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I have some problems in geometrically visualization/ interpretation of the morphism $$f: {\displaystyle {\text{Spec}}(k[t,x]/(x^{n}-t))\to {\text{Spec}}(k[t])}$$ that is induced by canonical inclusion $\varphi: k[t] \subset {\text{Spec}}(k[t,x]/(x^{n}-t))$ and is introduced as an example for a finite morphism in https://en.wikipedia.org/wiki/Finite_morphism.

Indeed the algebraic reason is clear to me, but the geometric statement that the morphism is an $n$-sheeted cover which ramifies at the origin seems not obviously to me.

In following I will use following notation: If $R$ is a ring and $X= Spec(R)$ is the corresponding spectrum, then the prime ideals $p \subset R$ correspond to the points are coventionelly denoted by $x_p \in X$. In following I will abuse this notation by interpreting $p$ instantly as the prime ideal but also as the corresponding point $x_p \in X$.

Back to the question:

The statement is espectially that set-theoretically for every point $p \in Spec(k[t])$ holds: The fiber $f^{-1}(p)$ has exactly $n$ element iff $p \neq (0)$. Now, by definition, for a point $q \subset Spec(k[t,x]/(x^{n}-t))$ the map $f$ is defined via $f(q):= \varphi^{-1}(q) = q \cap k[t]$ (the last one holds since $\varphi$ is the canonical inclusion).

So I don’t know how to show that in case of a not neccessary algebraically closed field for $p \neq (0)$ we have $\vert f^{-1}(p) \vert = n$?

Naively $f^{-1}(p) = \{q \subset k[t,x]/(x^{n}-t) \vert q \cap k[t] = p \}$. I don’t see that for $p \neq (0)$ it consists of exactly $n$ elements (=prime ideals). Specially what shape have the ideals of $k[t,x]/(x^{n}-t)$ exept of the fact that they contain the prime ideal $(x^{n}-t) $?

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    $\begingroup$ If $k$ is algebraically closed, then $x^n-t_0$ for a fixed $t_0\in k$ splits in to linear factors, and each of these linear factors generate exactly the maximal ideals you're looking for. Try some examples to see it in action! $\endgroup$ – KReiser Dec 3 '18 at 1:35
  • $\begingroup$ @KReiser: The case that $k$ is alg closed is indeed trivial. The example in the reference seemingly works with an arbitrary field... $\endgroup$ – KarlPeter Dec 3 '18 at 1:41
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    $\begingroup$ You should add to the question the fact that you're specifically struggling with the case of a non-algebraically closed field, then. In the case of a non-algebraically closed field, one counts points by degree: if $x^n-t_0$ splits as a product of irreducible polynomials of (possibly different) degrees $d_i$, then each of those polynomials defines a point with residue field of degree $d_i$ over $k$, and it is immediate that $\sum d_i = n$. $\endgroup$ – KReiser Dec 3 '18 at 1:44
  • $\begingroup$ @KReiser: When you talking about $x^n -t_0$ for fixed $t_0 \in k$ you refer probably to the fiber over the fixed prime ideal of the shape $(t - t_0)$, right? But here, in case of non alg closed field $k$ I encounter following problems by considering this argument: In $k[t]$ for non alg cl $k$ there could occure a prime ideal not of the shape $(t - t_0)$ so does does the splitting of $(x^n -t_0)$ into irreducible components for fixed $t_0 \in k$ really make sense in order to calculate the order of an arbitrary fiber? $\endgroup$ – KarlPeter Dec 3 '18 at 12:08
  • $\begingroup$ Or did you splitted in your second post concretely $x^n -t$ as polynomial in two variables without really fixing a $t_0 \in k$? In this case $x^n -t$ factors indeed into a product of irreducible $f_i \in k[t,x]$ with $deg(f_i) =d_i$. Here I don't understand why the obvious fact that $\sum_{i=1} ^s d_i = n $ holds provides that every fiber of has $n$ elements. Although every $f_i$ gives a residue field over $k$ of degree $d_i$ nevertheless every $f_i$ defines only one point in the fiber $f^{-1}(p)$ for arbitrary prime $p \subset k[t]$. $\endgroup$ – KarlPeter Dec 3 '18 at 12:10
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Maybe an example can be helpful: consider the case $\varphi: \mathbb{R}[y] \subset \mathbb{R}[x,y]/(x^2 -y)$. Now, if you try to draw the induced morphism

