# Atiyah and MacDonald, Proposition 2.4

Let $M$ be a finitely generated $R$-module, $\mathfrak a \lhd R$ an ideal and $\phi:M\to M$ an $R$-linear map such that $\phi(M)\subseteq \mathfrak a M$. Then $\phi^n+a_1 \phi^{n-1}+\cdots+a_n=0$.
(Atiyah/MacDonald, Proposition 2.4, page 21)

The proof goes as follows:

Let $x_1,...,x_n$ be a set of generators of $M$, then each $\phi(x_i)\in\mathfrak a M$ can be written as $\phi(x_i)=\sum_{j=1}^n a_{ij} x_j$ with some $a_{ij}\in\mathfrak a$, i.e. $$\sum_{j=1}^n(\delta_{ij}\phi-a_{ij})x_j=0$$

By multiplying on the left by the adjoint of the matrix $(\delta_{ij}\phi-a_{ij})$ it follows that $\det(\delta_{ij}\phi-a_{ij})$ annihilates each $x_i$, hence is the zero endomorphism of $M$. Expanding out the determinant, we have an equation of the required form.

I understand the first part, but after "by multiplying on the left by the adjoint" I am not really sure what is happening anymore. Would somebody be so kind and shed some light on this?

Once you have: $$\sum_{j=1}^n(\delta_{ij}\phi-a_{ij})x_j=0$$ you can rewrite this as $(I\phi-A)X=0$, where $A=(a_{ij})$, $I=(\delta_{ij})$, and $X=(x_1,\dots,x_n)^T$. Since $\text{det}(I\phi-A)I=[\text{adj}(I\phi-A)](I\phi-A)$, multiplying by $\text{adj}(I\phi-A)$, whose entries are all in $\text{End}_R(M)$, gives: $$\text{det}(I\phi-A)x_1=\cdots=\text{det}(I\phi-A)x_n=0$$ and $\text{det}(I\phi-A)=0\in\text{End}_R(M)$.

• And $\det(I\phi−A)$ gives us the desired polynomial? Commented Apr 24, 2013 at 1:04
• Yes, because $\text{det}(I\phi-A)$ is the characteristic polynomial evaluated at $\phi$, i.e. $\phi^n+a_1\phi^{n-1}+\cdots+a_n=0$. Commented Apr 24, 2013 at 1:27
• A subtlety: at this point the text only defines $\operatorname{Hom}_A(M,M)$ as a module, so the multiplication $\phi^n$ isn't defined. But pointwise multiplication for modules is not defined for arbitrary modules, so $\phi^2(x) \neq \phi(x)\phi(x)$, since the expression on the RHS isn't defined in general. Instead, $\phi^2 = \phi \circ \phi$. Commented Jun 1, 2021 at 15:37
• @Anakhand First, notation, the ring was $R$ not $A$. I don't think that's an endomorphism, since $rm+x_i = x_i(rm) = r x_i(m) = r(m+x_i) = rm + rx_i$ for $r\in R$ and $m\in M$. In fact, $X$ isn't a vector of $A$-module endomorphisms. We're acting $I\phi-A$ on $X$, and $I\phi-A$ has entries in $\operatorname{End}_R(M)$ so it suffices for $X$ to just have entries in $M$. It's not true matrix multiplication, but a group action. After all at the end of the day, the point of $X$ is to show that $\det(I\phi - A) = 0$ in $\operatorname{End}_R(M)$, which we do by showing it's $0$ on a basis of $M$ Commented Jan 1 at 17:53
• @WanderingMathematician A further subtelty which I read on Wikipedia: we must consider the matrices as having entries in the subring $R[\phi]$ of $\mathrm{End}_R(M)$ (where scalars in $R$ are seen as the multiplication map), because the latter is not commutative, so we can't define the determinant, but the former is commutative. Commented Jan 2 at 10:11

The "adjoint" here means "classical adjoint". For a matrix $A$, the classical adjoint is the matrix whose $i,j$ entry is $(-1)^{i+j} A_{ji}$, where $A_{ji}$ means the $j,i$ cofactor---the determinant of the matrix obtained by omitting row $j$ and column $i$. The point is that the product of the classical adjoint and $A$ is a diagonal matrix with $\mathrm{det}(A)$ on the diagonal.

