The question is what's the meaning of the symbol $\phi$?
If it just a mapping,what's the mean of the equation?
I guess it's $\phi(x)$, $x$ is the element of $M$. Then $\phi(x)$ is the element of $M$ too,the equation will mean a element of $M$ plus a element of ideal a equal to zero.
We haven't define the sum.The binary operation of module should be $A \times M \to M$. The symbol $\phi$ is not $\phi(x)$.
Could you tell me what's meaning of a symbol in the proposition or what's the mean of the equation.
I'm sorry for my poor english.
The set of endomorphisms of an $A$-module $M$, just like the set of linear mappings from a vector space to itself, form a ring, if you define the operations pointwise. In other words, if $\phi,\psi$ are a $A$-module endomorphism, define $$\begin{eqnarray} (\phi \mathbf{+} \psi) &\,:\,& M \to M &\,:\,& x \mapsto \phi(x) + \psi(x) \\ (\phi \circ \psi) &\,:\,& M \to M &\,:\,& x \mapsto \phi(\psi(x)) \\ \mathbf{1} &\,:\,& M \to M &\,:\,& x \mapsto x \\ \mathbf{0} &\,:\,& M \to M &\,:\,& x \mapsto 0 \text{.} \end{eqnarray}$$ With that $\phi^n$ simply means $\underbrace{\phi\cdot\ldots\cdot\phi}_{\textrm{$n$ times}}$, i.e. $\phi^n(x) = \phi(\ldots\phi(x)\ldots)$. Just like as for numbers, we set $$ \phi^0 := \mathbf{1} \text{,} $$ i.e. for every endomorphism $\phi$, we define $\phi^0$ to be the identical mapping $x \mapsto x$.
Additionally, since you can multiply elements of an $A$-module with elements of $A$ (just like you can multiply and vector in a vector space with a scalar), every element $c \in A$ naturally corresponds to the $A$-module endomorphism $$ \mathbf{c} \,:\, M \to M \,:\, x \mapsto c\cdot x \text{.} $$
This allows you to evaluate poylnomials $p \in A[t]$ for arbitrary endomorphisms $\phi$. Let $$ p(t) = a_n t^n + \ldots + a_1 t + a_0 \in A[t] \text{.} $$ Then, evaluating $p$ at $\phi$, yields $$ p(\phi) = \mathbf{a_n} \circ \phi^n \mathbf{+} \ldots \mathbf{+} \mathbf{a_1}\circ \phi \mathbf{+} \mathbf{a_0} \text{,} $$ where $\mathbf{a_i}$ stands for the endomorphism corresponding to $a_i$, i.e. for the map $x \mapsto a_i\cdot x$. The expression $p(\phi)$ is thus endomorphism defined as the sum of producs of endomorphisms. Let's now apply an $x$ to that endomorphism, i.e. compute $$\begin{eqnarray} p(\phi)(x) &=& \left(\mathbf{a_n} \circ\phi^n \mathbf{+} \ldots \mathbf{+} \mathbf{a_1}\circ \phi \mathbf{+} \mathbf{a_0}\right)(x) \\ &=& \mathbf{a_n} \circ\phi^n(x) + \ldots + \mathbf{a_1}\circ \phi(x) + \mathbf{a_0}(x) \\ &=& a_n \cdot\phi^n(x) + \ldots + a_1 \cdot\phi(x) + a_0\cdot x \text{.} \end{eqnarray}$$ Note that bold symbols (e.g. $\mathbf{a_i}$ and $\mathbf{+}$) and $\circ$ represent operations on endomorphisms, while their regular counterparts represent the operations on the $A$-module $M$. It follows that $p(\phi)$ is simply the endomorphism $$ p(\phi) \,:\, M \to M \,:\, x \mapsto a_n\cdot \phi^n(x) + \ldots + a_1\cdot \phi(x) + a_0\cdot x \text{.} $$