Cfrgtkky

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:

1. 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 $k[t]oplus k[t]xopluscdotsoplus k[t]x^{n-1}$ as $k[t]$-modules. That is all you need.

Remember the definition in Hartshorne: $f:Xto Y$ is a finite morphism if there is an affine covering of $Y$, write $Y=bigcup U_i$, with $U_isimeq mbox{Spec} B_i$ such that $f^{-1}(U_i)=mbox{Spec}(A_i)$ with $A_i$ a f.g. $B_i$-module.