In a paper, I found a proof that any is non-constant element of the rational functions $F(x)$ over the field $F$ is transcendental with respect to the base field $F$. However, there is something in this proof that I do not understand and I was hoping that someone could explain it to me.
The proof goes as follows:
Let $q = \frac{f(X)}{g(X}$ be a non-constant rational function reduced to lowest terms and let the greater of the degrees of its numerator and denominator be $n$.
The element $X$ satisfies the equation $q g(X) - f(X) = 0$, the coefficients of which are in the field $F(q)$. If all these coefficients were zero, we could take one non-zero coefficient $b_{\nu}$ in $g(X)$ and the coefficient $a_{\nu}$ of the same power of $X$ in $f(X)$, and then we would have that $qb_{\nu}-a_{\nu} = 0$.
This would mean that $\displaystyle q = \frac{a_{\nu}}{b_{\nu}}$ equals a constant, contrary to the supposition.
Thus, at least one coefficient in $qg(X) - f(X) = 0$ differs from zero, and we conclude that $X$ is algebraic with respect to $F(q)$.
If $F(q)$ were algebraic with respect to $F$, then also $X$ should be algebraic with respect to $F$, which it is not. Therefore, we see that $F(q)$ is transcendental.
The things I don't understand are the things I wrote in bold face.
- I don't understand how if all the coefficients of $qg(X) - f(X) = 0$ are zero we can take one non-zero coefficient $b_{\nu}\in g(X)$ and the coefficient $a_{\nu}$ of the same power of $X$ in $f(X)$. Didn't they just say all the coefficients were zero??
- I don't understand how what I mentioned in bullet point 1 allows us to turn $qg(X) - f(X) = 0$ into $gb_{\nu}-a_{\nu}=0$.
- I don't understand how showing that $F(q)$ is transcendental tells us that $q$ itself is transcendental with respect to $F$, although I have the feeling this might be due to some theorem.
If someone could please explain these three things to me, I would be very much appreciative. What it's saying is pretty much exactly what I need, but it is useless to me if I can't wrap my head around it, or if it is incorrect!
Thank you ahead of time for your time and patience.
They said that the coefficients of the polynomial $h(X)=qg(X)-f(X)$ are zero, not that the coefficients of $g(X)$ and $f(X)$ are zero. Since $g(X)$ is a nonzero polynomial (it's the denominator of a rational function), it must have some coefficient which is nonzero.
We are considering $h(X)=qg(X)-f(X)$ as a polynomial with coefficients in $F(q)$. If $g(X)=\sum_{n=0}^N b_nX^n$ and $f(X)=\sum_{n=0}^N a_n X^n$ (here $N=\max(\deg f, \deg g)$), then $$h(X)=\sum_{n=0}^N(qb_n-a_n)X^n.$$ If $h(X)=0$, that means all its coefficients are $0$, so $qb_n-a_n=0$ for all $n$.
If $q$ were algebraic over $F$, then $F(q)$ would be algebraic over $F$ (a field extension generated by algebraic elements is algebraic). Since $F(q)$ is not algebraic over $F$, $q$ cannot be algebraic over $F$.