# $K/k$ finite extension, then $K(X)/k(X)$ is finite

This is problem 3.5 of Falko, Algebra I Fields and Galois Theory.

Denote $K(X)$ the field of rational functions in one variable $X$ over field $K$. Suppose $K/k$ is a finite extension of degree $n$. Then $K(X)/k(X)$ is finite extension of degree $n$.

I am trying to see why $K(X)/k(X)$ would be finite. So it suffice to check degree $2$ extension for $K/k$. Suppose this is the case. Then every $f\in K(X)$ can be written into ratio of $g$ and $h$ where $g,h\in K[x]$. Since $K/k$ is degree 2, we can label every $c\in K$ as $c=c_1e_1+c_2e_2$ for $e_1,e_2$ basis of $K$ as $k$ vector space. Then $h=h_1e_1+h_2e_2,g=g_1e_1+g_2e_2$, $h_i,g_i\in k(X)$. Say $f=\frac{h}{g}=\frac{e_1+\frac{h_2}{h_1}e_2}{e_1+\frac{g_2}{g_1}e_2}\frac{h_1}{g_1}=\frac{e_1+c'_1e_2}{e_1+c'_2e_2}c'_3$ where $c'_i\in k(X)$. However I cannot get rid of denominator. However from my intuition $K(X)=K\otimes_k k(X)$. Thus it will become 2 dimension vector space automatically.

Question: How do I write down such a basis explicitly say for $n=2$?

• Commented Nov 26, 2018 at 10:11

Assume $K = k(\alpha)$. Then $k(\alpha)(X) = k(X)(\alpha)$. So $1,\alpha,\ldots,\alpha^{n-1}$ is a basis of the $k(X)$-vector space $K(X)$, with $n = [k(\alpha):k]$.
In your example $K = k(\alpha), [k(\alpha):k] = 2$ so $K(X)=\{\frac{u_1(X)}{v_1(X)}+\alpha\frac{u_2(X)}{v_2(X)}, u_i,v_i \in k[X]\}$.
If you have $\frac{a(X)+\alpha b(X)}{c(X)+\alpha d(X)}, a,b,c,d \in k[X]$ then $$\frac{a(X)+\alpha b(X)}{c(X)+\alpha d(X)} =\frac{(a(X)+\alpha b(X))(c(X)+\overline{\alpha} d(X))}{(c(X)+\alpha d(X))(c(X)+\overline{\alpha} d(X))}= \frac{w_1(X)+\alpha w_2(X)}{z(X)}$$ where $\overline{\alpha} = e+\alpha f$ is the other root of the minimal polynomial of $\alpha$ over $k$
• $K/k$ might not be generated by a single element. Commented Aug 6, 2017 at 4:28
• @reuns Is it clear that given $[K:k]=n$ extension which may not be simple that you can always find a norm map which sends denominator of $K(X)$'s element to $k(X)$? For quadratic case, it is clear. For arbitrary case, it seems a bit over stretching? Commented Aug 6, 2017 at 4:46
• @user45765 That's what proves $k(\alpha)(X)=k(X)(\alpha)$. If $K/k$ is separable then take $N/k$ its Galois closure, then $f \in k(\alpha)[X]$ implies $\prod_{\sigma \in Gal(N/k)} f^\sigma(X) \in k[X]$ where $\sigma$ acts on the coefficients of $f$. If $K/k$ is not separable then be careful with the multiplicities of the minimal polynomial of $\alpha$. Commented Aug 6, 2017 at 4:50