Exercise 3.1.13 (pth root)

Question

Let $p$ be a prime and consider the field $\mathbb{F}_p(t)$ of
rational expressions over $\mathbb{F}_p$. Show that $t$ has no $p$th root in $\mathbb{F}_p(t)$.
(Hint: consider degrees of polynomials.)

Hassaan Naeem · Accepted Answer

extbf{Solution:}
A rational expression over $K$ is $\frac{f(t)}{g(t)}$ where $f(t), g(t) \in K[t]$ with $g 
eq 0$.
For any $\frac{f(t)}{g(t)} \in \mathbb{F}_p(t)$ where $f(t), g(t) \in \mathbb{F}_p[t]$, suppose we have
have that $\left( \frac{f(t)}{g(t)} ight)^p = t$. We then have that $f^p = tg^p$. Then $deg(f^p) = np$ where $n=deg(f)$ and
$deg(tg^p) = deg(t)+deg(g^p) = 1 + mp$ where $m=deg(g)$, hence we have $np = mp+1 \leadsto p = \frac{1}{n-m}$.
But this is impossible since $p$ is prime, hence a contradiction, hence $t$ has no $p$th root in $\mathbb{F}_p(t)$.

Exercise 3.1.13 (pth root)

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Exercise 3.1.13 (pth root)

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