Unique expression of a polynomial under quotient mapping? I have a weird feeling about something I'm reading. 
Suppose $f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ is a polynomial over a field $F$. Let $y=x+(f(x))$ be the image of $x$ in the quotient $F[x]/(f(x))$. Then every element of $F[y]$ can be uniquely expressed in form
$$
b_0+b_1y+\cdots+b_{n-1}y^{n-1}
$$
with $b_i\in F$.
I see that 
$$y^n = x^n+(f(x))
= -(a_{n-1}x^{n-1}+\cdots+a_1x+a_0)+(f(x))\\
= -a_{n-1}x^{n-1}+(f(x))+\cdots+-a_1x+(f(x))+-a_0+(f(x))
$$
so it looks like any power of $y$ greater or equal to $n$ can be written in terms of lower powers of $y$ if I could pull out the coefficients. However, why would the coefficients still be in $F$? Wouldn't they be somewhere else? And does uniqueness of this expression follow simply because the expressions $b_0+b_1y+\cdots+b_{n-1}y^{n-1}$ are polynomials, or is there something more to it than that? Thanks.
 A: 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 $\rm\,f\neq 0)$
A: 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 $y^k$.  
A: The map
$$\pi:F[x]\longrightarrow F[x]/(f(x))\,\,,\,\,\pi(g(x)):=g(x)+(f(x))$$
is a ring homomorphism under which the field $\,F\,$ is embedded in the field
$$\overline F:=\{a+(f(x))\;\:\;a\in F\}$$
Thus, we can, and we actually do, identify the elements of both fields, and under this agreement is that we can say the coefficients you ask about are in $\,F\,$, i.e. in $\,\overline F\,$.
About uniqueness: we know we can divide  $\,h(x)\in F[x]\,$, by $\,f(x)\,$ with residue (since we're in an Euclidean domain we can always do this!):
$$h(x)=q(x)f(x)+r(x)\,\,,\,\,r(x)=0\,\,or\,\,\deg r<\deg f\,\,\,\,(**)$$
Uniqueness follows from degree considerations: 
$$q(x)f(x)+r(x)=q'(x)f(x)+r'(x)\Longrightarrow (q(x)-q'(x))f(x)=r'(x)-r(x)$$
but if $\,q(x)-q'(x)\neq 0\,$ then $\,\deg\left(q'(x)-q(x)\right)f(x)\geq f(x)>\deg(r'(x)-r(x))\,$ which, of course, contradicts $\,(**)\,$ above.
The only thing left to do is to observe that 
$$g(x)+(f(x))=r(x)+(f(x))$$
