Why is $\alpha$ transcendental over $K$ 
Let $K$ be a field and let $\alpha$ be an element of the field $K(T)$ of rational functions, with $\alpha\not\in K$. Prove that $\alpha$ is transcendental over $K$.

In this case $\alpha$ is of the form $\displaystyle\frac{f(T)}{g(T)}$, where $f,g\in K[T]$ and $g$ is not the zero polynomial.
Being algebraic over $K$ means that $\exists P\in K[T]$ s.t. $P\left(\frac{f(T)}{g(T)}\right)=0$.
Now what is the problem here, why does it hold only if the fraction is in $K$? Is it true because $T$ is an indeterminate ?
 A: Suppose $\alpha\in K(T)\text{ \ } K$
Then $\alpha={f(T)\over g(T)}$ for some rational functions $f$ and $g\ne0 $
WLOG assume that $(f,g)=1$
Suppose on contrary that $\alpha$ is algebraic, then $K(\alpha)$ is a field and $K(\alpha)=K(1/\alpha)=K(g/f)$
Thus, ${f\over g}\in K(g/f)$
${f\over g}=a_0+a_1{g\over f}+\cdots+a_n({g \over f})^n$
$f^{n+1}=a_0f^ng+a_1f^{n-1}g^2+\cdots+a_ng^{n+1}$
$g|f^{n+1}\Rightarrow g|f$ (Since, $(f,g)=1$)
Similarly it can be shown that $f|g$
We have $\alpha={f\over g}$ is a unit in $K$. 
Contradiction.
A: Assume $\alpha$ is algebraic, say
$$ a_n\frac{f^n(T)}{g^n(T)}+a_{n-1}\frac{f^{n-1}(T)}{g^{n-1}(T)}+\ldots+a_1\frac{f(T)}{g(T)}+a_0=0$$
with $n\ge 1$.
We may assume wlog that $a_0\ne 0$ (else divide $P(X)$ by $X$) and that $\deg f\ne \deg g$ (else subtract the quotient of the leading coefficients, which is in $K$). Then 
$$ a_nf^n(T)+a_{n-1}f^{n-1}(T)g(T)+\ldots+a_1f(T)g^{n-1}(T)+a_0g^n(T)=0,$$
where the left hand side is a polynomial of degree $$n\max\{\deg f,\deg g\}\ge\max\{\deg f,\deg g\}\ge 1,$$
where the last inequality follows because at least one of $f,g$ must be nonconstant to have $\alpha\notin K$.
But the nonzero polynomials in $K[T]$ are, well, nonzero. 
A: Less elementary, but quicker: $T$ is algebraic over $K(\alpha)$ (why?), so if $\alpha$ is algebraic over $K$ we get that $T$ is algebraic over $K$, a contradiction.
