Show that rational functions are unique after canceling common divisors There is a proof in my book that I don't understand. I hope somebody can help me with this part:
If $R=\frac{f}{g}$ where $f$ and $g$ are polynomials with no common Divisor i.e. if $f=qd$ and $g=q'd$ then $d$ is a constant then the following implication is true:
$$R=\frac{F}{G}\wedge F,G \text{ have no common divisors} \Longrightarrow F=cf\wedge G=cg,c\in\mathbb{C}^*$$
And also the Domain where the function $R$ is defined is unique.
The proof:
Let $\frac{f}{g}=\frac{F}{G}=R$. Then by Definition of a rational function there exists infinite $z\in\mathbb{C}$ such that $g(z)\neq 0$ and $G(z)\neq 0$. For the Domain where this property is given we have.
$$F(z)g(z)=G(z)f(z)$$
and then by the identity Theorem $Fg=Gf$. 
I am not sure why the identity Theorem can be applied here. The identity Theorem says that if we have two polynomials
$$f(x)=a_nx^n+...+a_0$$
$$g(x)=b_nx^n+...+b_0$$
i.e two polynomials of degree n then if $f(x)$ and $g(x)$ is equal for $n+1\,x's$ then $f=g$
My first Question is therefore how do we know that $F(z)g(z)$ and $G(z)f(z)$ are of the same degree? Or more generaly how do we know if $a(x)=b(x)$ for all $x$ then $a=b$
The other Thing that I did not understand was that after we have conclude that $Fg=Gf$ then because of the fact that there are no common Divisors (it was not specified whether they are Talking About $(f,g)$ or $(F,G)$) we have $G=cg$ and $F=cf$.
The last Thing I do not understand is why the Domain must be unique in the above proof we have found out that for $g(z)\neq 0$ and $G(z)\neq 0$ we can conclude that $F=cf$ and $G=cg$. But what happens if $G(z)\neq 0$ but $g(z)=0$ and the other way round?
Thank you for your time
EDIT:
I could anwer the first Question by myself but I still don't know what Comes after I have deduced $F(z)g(z)=G(z)f(z)$
 A: We have $F(z)g(z)=f(z)G(z)$ for all $z$ where boteh $g(z)\ne 0 $ and $G(z)\ne 0$. As neither of $g,G$ is the zero polynomial, they have at most finitely many roots each. After removing these finitely many points from $\Bbb C$, we are still left with infinitely many $z$ where the equality holds. Thus the identity theorem is applicable.
Note that in the identity theorem, we do not need that $f,g$ have degree $n$, instead we need that they have degree at most $n$. Indeed, all we need is that they can be written as $f(x)=a_nx^n+\ldots$ etc., i.e., $a_n=0$ and/or $b_n=0$ is allowed.
Now that we know that $Fg=fG$, we make use of $\gcd(f,g)=1$. By Bezout, there are polynomials $u,v$ such that $uf+vg=1$. So 
$$G= ufG+vgG=uFg+vgG=(uF+vG)g.$$
Hence $G=cg$ where $c$ is the polynomial $uF+vG$.
Similarly, we find $F=\tilde cf$ for some polynomial $\tilde c$. 
With this,  $Fg=fG$ becomes $\tilde cfg=cfg$, or:
$$(\tilde c-c)fg=0.$$
As above, we conclude that $\tilde c(x)=c(x)$ for infinitely many $x$ and hence $\tilde c=c$.
In other words, there exists apolynomial $c$ such that $F=cf$ and $G=cg$.
As $c$ is clearly a common divisor of $F$ and $G$, we conclude that $c$ is in fact a (non-zero) constant.
A: The first thing is that the identity theorem applies to two polynomials whose degrees are both bounded by some number $n$, i.e., if you have a quadratic equal to a cubic (so $n = 3$) and they agree at 4 points, then they're equal (hence the leading coeff of the cubic is zero). To put it differently, in the identity theorem, it's perfectly OK to say that, for instance, $z^2 + z + 1$ is a cubic...with leading coefficient zero...and then apply the theorem. The only point of having a common degree $n$ is to tell you at how many points the polynomials must agree before you can say that they're the same. 
For the second statement, you know that $Fg = Gf$; that means (by factoring everything) that any irreducible factor of $f$ is a factor of $Fg$, and therefore a factor of either $F$ or $g$...but it cannot be $g$ because $f$ and $g$ have no common factors. So every factor of $f$ is a factor of $F$, hence $F = f \cdot \text{something}$. Make the same argument about factors of $F$ to get that $f = F \cdot \text{something else}$. Conclude that the two somethings must both be constants. 
