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

  • 2
    $\begingroup$ Note that $a(x)/b(x) = a(x)b(x)^{p-1} / b(x)^p$, and try applying your basis argument to the numerator. $\endgroup$ – user125932 Jul 6 at 23:59
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    $\begingroup$ 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 $K(u_1,\ldots,u_n)^p = K^p(u_1^p,\ldots,u_n^p)$. $\endgroup$ – reuns Jul 7 at 1:20
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    $\begingroup$ 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 $\left(\sum_{i=1}^na_ix^{i-1}\right)^p=\sum_{i=1}^na_ix^{(i-1)p}$, $(a(x)/b(x))^p = a(x^p)/b(x^p)$ $\endgroup$ – reuns Jul 7 at 2:06

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 $k_i<p$. 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, $[F:F^p]=p$.


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