I am reading "Introduction to Commutative Algebra" written by Michael Atiyah; In the Finitely Generated Modules section, there's a corollary that from Proposition 2.4 which I don't understand its proof:
Let $M$ be a finitely generated $A$-module and let $\mathfrak{a}$ be an ideal of $A$ such that $\mathfrak{a} M = M$. Then there exists $x \equiv 1 \pmod{a}$ such that $xM=0$.
Suppose $\iota$ is the identity function on $M$.
There exists $a_1, \dotsc , a_n \in \mathfrak{a}$ with $\iota^n + a_1 \iota^{n-1} + \dotsb + a_n = 0$.
Hence, since $\iota^n(m) = \iota(m)$, we have $\iota + a_1 \iota + a_2 \iota + \dotsb + a_n = 0$.
Therefore we have for every $m \in M$ that $m + a_1 m + a_2 m + \dotsb + a_n = 0$.
In the book's own proof $1 + a_1 + \dotsb + a_n$ is the desired $x$. I don't know what my mistake is, what I have obtained is $m + a_1 m + a_2 m + \dotsb + a_n = 0$. The coefficient $a_n$ has not been multiplicated in $m$.
Best Answer
The proposition in $2.4$ states that there is a module endomorphism $\phi : M \rightarrow M$ and $a_{1} , \dots , a_{n} \in \mathfrak{a}$ such that $\phi^{n} + a_{1} \phi^{n-1} + \dots + a_{n} \equiv 0$. Note that $a_{0} = a_{0} \cdot i$, where $i : M \rightarrow M$ is the identity map. This is a sort of Cayley Hamilton analogue, if $f(t)$ is a characteristic polynomial of the matrix $A$, then $f(A) = 0$, and we consider the constant term $c_{0}$ as $c_{0} \cdot I$ for the equation to hold true. So: for $\phi = i$ $$ (\phi^{n} + a_{1} \phi^{n-1} + \dots + a_{n})(m) = (1 + a_{1} + \dots + a_{n})(m) = 0 , \forall m \in M \\ \implies x = (1 + a_{1} + \dots + a_{n}) \equiv 1 \pmod{\mathfrak{a}} $$