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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?

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3 Answers 3

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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 $\text{det}(I\phi-A)=0\in\text{End}_R(M)$.

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  • $\begingroup$ And $\det(I\phi−A)$ gives us the desired polynomial? $\endgroup$
    – Phil-ZXX
    Commented Apr 24, 2013 at 1:04
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    $\begingroup$ Yes, because $\text{det}(I\phi-A)$ is the characteristic polynomial evaluated at $\phi$, i.e. $\phi^n+a_1\phi^{n-1}+\cdots+a_n=0$. $\endgroup$ Commented Apr 24, 2013 at 1:27
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    $\begingroup$ 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, $\phi^2 = \phi \circ \phi$. $\endgroup$ Commented Jun 1, 2021 at 15:37
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    $\begingroup$ @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 $M$ $\endgroup$ Commented Jan 1 at 17:53
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    $\begingroup$ @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. $\endgroup$
    – Anakhand
    Commented Jan 2 at 10:11
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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.

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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 \begin{equation} \sum_{i=1}^n \psi_{ki}(x_i)=0,\, 1\le k\le n. \tag{$\star$} \end{equation} 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 \begin{equation} \sum_{i=1}^n \psi_{ki}x_i=0,\, 1\le k\le n. \tag{$\star\star$} \end{equation} 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*}

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