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Let $K$ be a field and $K(X)$ the field of rational functions with one variable over $K$. If $T(X) \in K(X)$ show that the field extension $K(X)/K(T)$ is finite.


The definition of the field of rational function was:

$$K(X):=\{\frac{f}{g} | f,g \in K[X], g \neq 0\}$$

The field operations for $K(X)$ are:

$$\frac{f_1}{g_1} + \frac{f_1}{g_1} = \frac{f_1 g_2 + f_2 g_1}{g_1 g_2}$$ $$\frac{f_1}{g_1} + \frac{f_1}{g_1} = \frac{f_1 f_2}{g_1 g_2}$$


I have the feeling it has something to do with separability. There is a subfield of $K(T)$ which is isomorph to $\mathbb{Q}$, therefore the charateristic of $K(T)$ is $0$, which means it is a perfect field. Then $K(X)/K(T)$ is seperable.

But I have no idea how to connect this to $[K(X):K(T)]<\infty$. Maybe I have a completely wrong approach. Any help would be appreciated and I'm sorry if this is a duplicate.

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  • $\begingroup$ I meant there is a subfield of $K(T)$ such that it is isomorphic to $\mathbb{Q}$. I will correct it. $\endgroup$ – matt Nov 29 '18 at 23:05
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    $\begingroup$ Just show that $X$ is algebraic over $K(T)$ $\endgroup$ – user8268 Nov 29 '18 at 23:59
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    $\begingroup$ Is $T(X)$ the same thing as $T$? $\endgroup$ – Inactive - avoiding CoC Nov 30 '18 at 0:30
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    $\begingroup$ $T(X) =\frac{\sum_{j=0}^J a_j X^j}{\sum_{l=0}^L b_l X^l}$ then $X$ is a root of $T(X)\sum_{l=0}^L b_l z^l-\sum_{j=0}^J a_j z^j\in K(T(X))[z]$. If $\gcd(\sum_{j=0}^J a_j X^j,\sum_{l=0}^L b_l X^l) = 1$ then it is its minimal polynomial. $\endgroup$ – reuns Nov 30 '18 at 1:13
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Let $f,g\in K[X]$ be such that $T=\frac{f}{g}$, and consider the polynomial $$h:=Tg(Y)-f(Y)\in (K(T))[Y].$$ Note that $h=0$ if and only if $f=g$, in which case $T\in K$ is constant and the extension $K(X)/K(T)$ is in fact not finite.

If $h\neq0$ then the fact that $h(X)=0$ implies that $X$ is algebraic over $K(T)$, and hence that $K(X)$ is a finite extension of $K(T)$.

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