Consider the quotient ring $A=k[x_1,x_2,...,x_n]/(F)$, that is, the ring of residue classes modulo the ideal generated by $0 \not= F \in k[x_1,x_2,...,x_n]$.
Im not quite sure i understand this quotient ring, I have always thought about $A=k[x_1,x_2,...,x_n]/(F)$ as the set of all congruence classes containing polynomials congruent to each other modulo $F$. Is this wrong?
I am reading undergraduate commutative algebra by Miles Reid, I have a feeling he treats this quotient ring slightly different (I just started reading the book).
If $R$ is a ring and $I$ is an ideal of $R$, you should think of $R/I$ as a natural ring equipped with a surjective homomorphism $p : R \to R/I$ with $\ker(p)=I$. This means in particular that $p(a)=p(b)$ iff $a-b \in I$. Once $R/I$ with this property is constructed, you may forget about cosets completely. You will never need them again. Far more important is the fundamental theorem on homomorphisms: Any homomorphism $f : R \to S$ with $f|_I = 0$ extends uniquely to a homomorphism $\overline{f} : R/I \to S$. This means that $R/I$ is the "universal ring over $R$ which kills $I$". In particular, in your example, $k[x_1,\dotsc,x_n]/(F)$ is the universal $k$-algebra containing $n$ elements $x_1,\dotsc,x_n$ such that $F(x_1,\dotsc,x_n)=0$ (here, $p(x_i)$ are just named $x_i$). Thus, it is the universal $k$-algebra which has a root of $F$. For $n=1$ this basic construction appears in field theory, for $n>1$ it appears in algebraic geometry. In fact, $k[x_1,\dotsc,x_n]/(F)$ is the coordinate ring of the affine variety $V(F) \subseteq \mathbb{A}^n_k$ which is cut out by the single equation $F=0$. For example, look at $k[x,y]/(x^2+y^2-1)$, the coordinate ring of the "circle". A $k$-basis is given by $1,x,x^2,\dotsc,y,xy,x^2 y,\dotsc$. Of course $y^2$ is also contained in this algebra, but you can express it as $y^2=1-x^2$.