Field extensions - if $(E : F) = n$ then $(E(x) : F(x)) = n$

Question

Well, pretty much everything is in the title - I'm looking for the proof of the following statement: if we have a field extension $F \subset E$ then the degree of the extension $F(x) \subset E(x)$ [fields of rational functions] is the same as that of $E$ over $F$. Will be grateful for any hints!

Answer 1

This is a continuation of the other answer I wrote; we'll show that $v_1, ... , v_n$ is a spanning set. I am not sure whether giving an explicit formula for how to write an arbitrary member of $E(X)$ as an $F(X)$-linear combination of the $v_i$ is possible, unless more is known about the basis elements $v_i$.

Here is a "proof by shenanigans." We know that $E = F[v_1, ... , v_n]$. So $$E(X) = F[v_1, ... , v_n](X)$$ We want to first of all show this is equal to $$F(X)[v_1, ... , v_n]$$ Now, the first set I've written is by definition a field, and a subset of every field which contains $F[v_1, ... , v_n]$ and $X$ (it is the "smallest" such field). The second set is by definition a ring, and contained in every ring which contains $F(X)$ and $v_1, ... , v_n$ (it is the "smallest" such ring).

It's clear that the second set is contained in the first. That the first set is contained in the second follows from the fact that $F(X)[v_1, ... , v_n]$ is a field (a field adjoined with any number of algebraic elements remains a field).

So $E(X)$ is equal to $F(X)$ adjoined with the elements $v_1, ... , v_n$. Thus an arbitrary member of $E(X)$ is equal to a finite sum $\sum\limits_i \frac{f_i(X)}{g_i(X)} v_1^{n_{i_1}} \cdots v_n^{n_{i_n}}$, for $n_{i_j} \geq 0$ and $\frac{f_i(X)}{g_i(X)} \in F(X)$. But we are now done, because each product $v_1^{i_1} \cdots v_n^{i_n}$ is equal to an $F$-linear combination of $v_1, ... , v_n$.

Answer 2

You could prove this by proving that the natural $F(x)$-algebra map $E\otimes_FF(x)\to E(x)$ is an isomorphism. This is not a trivial thing in the sense that, if $E/F$ were transcendental, this map would not be an isomorphism (indeed if $E=F(x)$ itself, for example, then the tensor product $F(x)\otimes_FF(x)$ is not a field). Anyway, first note that $E\otimes_FF(x)=(E\otimes_FF[x])\otimes_{F[x]}F(x)=E[x]\otimes_{F[x]}F(x)$. This is a localization of the domain $E[x]$, hence a domain, and it is also a finite-dimensional $F(x)$-algebra. Such a ring (a domain of finite dimension over a field) must be a field. Ring maps out of fields are injective.

So $E\otimes_FF(x)=E[x]\otimes_{F[x]}F(x)\to E(x)$ is an injective map whose image is a subfield of $E(x)$ containing $E[x]$ (since $E[x]\otimes_{F[x]}F(x)$ contains $E[x]$ via $f(x)\mapsto f(x)\otimes 1$, and this is mapped to $f(x)\in E(x)$). It follows that the image contain all of, and hence equals, $E(x)$.

So we have an $F(x)$-algebra (and in particular an $F(x)$-linear) isomorphism $E\otimes_FF(x)\simeq E(x)$. The tensor product visibly has dimension over $F(x)$ equal to $[E:F]$.

This argument works more generally if $E/F$ is algebraic. In deducing that $E[x]\otimes_{F[x]}F(x)$ was a field because it was a domain of finite dimension over a field, it would have been enough to know that it was integral over a field, and this will be true if $E/F$ is algebraic.

As I alluded to above, if $E/F$ is not algebraic, this map can't be injective. Namely, we can reduce to $E=F(x)$, and in this case, $x\otimes 1-1\otimes x$ is a non-zero element of the kernel of $E\otimes_FF(x)\to E(x)$.

Answer 3

Let $v_1, ... , v_n \in E$ be a basis for $E/F$. First let's show these elements are linearly independent over $F(X)$. Suppose $$\frac{f_1(X)}{g_1(X)}v_1 + \cdots + \frac{f_n(X)}{g_n(X)}v_n = 0$$ where $f_i(X)$ and $0 \neq g_i(X)$ are in $ F[X]$. Let $G(X) = g_1(X) \cdots g_n(X)$, and $h_i(X) = \frac{G(X)}{g_i(X)} = g_1(X) \cdots g_{i-1}(X)g_{i+1}(X)$. Multiply both sides of the above equation by $G(X)$ to get $$h_1(X)v_1 + \cdots + h_n(X)v_n = 0$$ By pairing up the coefficients belonging to the terms $1, X, X^2$ etc. its easy to see that the coefficients of each polynomial $h_i(X)$ are all $0$.

Therefore $0 = f_i(X)g_1(X) \cdots g_{i-1}(X)g_{i+1}(X) \cdots g_n(X)$, which implies $f_i(X) = 0$ because none of the polynomials $g_i(X)$ are $0$.

I'll leave it to you to show that $v_1, ... , v_n$ span $E(X)/F(X)$.