Degree of field extension and rational function field

Question

Suppose that $k$ and $k'$ are fields such that $k\subset k'$ and $[k':k]=n$, where $n$ is an positive integer.

Do we have $[k'(x):k(x)]=[k':k]$? Why?

Thanks for your help!

Answer 1

As you tagged the question with rational-functions I assume that $x$ is an indeterminate (i.e. an element that is transcendental over $k$ and hence also over $k'$).

Yes. It probably is not a surpise that a $k$-basis $\mathcal{B}$ of $k'$ is also a $k(x)$ basis of $k'(x)$. We get there for example as follows.

1. Writing individual coefficients of polynomials $\in k'[x]$ in terms of $\mathcal{B}$ shows that $k'[x]$ is a free $k[x]$-module with basis $\mathcal{B}$.
2. Consequently $\mathcal{B}$ is linearly independent as a subset of $k'(x)$ viewed as a vector space over the subfield $k(x)$. This is because we can clear the denominators of the coefficients of an eventual linear dependency relation reducing this to the case where there would be a linear dependency relation with coefficients from $k[x]$. In violation of step 1.
3. If $f(x)$ is an arbitrary non-zero polynomial $\in k'[x]$, I claim that $f(x)$ is a factor of a polynomial from $k[x]$. Consider the quotient $V=k'[x]/\langle f(x)\rangle$ as a module over both $k[x]$ and $k'[x]$. Because $\dim_k V<\infty$ the annihilator ideal $$I=\{a(x)\in k[x]\mid a(x)V=0\}$$ is non-trivial. As $k[x]$ is a PID $I$ is generated by a polynomial $g(x)$. The annihilator $I'$ of $V$ as a $k'[x]$-module is generated by $f(x)$. As $I\subset I'$ we must have $f(x)\mid g(x)$.
4. Let $z=p(x)/q(x)\in k'(x), p,q\in k'[x]$, be arbitrary. By step 3 there exists polynomials $r(x)\in k'[x]$ and $g(x)\in k[x]$ such that $r(x)q(x)=g(x)$. We can write $$ z=\frac{p(x)r(x)}{g(x)}=\frac{h(x)}{g(x)}, $$ where $h(x)\in k'[x]$ and $g(x)\in k[x]$. By step 1 $h[x]$ is a $k[x]$-linear combination of elements of $\mathcal{B}$. Consequently $z$ is a $k(x)$-linear combination of elements of $\mathcal{B}$.
5. Combining items 2. and 4. shows that $\mathcal{B}$ is a basis of $k'(x)/k(x)$ proving the claim.