If K/F is a finite field extension of degree n, so is K(x)/F(x), where K(x) is the field of rational functions in one variable over K, idem F(x).
I could prove K(x)/F(x) is a finite extension, but I cannot prove the degree is n. It helped me to have found that K(x) = K(p), where p is the polynomial p(x) = x. Analogously, F(x) = F(p). Any hint will be welcome. Thanks for reading.
November 5th 2009, 04:43 PM
is finite iff it's algebraic and finitely generated. Let be generators of the extension try showing these also generate over (I'm not really sure it works but seems a good place to start).
November 6th 2009, 02:17 AM
It's what I did. I proceeded like this: let K:F=n. Let a_1,...,a_n be a basis of K over F. Then K= F(a_1,...,a_n). I know that
K(x)= K(p) and F(x)= F(p), (1)
where p belonging to F[x] is given by p(x)= x. Substituting,
K(p)= F(a_1,...,a_n)(p)= F(p)(a_1,...a_n). (2)
Because K/F finite, K/F algebraic and because a_i belongs to K, a_i algebraic over F and, all the more so, algebraic over F(p). Hence, F(p)(a_1,...,a_n)/F(p) is a finite extension. Keeping in mind (1) and (2), K(x)/F(x) is finite.
This done, I tried to prove a_1,...,a_n generate K(x) over F(x). Or that they are linearly independent over F(x). But in vain. Anyways, thank you for your post.