Finite power map of the affine line

For $n\ge 1$, every exponent $m\ge 0$ has a unique form $m=qn+i$ with $q\ge 0$ and $0\le i<n$. Thus each polynomial in $k[t]$ has a unique expression ${\sum }_{i=0}^{n-1}{f}_i(t^n)t^i$ with $f_i\in k[s]$. Existence follows by grouping its monomials by their remainder modulo $n$; uniqueness follows because the exponents $qn+i$ are distinct for distinct pairs $(q,i)$. Taking only the $i=0$ term also shows that the ring map $k[s]\to k[t]$, $s\mapsto t^n$, is injective. Therefore $1,t,\dots ,t^{n-1}$ is a free $k[s]$-basis of rank $n$. [F1, algebra] 1.2 Suppose $\mathrm{char}(k)=p>0$ and $p\mid n$. By [F13], $p\cdot 1_k=0$, and by [F14], $p$ is prime; set $r=n\mathrm{/}p$, so $0<r<n$. In the generic extension $k(s)\subseteq k(t)$, let $\alpha =t^r$. It satisfies ${\alpha }^p=s$, hence is algebraic by [F15]. It is not in $k(s)$: if $t^r=P(t^n)\mathrm{/}Q(t^n)$ for $P,Q\in k[s]$ and $Q\mathrm{\ne }0$, then $t^{r}Q(t^n)=P(t^n)$. Every exponent on the left is congruent to $r$ modulo $n$, while every exponent on the right is divisible by $n$; since $0<r<n$, equality is impossible. Let $m_{\alpha }$ be its minimal polynomial. By [F16], $m_{\alpha }\mid X^p-s$. In an algebraic closure, $X^p-s=(X-\alpha {)}^p$ because ${\alpha }^p=s$. Since $\alpha \mathrm{\notin }k(s)$, $m_{\alpha }$ has degree at least $2$; all its roots are $\alpha$, so it has a repeated root. By [F17] and [F18], $\alpha$ is inseparable over $k(s)$, so the generic extension is not separable. [F12, F13, F14, F15, F16, F17, F18, algebra] 1.3 For $n=0$, the coordinate map is $k[s]\to k[t]$, $s\mapsto 1$. The $k[s]$-module action on $k[t]$ factors through $k[s]\mathrm{/}(s-1)\cong k$, so if it were finitely generated as a $k[s]$-module then $k[t]$ would be finite-dimensional over $k$. This is impossible because $1,t,t^2,\dots$ are linearly independent over $k$. The finite-morphism condition already fails on the whole target affine open, so the constant map is not finite. [F3, F4, algebra] 2.1 Let $U\subseteq \mathrm{Spec}k[s]$ be any affine open. Its closed complement is $V(I)$ for an ideal $I$ by [F6], and [F5] writes $I=(g)$, so $U=D(g)$ by [F7]. This includes $U=\mathrm{\varnothing }$ with $g=0$ and the whole target with $g=1$. For $g\mathrm{\ne }0$, [F2] and [F8] identify its inverse image with $D(g(t^n))=\mathrm{Spec}k[t{]}_{g(t^n)}$ and its coordinate map with $k[s{]}_g\to k[t{]}_{g(t^n)}$. The image $g(t^n)$ is nonzero by step 1.1. Every localized element is $h(t)\mathrm{/}g(t^n{)}^N$ by [F9]; writing $h$ in the basis from step 1.1 shows that the localized basis spans over $k[s{]}_g$. To prove independence, clear the coefficient denominators in a relation. By [F11], some power of $g(t^n)$ then kills the resulting numerator. The ring $k[t]$ is a domain, since leading coefficients of nonzero polynomials over $k$ multiply to a nonzero coefficient, so this power can be cancelled. The original basis independence then makes every coefficient zero. Thus the localized algebra is module-finite over $k[s{]}_g$. If $g=0$, [F10] gives the zero coordinate ring on the empty inverse image and empty target open, and the zero module is finite. By [F3] and [F4], these checks on every affine target open prove finiteness. [F2, F3, F4, F5, F6, F7, F8, F9, F10, F11, step 1.1, algebra] 3.1 At $n=1$, the basis is $\{1\}$ and the map is the identity; the generic extension is $k(t)\mathrm{/}k(t)$ and is separable. The proof uses no choice: the basis is explicit, an arbitrary affine target open is handled one at a time, and the characteristic argument uses one explicit element. The empty target open is handled in step 2.1, and there is no empty-source case or interval endpoint. The source is nonempty because $k[t]$ is a domain and $(0)$ is prime. [F12, step 1.1, step 2.1, step 1.2, step 1.3] $\mathrm{\square }$