# Field extension

• Feb 19th 2010, 03:31 AM
dangkhoa
Field extension
Let K(s) be the field of rational functions in one variable s, i.e the field of fractions of K[s]. Then K(s) is a field extension of K(s^n).

Prove that [K(s):K(s^n)] = n. Hence show that the minimum polynomial of s over K(s^n) is (t^n) - (s^n)

How do you prove it?

Thank you very much
• Feb 19th 2010, 07:12 PM
aliceinwonderland
Quote:

Originally Posted by dangkhoa
Let K(s) be the field of rational functions in one variable s, i.e the field of fractions of K[s]. Then K(s) is a field extension of K(s^n).

Prove that [K(s):K(s^n)] = n. Hence show that the minimum polynomial of s over K(s^n) is (t^n) - (s^n)

How do you prove it?

Thank you very much

Let $\displaystyle F=K(s^n)$. Then, $\displaystyle \{1_{F}, s, s^2, \cdots, s^{n-1}\}$ is a basis of vector space $\displaystyle K(s)$ over $\displaystyle F=K(s^n)$, which implies that every element of $\displaystyle K(s)$ can be written uniquely in the form of $\displaystyle c_0 + c_1s \cdots c_{n-1}s^{n-1} (c_i \in F)$. Thus $\displaystyle [K(s):K(s^n)] = n$.

$\displaystyle f(t)=t^n-s^n \in K(s^n)[t]$ is an irreducible monic polynomial of degree n over $\displaystyle K(s^n)$ such that $\displaystyle f(s)=0$.