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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 ?

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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.

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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.

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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.

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  • $\begingroup$ $T$ is algebraic over $K(\alpha)$, it is the root of $X-T\in K(\alpha)[X]$. and $T$ is not algebraic over $K$, assume $\exists P$, $P(T)=0$ ''should imply'' $P=0$, since $1,T,T^2,...$ are linearly independent, right ? Now you take $g=1, f=id$, but is it sufficient to generealize ? $\endgroup$ – derivative Nov 30 '14 at 11:57
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    $\begingroup$ @derivative $T$ is a root of the polynomial $f(X)-\alpha g(X)\in K(\alpha)[X]$. (I don't get the last sentence of your comment. Maybe you need to recall that a tower of algebraic extensions gives rise to an algebraic extension?) $\endgroup$ – user26857 Nov 30 '14 at 12:00
  • $\begingroup$ why $\alpha$ algebraic over $K\implies T$ algebraic over $K$ ? If $f=id, g=1$ then $\alpha=\frac fg=T$ then the implication is correct, otherwise ? $\endgroup$ – derivative Nov 30 '14 at 12:08
  • $\begingroup$ @derivative Since $\alpha$ is algebraic over $K$ we have $K\subset K(\alpha)$ algebraic extension and combined with $K(\alpha)\subset K(T)$ algebraic we get that $K\subset K(T)$ is also an algebraic extension. (Btw, if you are not aware of these basic facts, then maybe you should focus on the other answer.) $\endgroup$ – user26857 Nov 30 '14 at 12:10

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