Corollary 2.5 from Atiyah-Macdonald small question

cayley-hamiltoncommutative-algebramodulesproof-explanation

I'm having a doubt in the proof of Corollary 2.5 from Atiyah-Macdonald's Introduction to Commutative Algebra (page 21).

The proof is simple: apply Proposition 2.4 (basically a version of Cayley-Hamilton's theorem) with the identity module homomorphism: from
$$
\phi^n+a_1\phi^{n-1}+…+a_n \text{ and } \phi(x)=x,
$$
we get
$$
x + a_1 x + a_2 x + … + a_{n-1}x+a_n = x(1+a_1+…+a_{n-1})+a_n.
$$

Proof follows by picking $x=1+a_1+…+a_n$ as if you could put in evidence the factor $x$ from the previous expression. But you can't. There is no $x$ multiplying the last $a_n$.

Can someone go through some trouble and explain what I'm missing? Thanks in advance.

Best Answer

I'm having a bit of trouble parsing what you're saying. I did look up the version in A-M. in A-M, $x$ is an element of $A$, defined to be the sum of the coefficients of the characteristic polynomial of $\phi=\textrm{id}_M$.

In detail the proof goes as follows:

Let $A$ be a commutative ring, $M$ a finitely generated $A$ module. If $I\subseteq A$ is an ideal such that $IM=M$, then there exists $x\in A$, $x\equiv 1 \pmod{I}$ such that $xM=0$.

Proof

Let $\phi=\textrm{id}_M$. $\phi(M)=M=IM$, so $\phi(M)\subseteq IM$. Then by 2.4 (generalized Cayley-Hamilton kinda) $\phi$ satisfies a polynomial equation $$\phi^n + a_1\phi^{n-1}+\cdots + a_n=0,$$ with $a_j\in I^j$. Let $x=1+a_1+\cdots + a_n$. Certainly $x\equiv 1\pmod{I}$. Then if $m\in M$, $$xm = (1+a_1+\cdots a_n)m = m + a_1m + \cdots a_nm = \phi^n(m) + a_1\phi^{n-1}(m)+\cdots a_nm=0.$$ Hence $x$ annihilates $M$, as desired.

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