Let $F$ be a field and
$f_1=x^2+y^2, \ \ f_2=x^2y^2, \ \ f_3=x^3y-xy^3$
Denote by $A$ the $F$-algebra generated by $f_1, f_2$ and $f_3$. Writing $A$ as a module over $F[f_1,f_2]$ gives us
$$A=F[f_1,f_2] \oplus F[f_1,f_2] f_3,$$
because $f_3^2=f_1^2f_2-3f_2^2$.
I've read this in a book. My question is, how do we get the direct sum $A=F[f_1,f_2] \oplus F[f_1,f_2] f_3$ ? is the relation $f_3^2=f_1^2f_2-3f_2^2$ important?
Thanks for help.
To say that we can write $A$ as a direct sum $A=F[f_1,f_2] \oplus F[f_1,f_2]f_3$ means that $A$ has a basis given by $1,f_3$, so any element of $A$ can be written uniquely as a linear combination of $1$ and $f_3$ with coefficients from $F[f_1,f_2]$.
The point is that, given any element of $A$, you can write it as a polynomial in $f_3$ where the coefficients come from $B=F[f_1,f_2]$. (Simply collect terms which have $\ f_3$ appearing to the same degree.) Thus, any element $\alpha \in A$ can be written
$\alpha = a_1 + a_2 f_3 + a_3 f_3^2 + \cdots + a_n f_3^n$
for some elements $a_1,\ldots,a_n \in B$. But we can use the relation
$f_3^2 = f_1^2 f_2 - 3 f_2^2 \in B$
to eliminate any instance of $f_3^i$ where $i \geq 2$. Thus, we may write everything in terms of just $1$ and $f_3$, and its easy to check that such a representation will be unique.