# If a field $F$ is an algebraic extension of a field $K$ then $(F:K)=(F(x):K(x))$

Suppose $K$ is a field and $F$ is an algebraic extension of some degree $n=(F:K)$. It is stated that the field of rational functions $F(x)$ is in fact an algebraic extension of the field $K(x)$ and moreover $(F(x):K(x))=n$.

How do I approach this exercise? I’m new in this area so any help would be greatly appreciated!

#### Solutions Collecting From Web of "If a field $F$ is an algebraic extension of a field $K$ then $(F:K)=(F(x):K(x))$"

Let $a_1,a_2,\dots,a_n$ be a basis for $F$ over $K$. Let’s first show that they are linearly independent as elements of $F(x)$ over $K(x)$. So assume
$$\sum_{i=1}^n a_i\frac{f_i(x)}{g(x)}=0$$
where $f_1,\dots,f_n,g\in K[x]$ (it’s not restrictive to assume the denominators are the same). This implies
$$\sum_{i=1}^n a_if_i(x)=0$$
If we have $f_i(x)=\sum_{j=0}^k b_{ij}x^j$, we deduce
$$\sum_{i=1}^n a_ib_{ij}=0,\quad j=0,1,\dots,k$$
so all polynomials are $0$.

Now the task is to show that $F(x)=K(x)[a_1,a_2,\dots,a_n]$ and we can reduce this to showing that, if $a$ is algebraic over $K$, then
$$K[a](x)=K(x)[a]$$
One inclusion is obvious, namely $K(x)[a]\subseteq K[a](x)$. In order to show the converse inclusion we just need to prove that if $f(x)\in K[a](x)$, then $f(x)\in K(x)[a]$, because the latter is a field. This is trivial, by considering polynomials of the form $cx^m$, where $c\in K[a]$. Just write $c=d_0+d_1a+\dots+d_ra^r$, where $r+1$ is the degree of $a$ over $K$ and then $cx^m=\sum_{j=0}^r a^j(d_jx^m)\in K(x)[a]$.