Going Up Theorem - Examples? Witnesses? I am currently in Chapter 5 - Integral Dependence and Valuations of the text Introduction to Commutative Algebra by Atiyah - Macdonald. I am in particular studying the `Going-up theorem', and I have a request/question that the community here may be able to help with. First I should mention that I know there are many other questions on MSE regarding the Going-up theorem, and none of them address my question. I have also spent several hours scouring google, and still have nothing.
I currently feel confident that I understand both the statement and proof of the Going-up theorem, but I feel I am missing the importance. It has been noted in  this question that the algebro-geometric interpretation is a more compelling account of the result. What I believe would solidify my understanding is to see concrete examples and applications of the Going-up theorem. I am also familiar with the results in the exercises from Chapter 5 that give the geometric equivalent of the theorem, i.e. the conditions of the map $f^*:\mathrm{Spec}(B) \to \mathrm{Spec}(A)$ when the map $A \to B$ is an integral morphism, or satisfies the Going-up property, so geometric examples that transpire from the algebraic theorem would also be nice.
Working out of Atiyah-Macdonald is currently my only exposure to commutative algebra, I am familiar with nearly all of the content in the previous chapters and have done the majority of the exercises from the previous chapters. However, as most know the book is very terse and general, in my opinion, in the sense that there are not many explicit examples included, which is fine, but filling in the blanks with explicit examples has been one of my main difficulties while working out of the text. In case it is relevant, I also have exposure to elementary algebraic geometry - so feel free to provide an example relating to varieties, coordinate rings, etc.
To be explicit, my question is:

Can someone provide explicit examples of any of the following:
  
  
*
  
*Explicit (preferably interesting) rings $A$ and $B$ such that $B$ is integral over $A$.
  
*An explicit chain of prime ideals $\mathfrak{p_1} \subseteq \dots \subseteq \mathfrak{p_n}$ of $A$ and a chain $\mathfrak{q_1} \subseteq
> \dots \subseteq \mathfrak{q_m}$ ($m < n$) such that $\mathfrak{q_i}
> \cap A = \mathfrak{p}_1$ and explicitly find the extended chain 
  $\mathfrak{q}_1 \subseteq \dots \subseteq \mathfrak{q}_m$
  
*$\mathfrak{q}_1 \subseteq \dots \subseteq \mathfrak{q}_n$
  
*Show explicitly what geometric information can be learned from $f \colon A \to B$ being an integral homomorphism implying that
  $\mathrm{Spec}(B) \to \mathrm{Spec}(A)$ being a closed map.
  

One of my thoughts was that coordinate rings of varieties might lead to good examples?
Last, I must mention that I know this is not a typical MSE question, but after reviewing the criteria of acceptable questions for MSE, I think it satisfies since it is mathematical and it is answerable, that is, there are objectively correct answers containing the explicit examples I seek. That said, I am entirely open to suggestions if anyone objects or if anyone has suggestions as to how to make this question better suited for the community.
Thank you for any help.
 A: OK, so you're asking quite a few questions. Let me try my best to satisfy your curiosity. First an example:
Example: Let $A=k[x]$ for some algebraically closed field $k$ and $B=k[x,y]/(f(y))$ where $f(y)=y^n+b_1y^{n-1}+\cdots+b_n=\prod_{i=1}^n(y-\alpha_i)$. For simplicity assume all roots $\alpha_i$ are also distinct. Clearly $A\subset B$ and I leave it as an exercise that $B$ is integral over $A$ (you can prove this by what you learned in Atiyah and Macdonald easily).
Now let us understand what is happening here. Let $C=k[y]/(f(y))$. First of all $B=k[x]\otimes k[y]/(f(y))=A\otimes C$. Let $X=\mathrm{Spec}(A)$, $Z=\mathrm{Spec}(C)$ and $Y=\mathrm{Spec}(B)$. The space $Z$ is actually just $n$ (distinct) points. If you think about it $Y=X\times Z$, so $Y$ is in fact just $n$ copies of $X$.
What you witnessed here is a special case of Noether normalization lemma:

Let $k$ be a field, $A$ any finitely generated $k$-algebra. Then there exists an integer $d\geq 0$ and algebraically independent elements $y_1, \cdots, y_d\in A$, such that $k[y_1, \cdots, y_d]\subset A$ is an integral extension. [the set $y_1, \cdots, y_d$ is algebraically independent if there are no polynomial relations of the form $F(y_1, \cdots, y_d)=0$.]

Geometrically any affine variety is $X=\mathrm{Spec}(A)$ with $A$ a finitely generated $k$-algebra (also sometimes it is required for $A$ to be reduced, but don't worry about that now). Noether normalization, therefore, gives us an integral (actually finite) morphism $p: X\to \mathbb{A}^d$. Given any point $x\in \mathbb{A}^{d}$ the fiber $p^{-1}(x)$ is a finite set. The morphism $p$ should be thought as a projection. See the article about Ramification points to get a bit more intuition about this. Needless to say, Noether normalization lemma is used all the time in algebraic geometry since it a semi-parameterization for our variety.
I'd recommend reading a little bit about Noether normalization lemma. I hope this is not too advanced and helpful for example. As you'll see on that note, among other things, normalization lemma is used to prove "Hilbert’s Nullstellensatz" which is absolutely essential.

The second important connection one should make when they first learn about going up theorem is to the concept of dimension. You need to wait until the last chapter of Atiyah and Macdonald to learn about this. However, let me give a little taste here.
Definition: Let $R$ be a ring. Consider a chains of prime ideals of $R$, i.e. $\mathfrak{p}_0\subset \mathfrak{p}_1\subsetneq \mathfrak{p}_2\subsetneq \cdots \mathfrak{p}_n$. The length of such chain is defined to be $n$. The Krull dimension of $R$ is the supremum of the lengths of all chains.
To understand this definition, let's again use geometry. First consider $R=k[x_1, \cdots, x_d]$ with $k$ a field. One can show the chain 
$$(0)\subset (x_1) \subset (x_1, x_2)\subset \cdots \subset (x_1, \cdots, x_d)$$
is a chain of prime ideals of maximal possible length. So the Krull dimension of $k[x_1, \cdots, x_d]$ is in fact, unsuprisingly, $d$. Geometrically this chain of prime ideals can be though of a chain of closed subsets of $\mathbb{A}^d$
$$
\mathbb{A}^d\supset V(x_1)\simeq \mathbb{A}^{d-1}\supset V(x_1,x_2)\simeq \mathbb{A}^{d-2}\supset \cdots\supset V(x_1, \cdots, x_d)\simeq \mathbb{A}^{0}
$$
in other words, a point included in a line, included in a surface, ..., included in $\mathbb{A}^d$. Roughly speaking we are thinking of dimension of $X$ as the capacity of $X$ for including a nested (irreducible) closed subsets.
Hopefully, you have at least a gut feeling about what Krull dimension is all about now. Now let's get back Going Up property. Let $A\subset B$ be an integral extension. What going up implies is that Krull dimension of $A$ and $B$ are equal (can you prove this?).
Geometrically then an integral morphism $\phi:\mathrm{Spec}(B)\to \mathrm{Spec}(A)$ means 1) $\dim \mathrm{Spec}(A) = \dim \mathrm{Spec}(B)$ and 2) $\phi:\mathrm{Spec}(B)\to \mathrm{Spec}(A)$ is a ramified covering map.
Hope this helps.