I also got stuck quite a well while understanding this proposition and the proof when I was new to commutative algebra. Therefore I have written a detailed proof and tried to make it as readable and clear as possible:

Proof: We first clarify that by $$\phi^n+a_1 \phi^{n-1}+\cdots+a_n=0$$ the author actually means $$\begin{equation*} \phi^n(x)+a_1 \phi^{n-1}(x)+\cdots+a_nx=0,\, x\in M. \end{equation*}$$

Let us continue with the following equation (where I substituted $$a_{ij}\gets -a_{ij}$$ for convenience): \begin{align*} (\phi(x_1)+a_{11}x_1)&+a_{12}x_2&+\cdots&+a_{1n}x_n&=0.\\ a_{21}x_1&+(\phi(x_2)+a_{22}x_2)&+\cdots&+a_{2n}x_n&=0.\\ \cdots\\ a_{n1}x_1&+a_{n2}x_2&+\cdots&+(\phi(x_n)+a_{nn}x_n)&=0. \end{align*} Given $$a\in A$$, we define $$l_a:M\to M, m\mapsto am$$. We note that $$l_a$$ is actually an $$A$$-module endomorphism of $$M$$. If we set $$\begin{equation*} \psi_{ij}=\delta_{ij}\phi+l_{a_{ij}}=\begin{cases} l_{a_{ij}}, &i\ne j\\ \phi+l_{a_{ij}}, &i=j \end{cases}, \end{equation*}$$ then $$(\psi_{ij})$$ are also endomorphisms of $$M$$, and by hypothesis $$$$\sum_{i=1}^n \psi_{ki}(x_i)=0,\, 1\le k\le n. \tag{\star}$$$$ Note that $$\psi_{ij}\in \text{End}_R(M)$$ and $$M$$ naturally has an $$\text{End}_R(M)$$-module structure (the multiplication of the ring $$\text{End}_R(M)$$ is composition), the scalar multiplication of which is $$(\psi,m)\mapsto \psi(m)$$. We may abuse the notation and write this scalar multiplication as $$\psi m$$. Then ($$\star$$) becomes $$$$\sum_{i=1}^n \psi_{ki}x_i=0,\, 1\le k\le n. \tag{\star\star}$$$$ Note that $$\text{End}_R(M)$$ is not necessarily a commutative ring, but $$\psi_{ij}$$s commute: we just do a casework and verify $$\psi_{i_1j_1}\psi_{i_2j_2}=\psi_{i_2j_2}\psi_{i_1j_1}$$ when $$\begin{equation*} \begin{cases} i_1=j_1,\, i_2\ne j_2\\ i_1\ne j_1,\, i_2\ne j_2\\ i_1=j_1,\, i_2=j_2 \end{cases}. \end{equation*}$$ Therefore the subring $$S$$ generated by $$\psi_{ij}$$s (and $$1_{\text{End}_R(M)}$$) $$\begin{equation*} S=\{ \sum_{l=1}^N \xi_{l,1}\cdots \xi_{l,{k_l}}: \xi_{p,q}\in \{\psi_{ij}\}\cup {1_{\text{End}_R(M)}}, N,k_1,\cdots,k_N\in\mathbb{Z}_+ \}\cup\{0_{\text{End}_R(M)}\} \end{equation*}$$ is commutative, and $$M$$ is an $$S-$$module, and $$(\star\star)$$ can be viewed as an equation on $$S-$$module action.

Finally, that $$\det(\psi_{ij})=0_{\text{End}_R(M)}$$ can be inferred from the following lemma (applied to $$S-$$module $$M$$) and that $$M=\sum_{i=1}^n Ax_i$$:

Lemma: Let $$A$$ be a commutative ring and $$M$$ an $$A$$-module. If $$(a_{ij})_{1\le i,j\le n}\in A$$ and $$\sum_{i=1}^n a_{li}m_i=0$$ for each $$1\le l\le n$$, then $$\det(a_{ij})_{n\times n} m_l=0$$ for each $$1\le l\le n$$.

Proof of the Lemma: Set $$D=\det(a_{ij})_{n\times n}$$. We only prove $$Dm_1=0$$, proof of the other equalities being similar.

In fact, let $$c_{ij}$$ denote the cofactor of $$a_{ij}$$ in $$\det(a_{ij})_{n\times n}$$. Then $$\begin{equation*} 0=\sum_{l=1}^n c_{1l} \left( \sum_{i=1}^n a_{li}m_i\right) . \end{equation*}$$ Let us study the coefficient of $$m_i$$s in the above equation: the coefficient of $$m_1$$ is $$\sum_{l=1}^n c_{1l} a_{l1}=D$$, while for $$m_i$$ ($$i\ne 1$$), its coefficient is $$\sum_{l=1}^n c_{1l} a_{li}$$, which is the determinant of a matrix the first and $$i-$$th column of which are both $$(a_{1i},a_{2i},\cdots,a_{ni})^T$$, and is therefore $$0$$, i.e. $$Dm_1=0$$ as desired. $$\square$$

Recall that $$\psi_{ij}=\delta_{ij}\phi+l_{a_{ij}}$$. Clearly $$\det(\psi_{ij})=0_{\text{End}_R(M)}$$ implies the long-desired existence of an equaton of the form $$\begin{equation*} \phi^n(x)+a_1 \phi^{n-1}(x)+\cdots+a_nx=0.\, \square \end{equation*}$$