$A[x]$ is an $A$-module if $A$ is a commutative ring with unity? Problem: Show that if $A$ is a commutative ring with unity then the polynomial ring $A[x]$ is an $A$-module.
I have some confusion when showing that:
+) $a\in A,\; f(x), g(x) \in A[x]$ then
$a\, \bigl[ f(x) + g(x) \bigr] = a\,f(x) + a\,g(x)$.
with the $a\,f(x)$ is the multiplication of a number in $A$ and polynomial $f(x) \in A[x]$.
Isn't it obvious or should I consider $a$ to be an element of $A[x]$, then the statement holds for the distribution in $A[x]$?
The same questions go for the other 3 criteria. Thank you for any suggestions!
 A: It sounds like you want to explicitly define the action of the ring $A$ on the module $A[x]$. In other words, you want to define the map
\begin{align}
A \times A[x] &\to A[x] \\
\bigl( a,\, f(x) \bigr) &\mapsto a \cdot f(x)
\end{align}
When you check that $A[x]$ is an $A$-module, you are actually verifying properties of this map.
The trick in this case is to think of the base ring (a.k.a. ring of scalars) $A$ as the polynomials of degree $0$ in the natural way. That is, identify $a \in A$ with its image $\iota(a) \in A[x]$ under the natural inclusion
$$
\iota: A \hookrightarrow A[x].
$$
Then, you can think of $a \cdot f(x)$ as $\iota(a)\, f(x)$, where the latter is multiplication in the polynomial algebra. But this is actually overkill.
In order to define the action, just take an arbitrary scalar $a \in A$ and polynomial
$$
f(x) = b_0 + b_1x + \cdots b_nx^n,
$$
where $n \in \mathbb{Z}_{\geq 0}$ and $b_0, b_1, \dots, b_n \in A$, and define
$$
a \cdot \bigl( b_0 + b_1x + \cdots b_nx^n \bigr) 
= ab_0 + ab_1x + \cdots ab_nx^n. 
$$
Now use this definition to check the module properties.
