# Field Extension over the Field of Rational Functions is Finite

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.

• I meant there is a subfield of $K(T)$ such that it is isomorphic to $\mathbb{Q}$. I will correct it. – matt Nov 29 '18 at 23:05
• Just show that $X$ is algebraic over $K(T)$ – user8268 Nov 29 '18 at 23:59
• Is $T(X)$ the same thing as $T$? – Inactive - avoiding CoC Nov 30 '18 at 0:30
• $T(X) =\frac{\sum_{j=0}^J a_j X^j}{\sum_{l=0}^L b_l X^l}$ 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. – reuns Nov 30 '18 at 1:13

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 $$K(T)$$.