Why $(b \rightharpoonup c ) = b\cdot c$ when $b,c \in A$?

Question

Let $A$ be an algebra. We define $\forall f \in A^*, a \in A $ $$ a \rightharpoonup f: A \rightarrow \mathbb{k}, l \mapsto f(la) $$ This operation defines a module structure on $A^*$ so it is associative. I'm wondering why $b \rightharpoonup (c \rightharpoonup g) = (b \rightharpoonup c ) \rightharpoonup g = b\cdot c \rightharpoonup g $, where $g \in A^*$. How can I use $a \rightharpoonup f$ when $f$ is an element of A? Why $b \rightharpoonup c $ becomes a product in $A$? Maybe there is a property of associativity that I don't get.

Answer 1

The requirement (for modules) that you call associativity seems to have confused you because of ambiguous notation. People often write it as $a\cdot(b\cdot x)=(a\cdot b)\cdot x$ without mentioning that the two occurrences of $\cdot$ on the left side and the second occurrence on the right side refer to the module structure (multiplying a module element by an element of the ring of "scalars") whereas the first occurrence on the right is the multiplication within that ring of scalars.

In your situation, the former operation has been given the unusual symbol $\rightharpoonup$, while the latter is still written as multiplication. So the associative law reads, in this notation, $a\rightharpoonup(b\rightharpoonup x)=(a\cdot b)\rightharpoonup x$.

In your context, $\rightharpoonup$ between two elements of $A$ has no meaning; it has been defined only with an element of $A$ on its left and an element of $A^*$ on its right.

Answer 2

Once you have cleared up notation (see answer by @Andreas Blass), this is a one line calculation (after reading that answer try this yourself, as you may not need this):

$$ (b\rightharpoonup(c\rightharpoonup g))(l)= $$

$$(c\rightharpoonup g)(l b)=g(l b c)$$

$$=((b\cdot c)\rightharpoonup g)(l)$$