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Let $p$ be a prime. Try to show that $[\mathbb F_p (t): \mathbb F_p(t^p)] = p$.

I tried to search for relevant hints and come across with these posts.

Example for non separable elements?

Non-separable, infinite field extensions of non-zero characteristic

I am not sure why the minimal polynomial is $X^p - t^p = (X - t)^p$ and how minimal polynomial helps here.

I know that $\mathbb Z_p$ has characteristic $p$, which gives us for all $a, b \in \mathbb{F}_p$: $$(a + b)^p = a^p + b^p$$

But isn't $t$ here a formal variable?

Also, I know a similar theorem which states that if $p(x)$ is the minimal polynpmial of $\alpha$ over $F$, we have $[F(\alpha) : F] = \deg(p)$.

But the $\mathbb F_p (t)$ and $\mathbb F_p(t^p)$ does not quite fit the above theorem.

Benjamin
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    The minimal polynomial divides $X^p-t^p=(X-t)^p$. Thus it is of the form $(X-t)^n$. Clearly $n<p$ won't give a polynomial $\in \Bbb{F}_p(t^p)[X]$ – reuns Mar 09 '21 at 19:51
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    Yes $t$ is the formal variable of the polynomial ring $\Bbb{F}_p[t]$. But $(a+b)^p=a^p+b^p$ relies on that $p=0$ (and that $ab=ba$) not on that $a,b\in \Bbb{F}_p$, it stays true in your rings. – reuns Mar 09 '21 at 19:54
  • In fact, we can prove the following more general statement:

    Theorem:

    Let $F$ be a field and let $F(t)$ be the function field in one indeterminate $t$. Let $f\in F(t)$ be a non-constant rational function. Let $f(t)=\frac{p(t)}{q(t)}$ for polynomials $p,q$ in $F[t]$ such that $\gcd(p,q)=1$. Then we have the following $$[F(t):F(f)]=\max{\deg(p),\deg(q)}$$

     – ShBh Apr 03 '21 at 14:37

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Let $K=\mathbb{F}_p(t^p)$ and let $f(X)=X^p-t^p\in K[X]$. Note that $t$ is irreducible in $\mathbb{F}_p[t^p]$. Since $\mathbb{F}_p[t^p]$ is a PID, it follows that $(t^p)$ is a prime ideal, $1\not\in (t^p), t^p\not\in t^{2p}$ and so by Eisentstein's Criteria $X^p-t^p$ is irreducible in $\mathbb{F}_p[t^p][X]$. Then by Gauss's lemma it is irreducible in $\mathbb{F}_p(t^p)[X]=K[X]$. Hence $L = K[X]/(X^p-t^p)$ is an extension of degree $p$. As you pointed out, $X^p-t^p=(X-t)^p$ so $t$ is the only root of $f$. Therefore $t\in L$ and so $L=K(t)=\mathbb{F}_p(t)$.

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