Atiyah-Macdonald Proposition 2.4

Question

I don't understand what the above equation means. My interpretation is that given any element $x$ of $M$, the equation $\phi (x)^{n}+a_{1}\phi(x)^{n-1}+...+a_n=0$ holds. Am I correct? I don't think so, because modules aren't necessarily closed under multiplication.

Answer 1

Note that $\phi^{n} = \phi \circ \phi \cdots \circ \phi$, $n$ times. So, $\phi^{n} (x)=\phi(\phi(\cdots\phi(x)))$, $n$ times.

Then $$ \phi^{n}+a_{1}\phi^{n-1}+\cdots+a_1\phi+a_n =0 $$ means $$ \phi^{n} (x)+a_{1}\phi^{n-1} (x)+\cdots+a_1\phi(x)+a_n x=0 $$ for all $x \in M$. Just note that the $a_i$ do not depend on $x$.