My question refers to an example from https://en.wikipedia.org/wiki/Finite_morphism dealing with the morphism
$$f: {\displaystyle {\text{Spec}}(k[t,x]/(x^{n}-t))\to {\text{Spec}}(k[t])}$$
of affine schemes induced by canonic inclusion
$$\varphi: k[t] \hookrightarrow {\text{Spec}}(k[t,x]/(x^{n}-t))$$
of rings. $f$ is described as a ramified $n$-sheeted covering with only ramification at the the origin. In turn for every prime ideal $p \subset k[t]$ (= point of $Spec(k[t])$) with $p \neq (0)$ the fiber $f^{-1}(p)$ contains exactly $n$ points.
That is not clear to me, especially in the case that $k$ is not algebraically closed.
In my considerations I discuss two cases:
- case: $k$ is algebraically closed. Therefore every prime $p \neq (0)$ has the shape $p= (t- \lambda)$ for $\lambda \in k$. Then $$x^n - t = \prod _{j=1} ^n (x - \zeta^j \sqrt[n]{t})$$
and therefore
$$k[t,x]/(x^n-t) \cong \oplus _{j=1} ^n k[t,x]/(x-\zeta^j \sqrt[n]{t})$$
by CRT and we have obvoiusly exactly $n$ preimages
$$f^{-1}(p) = \{q \subset \oplus _{j=1} ^n k[t,x]/(x-\zeta^j \sqrt[n]{t}) \text{ } \vert \text{ } q \cap k[t] = p \} = \{\langle (t- \lambda), (x-\zeta^j \sqrt[n]{t})\rangle \vert 1 \le j \le n\}$$
as desired. Is this argumentation correct?
Remark: By definition for $q$ prime $f$ is defined via $f(q) = \varphi^{-1}(q) = k[t] \cap q$.
Now the case 2: $k$ is an arbitrary field.
Firstly, we can split $x^n-t$ in $s$ to irreducible factors $f_j \in k[t,x]$ each of degree $d_i$ with $\sum _{j=1} ^s d_i =n$. Using CRT as in case 1 we get the splitting
$$k[t,x]/(x^n-t) \cong \oplus _{j=1} ^s k[t,x]/f_j$$
Now consider an arbitrary prime ideal $p$ of $k[t]$. Since $k$ isn't algebraically closed we can't generally expect that $p$ has the shape $(t- \lambda)$. Motivated by case 1 I think that ideals $\rangle p \cup f_i \langle $ are good candidates for preimages of $p$ since
$$q \in f^{-1}(p) \Leftrightarrow q \cap k[t] =p$$
But here I see two problems:
Firstly: This gives only $s$ preimages but according to the example $p$ should have exactly $n$ preimages.
Secoundly: Are $\langle p \cup f_i \rangle $ prime? Why?
Remark: In this question I reduced a previous question of mine (Ramified Cover of Affine Scheme) to the language of commutative algebra and trying to deduce a formally correct argument for the statement above.
Well, in that example, it is obvious that $k[t,x]/(t-x^n)$ is isomorphic to $k[t]\oplus k[t]x\oplus\cdots\oplus k[t]x^{n-1}$ as $k[t]$-modules. That is all you need.
Remember the definition in Hartshorne: $f:X\to Y$ is a finite morphism if there is an affine covering of $Y$, write $Y=\bigcup U_i$, with $U_i\simeq \mbox{Spec} B_i$ such that $f^{-1}(U_i)=\mbox{Spec}(A_i)$ with $A_i$ a f.g. $B_i$-module.