Unique expression of a polynomial under quotient mapping?

Solution 1:

Hint: Use the Division Algorithm for polynomials. Any polynomial $g(x)$ is congruent modulo $f(x)$ to a unique polynomial $r(x)$, where $r(x)=0$ or $r(x)$ has degree less than the degree of $f(x)$.

Alternately and somewhat more painfully, you have shown the result for $y^n$. Then do an induction, by writing down explicitly the expression for $y^{k+1}$ from the assumed expression for $y^k$.

Solution 2:

Hint $\ $ By the polynomial (long) Division Algorithm in $\rm\:F[x]\:$ we can divide any $\rm\:g\in F[x]\:$ by $\rm\:f\:$ yielding $\rm\ g = q\, f + r\ $ with $\rm\: deg\ r < deg\ f.\ $ The uniqueness of the remainder $\rm\,r\,$ follows just as for integers: if $\rm\ q\, f + r = g = q'\, f + r'\,$ then $\rm\,(q-q')\,f = r-r'\:$ so $\rm\,r-r' = 0,\,$ else $\rm\:f\:$ would divide the smaller degree $\rm\,r-r'\ne 0.\,$ (Further $\,\rm (q-q')f = 0\,$ so $\rm\,q-q' = 0\,$ by cancelling $\rm\,f\neq 0)$