A question of algebraic geometry applied to field theory I’ve come across this question in a coding theory course, and it has stumped me.  Any hints and/or suggestions would be appreciated.
Let $F$ be a field (for our purposes, assumed to be finite of order $q$).  Let $F\left[x,y,z\right]$ denote the ring of polynomials in three indeterminates with coefficients in $F$.  Let $p\left(x,y,z\right)\in F\left[x,y,z\right]$ be a homogeneous polynomial of positive degree, and let $$\chi_{F}:=\left\{\left(x:y:z\right)\in FP(2)\Big|p\left(x,y,z\right)=0\right\}$$ denote the projective algebraic plane curve over $F$ generated by $p$, where $FP(2)$ denotes the projective plane coordinatized by $F$.
Let $\frac{f}{g},\frac{h}{k}\in F\left(x,y,z\right)$ be rational functions in three indeterminates with coefficients in $F$.  $\frac{f}{g}$ and $\frac{h}{k}$ are called $equivalent$ $under$ $\chi$ iff $p\mid\left(fk-hg\right)$, and we write $\frac{f}{g}\sim\frac{h}{k}$.  Moreover, a rational function $\frac{f}{g}$ is equivalent to $0$ iff $p\mid f$.  Define $$F\left(\chi\right):=\left(\left\{\frac{f}{g}\in F\left(x,y,z\right)\Big|f,g\hspace{4pt}\mathrm{homogeneous};\deg(f)=\deg(g);p\nmid g\right\}\cup\left\{0\right\}\right)\Big/\sim.$$
I am trying to show that $F\left(\chi\right)$ is a field containing $F$ as a subfield.  The latter is easy to show; we simply identify $F$ with the constant rational functions.  As for showing that it is a field, I am stumped.  I’ve tried to show that the equivalence relation $\sim$ generates a maximal ideal, but that hasn’t panned out.  I would appreciate any suggestions on how to proceed.
 A: Note: I think that you need to assume that the polynomial $p$ is irreducible for this to work, and I don't see that assumption in your question.

Well one way would be to show first that addition and multiplication of classes in $F(\chi)$ is well defined. Then you can just prove that the operations satisfy the field axioms, where addition and multiplication are defined by
$$
\left [ \frac{f}{g} \right] + \left [ \frac{h}{t} \right] = \left [ \frac{ft + gh}{gt} \right]
$$
and 
$$
\left [ \frac{f}{g} \right]\cdot \left [ \frac{h}{t} \right] = \left [ \frac{fh}{gt} \right]
$$
For instance, to show that a nonzero element has a multiplicative inverse. Let the class $[0] \neq\left[ \frac{f}{g} \right] \in F(\chi)$. Being nonzero in $F(\chi)$ means that under the equivalence relation $\sim$, we have $\frac{f}{g} \nsim \frac{0}{1}$, which in turn means that $p$ doesn't divide $f\cdot 1 - 0\cdot g = f$. Since $p \nmid f$ then $\left[ \frac{g}{f} \right] \in F(\chi)$. But then 
$$
\left [ \frac{f}{g} \right] \cdot \left [ \frac{g}{f} \right] = \left[ \frac{fg}{gf}\right] = [1] \in F(\chi)
$$
so $\left [ \frac{g}{f} \right] = \left [ \frac{f}{g} \right]^{-1}$.
And in this way you can prove that the other field axioms are satisfied.
If you want to know more, $F(\chi)$ as described above is just what is called the function field of the projective curve $\chi_F$ (in your notation), and you can find it defined as the field of fractions of the coordinate ring of the curve (or variety in general) in any algebraic geometry book.