$${\displaystyle {\text{Spec}}(\mathbb{R}[x,y]/(x^{2}-y))\to {\text{Spec}}(\mathbb{R}[y])},$$

you will visualize it as a parabola in the $(x,y)$ coordinates (that's why I changed the notation) projecting to the $y$-axis. Now take an element of the $y$-axis, say $y_0$, bigger than zero (which corresponds to the prime ideal $(y-y_0) \subset \mathbb{R}[y]$). Its preimage will be two points, which geometrically are obviously $(\pm \sqrt {y_0}, y_0)$. In more algebraic terms, the inclusion $\varphi$ maps $$y - y_0 \mapsto y - y_0 = x^2 - y_0 = (x + \sqrt {y_0})(x - \sqrt {y_0}),$$ and you see that the prime ideals $(x \pm \sqrt {y_0}) \subset \mathbb{R}[x,y]/(x^{2}-y)$ are different (that's why we assume $y_0 >0$). Note that both ideals are defined by polynomials of degree 1.

Now, if $y_0 = 0$, then the morphism ramifies, because instead of getting the product of two different prime ideals, you get the square of a prime ideal (this is the definition of ramification). Note that the polynomial defining the prime ideal $(x)$ has degree 1.

And what happens if $y_0 < 0$? Here the arithmetic enters in the picture. Geometrically (i.e. in your real drawing), there is nothing below the $x$-axis, so you could ask yourself if there is actually a preimage of the morphism. But here is where you can see the power of algebraic geometry, which is able to see further: if we look in the algebraic picture, we have that $$y - y_0 \mapsto y - y_0 = x^2 - y_0$$ and $x^2 - y_0$ is an irreducible polynomial in $\mathbb{R}[x,y]/(y-x^2)$ because the square root of a negative number is not real. Hence, only the point corresponding to the prime ideal $(x^2 - y_0) \subset \mathbb{R}[x,y]/(y-x^2)$ maps to the point corresponding to the prime ideal $(y - y_0) \subset \mathbb{R}[y]$.

When $y_0 >0$, we had two points mapping to $(y - y_0)$. When $y_0 = 0$, we had only one point mapping to $(y - 0)$, but counted twice. But if $y_0 < 0$, then we have a unique point mapping to $(y - y_0)$ counted once. What happened here?

Well, the key here is that now in the irreducible polynomial defining the prime ideal has degree two. This is what it is called inertia degree, and measures the degree of the extension of the residue fields (in this case $\mathbb{R}[x,y]/(y- x^2, y - y_0) \simeq \mathbb{C}$ over $\mathbb{R}[y]/(y - y_0) \simeq \mathbb{R}$ has degree 2). You can check that when $y_0 \geq 0$, the inertia degree is 1 (because the degree of the polynomials defining the prime ideals is one).

In general, if you consider the sum of the products of the inertia degree and the ramification degree each point (= prime ideal) in your preimage, you get a constant number, which is the degree of the morphism (in our case 2).

  • If $y_0 > 0$, you have two points with ramification and inertia index equal to 1.
  • If $y_0 = 0$, you have one point with ramification index equal to 2 and inertia index equal to 1.
  • If $y_0 < 0$, you have one point with ramification index equal to 1 and inertia index equal to 2.

