varphi \mathbf 1 - A) \in \mathrm{End}(M)$$ has as image $$(0) \subseteq M$$. i.e., this is the 0 module homomorphism. So $$\det(\varphi \mathbf 1 - A) = 0$$.
• I think my last point was a bit disconnected from the original question. My point here is that, in fact, you can have $a\cdot x = 0$ for nonzero elements while working with modules. But this is not the case when working with 'linear operators' like we did, exactly by the definition of what it means for an operator to be null. Mar 18 at 20:11