This question already has an answer here:

This is not a problem I've found stated anywhere, so I'm not sure how much generality I should assume. I will try to ask my question in such a way that answers on different levels of generality could be possible. I'm also not sure that this question isn't trivial.

Let $E\subset F$ be a fields (both can be finite if needed), and let $n$ be the degree of the field extension ($n$ can be finite if needed). Can we find the degree of the extension $E(x)\subset F(x)$ of rational function fields?

Say $E=\mathbb F_2$ and $F=\mathbb F_4$. Then $(F:E)=2.$ I can take $\{1,\xi\}$ to be an $E$-basis of $F$. Now let $f\in F(x),$ $$f(x)=\frac{a_kx^k+\cdots +a_0 } {b_lx^l+\cdots+b_0 }$$ for $a_0,\ldots a_k,b_0,\ldots,b_l\in F$

I can write $a_0,\ldots a_k,b_0,\ldots,b_l$ in the basis: $$\begin{eqnarray}&a_i&=p_i\xi+q_i\\&b_j&=r_j\xi+s_j\end{eqnarray}$$

But all I get is $$f(x)=\frac{p_k\xi x^k+\cdots+p_0+q_kx^k+\cdots+q_0} { r_k\xi x^k+\cdots+r_0+s_kx^k+\cdots+s_0},$$

and I have no idea what to do with this. On the other hand, my intuition is that the degree of the extension of rational function fields should only depend on the degree of the extension $E\subset F,$ even regardless of any finiteness conditions.


marked as duplicate by Watson, John B, Daniel W. Farlow, ncmathsadist, C. Falcon Feb 4 '17 at 2:13

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ It's the same as the degree of the original extension. Here is a hint to get you started. Suppose I claimed that $\{ 1, \sqrt{2} \}$ was a basis for $\mathbb{Q}(\sqrt{2}, x)$ over $\mathbb{Q}(x)$. How would I express, say, $\frac{1}{x - \sqrt{2}}$ in terms of this basis? Well, it's just $\frac{x}{x^2 - 2} + \sqrt{2} \frac{1}{x^2 - 2}$... now think about partial fraction decomposition. $\endgroup$ – Qiaochu Yuan May 5 '12 at 20:13
  • $\begingroup$ @QiaochuYuan Thank you. I have one question about this though. Is it correct to say that since the tuple $(1,\sqrt 2)$ is linearly independent over $\mathbb Q,$ it's also linearly independent over $\mathbb Q(x)$ because its gramian is the same regardless of the underlying field? $\endgroup$ – user23211 May 5 '12 at 21:41
  • $\begingroup$ I don't understand what you mean by "gramian" here. A short way to see the result you want is to see that a linear dependence over $\mathbb{Q}(x)$ gives a linear dependence over $\mathbb{Q}$ after plugging in a suitable value for $x$. This argument breaks down if $\mathbb{Q}$ is replaced by a finite field $F$ but I think one can salvage it by embedding $F(x)$ into the field of Laurent series $F((x))$. $\endgroup$ – Qiaochu Yuan May 5 '12 at 21:45
  • $\begingroup$ I see why the argument breaks down for finite fields, but why does using $F((x))$ help? $\endgroup$ – user23211 May 5 '12 at 21:57
  • $\begingroup$ Because, given a linear dependence over $F((x))$, we can restrict our attention to the term of minimal degree (in $E((x))$) to get a linear dependence over $F$. $\endgroup$ – Qiaochu Yuan May 5 '12 at 22:02

Suppose $F = E(\alpha)$, where $\alpha$ has minimal polynomial $f(t)$. Then $$F(x) = E(\alpha)(x) = E(\alpha, x) = E(x)(\alpha)$$ and $\alpha$ still has minimal polynomial $f(t)$. If you're not sure on that point, the roots of $f(t)$ in $\overline{F}$ are in $\overline{F}$. The roots of $f(t)$ in $\overline{F(x)}$ are still in $\overline{F}$. (you may want to prove a theorem that says this reasoning makes sense, and implies what we want it to imply!)

Every finite example is either this one, or can be formed by iterating this one finitely many times.

  • $\begingroup$ By the theorem, do you mean one that says that $\overline F\subset\overline {F(x)}$? $\endgroup$ – user23211 May 5 '12 at 21:21
  • $\begingroup$ There are a number of tiny subtle things in the proof that may or may not need explanation depending on the audience, such as "does $E(\alpha,x)$ really make sense?" along with the issue you mention. Embedding everything into an algebraic closure of $F(x)$ clears up most to all of them. $\endgroup$ – Hurkyl May 5 '12 at 21:26
  • $\begingroup$ Thanks! I think I've got it. Is it still true when the extension is infinite? I suspect it is, but this reasoning fails then. $\endgroup$ – user23211 May 5 '12 at 21:42
  • $\begingroup$ I think you can adapt this proof to show that a basis for $E$ is also a basis for $E(x)$. Add in a clause for the case where $\alpha$ is transcendental, apply transfinite induction, and done! $\endgroup$ – Hurkyl May 5 '12 at 22:03
  • $\begingroup$ I can't do transfinite induction so I think I'll pass for now, but it's good to know that it's true! (I don't need this case anyway now.) Thank you very much for your help! $\endgroup$ – user23211 May 5 '12 at 22:10

My proof is concerned with commutative algebra. Probably there is also field-theoretic one. I prove that: if $E \subseteq F$ is a finite field extension, then $[F(x) : E(x)] = [F \colon E]$.

This follows from: if $E \subseteq F$ is an algebraic field extension, then $F(x) \simeq E(x) \otimes_E F$ as $E(x)$-vector space.

The ring $F[x]$ is integral over $E[x]$. Consider the multiplicative subset $S = E[x] \setminus \{ 0 \}$ and the ring $A = S^{-1} F[x]$. Obviously $A$ is integral over the field $E(x)$, hence $A$ is a field (Atiyah-Macdonald 5.7); but $A$ is contained in $F(x)$ and contains $F[x]$, hence $A = F(x)$. Therefore $$ F(x) = S^{-1} F[x] = E(x) \otimes_{E[x]} F[x] = E(x) \otimes_E F. $$

  • $\begingroup$ Thank you very much, Andrea. I haven't studied tensor products yet, so I don't understand this reasoning. I will try to remember to come back to it. $\endgroup$ – user23211 May 5 '12 at 21:18