It is an easy exercise to check what numbers do you get with $n = 3$ and $n = 4$. It is also a nice exercise to see what happens with more general fields (for example over $\mathbb{Q}$ or over $\mathbb{Q}[i]$), since over $\mathbb{R}$ we can only have inertia degree at most 2.

If you start over an algebraically closed field, then all the residue fields (of the closed points) will be this field, and therefore you will not have any inertia degree. In algebraic geometry, sometimes we say "geometrically" to refer to the picture over an algebraically closed field, in contrast to the "arithmetic" picture. If you like these things, take a look on Silverman's "The arithmetic of elliptic curves", where he first studies the "geometry" of the elliptic curves (i.e. elliptic curves over algebraically closed fields) and afterwards goes to the "arithmetic" (i.e. elliptic curves over non algebraically closed fields).

I hope that this was what your were looking for!

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The morphism $f$ is called a branched cover of degree $n$---provided that $\text{char}\,k$ does not divide $n$---but it is not the case in general that over every point other than $t=0$ there are exactly $n$ points. This is the case if $k$ is algebraically closed, but for example if $k$ is the field of real numbers and $n=2$ then over the point $t=-1$ there is exactly one point, corresponding to the ideal $(t+1)=(x^2+1)$. The residue field there is the field of complex numbers, so this is a point of degree $2$.

In general, for a branched cover of degree $n$---i.e. a generically separable finite morphism of degree $n$---over a point $P$ there will be $r$ points $Q_1,\ldots,Q_r$ for some $r$ that is at most $n$, of ramification indices $e_1,\ldots,e_r$---which are equal to $1$ where it is not ramified---and residue field degrees $f_1,\ldots,f_r$; and the sum of the $r$ numbers $e_if_i$ is equal to $n$. When thinking about this, one often pictures the algebraically closed case, and that leads to thinking of it along the lines described in the discussion.

Another way to deal with this is to speak of "geometric points", which again recaptures the desired property that there are exactly $n$ points over each unramified point. A geometric point of an affine variety $V = \text{Spec}\,R$ is by definition a ring homomorphism from $R$ to a given algebraically closed field. For example, if $k$ is the field of real numbers, then the map from $k[t]$ to the complex numbers sending $t$ to $-1$ is a geometric point, corresponding to the closed point where $t=-1$. Over that point, there are two geometric points on the branched cover---taking $n=2$ again---one where $x$ is sent to $i$, and one where $x$ is sent to $-i$. Both lie over the degree $2$ closed point $(x^2+1)$.

When thinking geometrically, one typically either thinks of the case of an algebraically closed field $k$, or else thinks in terms of geometric points. On the other hand, thinking in terms of the numbers $e_i$ and $f_i$ is also useful, and is common in number theory. For example, consider the Gaussian integers as a degree $2$ extension of the integers, corresponding to a degree $2$ ramified cover between their $\text{Spec}$s. Over the prime $(5)$ of the integers, there are two points---primes---on the cover, given by $(2+i)$ and $(2-i)$. There $r=2=n$, $e_i=1$, $f_i=1$. But over the prime $(3)$ of the integers, there is just one point on the cover, given by the prime $(3)$ in the Gaussian integers. There $n=2$, $r=1$, $e_1=1$, $f_1=2$. If you have studied number theory, this example will be familiar, and can shed light on the situation that you are trying to understand.

So I don’t know how to show that in case of a not neccessary algebraically closed field for $p \neq (0)$ we have $\vert f^{-1}(p) \vert = n$?

This statement is false. It is not difficult to construct a counterexample---e.g. by considering the generic fiber of $f$. In particular, it is meaningless to attempt to show that this statement is true.

Or did you splitted in your second post concretely $x^n - t$ as polynomial in two variables without really fixing a $t_0 \in k$? In this case $x^n - t$ factors indeed into a product of irreducible $f_i \in k[t, x]$ with $\deg(f_i) = d_i$.

Again, it is not true that every fiber has cardinality $n$.

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