# Computing $[\mathbb{F}_p(x):\mathbb{F}_p(x)^p]$

If $$F=\mathbb{F}_p(x)$$ is the rational function field in one variable over $$\mathbb{F}_p$$, find $$[F:F^p]$$.*

Here's where I am at with this question: Using the multinomial theorem (and some tears), I can prove that $$\left(\sum_{i=1}^na_ix^i\right)^p=\sum_{i=1}^na_ix^{ip}.$$

This at least shows that $$\{1,x,x^2,...x^{p-1}\}$$ is a basis for $$\mathbb{F}_p[x]$$ over $$\mathbb{F}_p[x]^p$$. But I am not sure how to get from here to the field of fractions.

My other thought was to try to find the minimal polynomial of $$x$$ over $$F^p$$, which I guess would be $$t^p-x\in \mathbb{F}(x)^p[t]$$, but I am not certain if $$\mathbb{F}_p(x)$$ is generated over $$\mathbb{F}_p(x)^p$$ by $$x$$.

Are any of these approaches sensible? How do I deal with the nasty quotients?

*Patrick Morandi: Field and Galois Theory

• Note that $a(x)/b(x) = a(x)b(x)^{p-1} / b(x)^p$, and try applying your basis argument to the numerator. – user125932 Jul 6 at 23:59
• For $a,b \in \Bbb{F}_p[x]$ then $(a(x)/b(x))^p = a(x^p)/b(x^p)$ thus $\Bbb{F}_p(x)^p = \Bbb{F}_p(x^p)$. In general $K(u_1,\ldots,u_n)^p = K^p(u_1^p,\ldots,u_n^p)$. – reuns Jul 7 at 1:20
• Note it is enough to show (in characteristic $p$ commutative ring) $(u+v)^p = u^p+v^p$ to obtain that $p$ is a ring homomorphism, that here it is injective and $\left(\sum_{i=1}^na_ix^{i-1}\right)^p=\sum_{i=1}^na_ix^{(i-1)p}$, $(a(x)/b(x))^p = a(x^p)/b(x^p)$ – reuns Jul 7 at 2:06

## 1 Answer

First we will show that $$\left(\sum_{i=1}^na_ix^{i-1}\right)^p=\sum_{i=1}^na_ix^{(i-1)p}.$$ We will use (without proof) the multinomial theorem: $$\left(\sum_{i=1}^na_ix^{(i-1)}\right)^p=\sum_{k_1,k_2,...k_n}\binom{p}{k_1,k_2,...k_n}\prod_{t=1}^na_t^{k_t}x^{(i-1)k_t}$$ where the sum runs over all combinations of $$n$$ non-negative integers $$k_i$$ which sum to $$p$$. Now, if $$k_i\neq p$$ $$\binom{p}{k_1,k_2,...k_n}=\frac{p!}{k_1!k_2!...k_n!}, ~~~~~~~~\text{so}~~~~~~~~p\Big|\binom{p}{k_1,k_2,...k_n}$$ because the denominator cannot have $$p$$ as a prime factor since all the $$k_i. So, in a finite field $$\mathbb{F}_p$$, we are left with only the terms where one $$k_i$$ if $$p$$ and the others are zero, and there are $$n$$ such terms which look like: $$\sum_{0,0,...k_i,0,0...}\binom{p}{0,0,...k_i,0,0...}\prod_{t=1}^na_t^{k_t}x^{(i-1)k_t}=a_i^{k_i}x^{(i-1)k_i}=a_i^px^{(i-1)p}$$ $$\text{Therefore, }~~~~~~\left(\sum_{i=1}^na_ix^{i-1}\right)^p=\sum_{i=1}^na_ix^{(i-1)p}.$$ By this formula, $$\mathbb{F}_p[x]$$ is generated over $$\mathbb{F}_p[x]^p$$ by the basis $$\{1,x,x^2...x^{p-1}\}$$. Now, by user125932's observation, if $$\frac{f(x)}{g(x)}\in F$$, then $$\frac{f(x)}{g(x)}=\frac{f(x)g(x)^{p-1}}{g(x)^p}$$. By the above argument $$f(x)g(x)^{p-1}\in \mathbb{F}_p(x)$$, so it is generated over $$F^p$$ by a basis of p elements, and clearly $$g(x)^p \in F^p$$.

Therefore, $$[F:F^p]=p$$.