Computing the "lying over", "going up", "going down" ideals. For any commutative unital ring $R$ and an ideal $\mathfrak{a}$ of $R$, we shall denote 
$$\begin{align*}
\mathrm{Spec}(R)&:=\{\text{prime ideals of }R\},\\ 
\mathrm{Max}(R)&:=\{\text{maximal ideals of }R\},\text{ and}\\ 
\mathrm{minAss}(\mathfrak{a})&:=\{\mathfrak{p}\in \mathrm{Spec}(R);\, \mathfrak{a}\subseteq\mathfrak{p}, \nexists\mathfrak{p}' \in\mathrm{Spec}(R): \mathfrak{a}\subsetneq\mathfrak{p}'\subsetneq\!\mathfrak{p}\}\\
&\;=\{\text{minimal prime ideals over }\mathfrak{a}\}.
\end{align*}$$
The Krull dimension of $R$ is 
$\dim(R)\!:=\!\mathrm{max} \{n\!\in\!\mathbb{N}_0;\, \exists\mathfrak{p}_0,\ldots,\mathfrak{p}_n\!\in\!\mathrm{Spec}(R)\!: \mathfrak{p}_0\!\subsetneq\!\ldots\!\subsetneq\!\mathfrak{p}_n\}=$ length of the longest chain of prime ideals.
I'm trying to understand the following excerpt from A SINGULAR Introduction to Commutative Algebra (Greuel & Pfister - 2008), p.242-243:

Questions:
(1) What is a ring of finite type over $K$? The term is never defined in the book, even though ring and other elementary notions are. I'm guessing it's a ring of the form $K[x_1,\ldots,x_n]/I$ for some $I\unlhd K[\mathbf{x}]$. But are there any restrictions on $I$?
(2) If I am not mistaken, in general (i.e. for any commutative unital ring $R$ and $\mathfrak{a}\unlhd R$), we have $\mathrm{Spec}(R/\mathfrak{a})=\{\mathfrak{p}/\mathfrak{a};\;\mathfrak{p}\in\mathrm{Spec(R),\; \mathfrak{a}\subseteq\mathfrak{p}}\}$ and 
$\mathrm{Max}(R/\mathfrak{a})=\{\mathfrak{m}/\mathfrak{a};\;\mathfrak{m}\in\mathrm{Max(R),\; \mathfrak{a}\subseteq\mathfrak{m}}\}$. Correct?
(3) I'm trying to formulate Remark 3.5.14 more precisely. Is the following correct:
Computing the "Lying Over", "Going Up", "Going Down" Ideals: Suppose we have $\{x_1,\ldots,x_m\}\subseteq\{y_1,\ldots,y_n\}$, $\;I\unlhd K[x_1,\ldots,x_m] =K[\mathbf{x}]$, $\;A=K[\mathbf{x}]/I$, $\;J\unlhd K[y_1,\ldots,y_n]=K[\mathbf{y}]$, $\;B=K[\mathbf{y}]/J$, and $A\leq B$ via the identification $f(\mathbf{x})\!+\!I\mapsto f(\mathbf{x})\!+\!J$ (this map is injective iff $J\cap K[\mathbf{x}]\subseteq I$, which we assume). For any $\mathfrak{a}\!\unlhd\!A$, let $\mathfrak{a}B$ denote the ideal of $B$, generated by $\mathfrak{a}$. We investigate the situation where we have $\mathfrak{p}_0\subseteq\mathfrak{p}_1\subseteq\mathfrak{p}_2$, $\;\mathfrak{p}_i\!\in\!\mathrm{Spec}(A)$, $\;\mathfrak{q}_0\subseteq\mathfrak{q}_1\subseteq\mathfrak{q}_2$, $\;\mathfrak{q}_i\!\in\! \mathrm{Spec}(B)$, and $\mathfrak{q}_i\!\cap\!A=\mathfrak{p}_i$, for $i\!=\!0,1,2$.
$$\begin{array}{c @{\hspace{1mm}} c @{\hspace{1mm}} c @{\hspace{1mm}} c @{\hspace{1mm}} c @{\hspace{1mm}} c @{\hspace{1mm}} c}
\frak{q}_0 & \subseteq & \frak{q}_1 & \subseteq & \frak{q}_2 \\
\downarrow &           & \downarrow &           & \downarrow \\
\frak{p}_0 & \subseteq & \frak{p}_1 & \subseteq & \frak{p}_2\\
\end{array}$$
Lemma 3.5.13 says that if $A\leq B$  is a finite extension of affine rings, $\mathfrak{p}\in\mathrm{Spec}(A)$, $\;\mathfrak{q}\in\mathrm{Spec}(B)$, $\;\mathfrak{p}B\subseteq\mathfrak{q}$, $\;\dim(B/\mathfrak{p}B)=\dim(B/\mathfrak{q})$, then we have $\mathfrak{q}\cap A=\mathfrak{p}$. 
(3.1) Does the converse of 3.5.13 hold: if $\mathfrak{q}\!\cap\!A\!=\!\mathfrak{p}$, then $\dim(B/\mathfrak{p}B)\!=\!\dim(B/\mathfrak{q})$? I think so. It suffices to show that $\mathfrak{p}B\!\subseteq\!\mathfrak{q}'\!\subseteq\!\mathfrak{q}$ and $\mathfrak{q}'\!\in\!\mathrm{Spec}(B)$ imply $\mathfrak{q}'\!=\!\mathfrak{q}$. Indeed, we have:
$$\begin{array}{r @{\hspace{1mm}} c @{\hspace{1mm}} c @{\hspace{1mm}} c @{\hspace{1mm}} c @{\hspace{1mm}} c @{\hspace{1mm}} c}
\mathfrak{p}B & \subseteq & \mathfrak{q}' & \subseteq & \mathfrak{q} \\
\downarrow &           & \downarrow &           & \downarrow \\
\mathfrak{p}\!\subseteq\!\mathfrak{p}B\!\cap\!A & \subseteq & \mathfrak{q}'\!\cap\!A & \subseteq & \mathfrak{p}\\
\end{array},$$
so $\mathfrak{q}'\!\cap\!A\!=\!\mathfrak{p}$, which by the "incomparable" theorem implies $\mathfrak{q}'\!=\!\mathfrak{q}$. Correct? This means that the only candidates for the "lying over $\mathfrak{p}$" ideals are among $\mathrm{minAss}(\mathfrak{p}B)$.
(3.2) If I see correctly, with the hypotheses $\mathfrak{p}\in\mathrm{Spec}(A)$, $\;\mathfrak{q}\in\mathrm{Spec}(B)$, $\;\mathfrak{p}B\subseteq\mathfrak{q}$, the condition $\dim(B/\mathfrak{p}B)=\dim(B/\mathfrak{q})$ is equivalent to $\mathfrak{q}\in\mathrm{minAss}(\mathfrak{p}B)$, by (2) and the definition of $\dim$ and $\mathrm{minAss}$. Yes?  Why do we then have to check in 3.5.14 that the dimension is right?
 A: I finally figured it out. The notation below is consistent with the book A Singular Introduction to Commutative Algebra. All the book references are to this book.
An extension of commutative unital rings $A\!\leq\!B$ is finite / of finite type / integral, when $B$ is a finitely generated $R$-module / when $B$ is a finitely generated $A$-algebra / when $\forall b\!\in\!B$ $\exists \text{monic} f\!\in\!A[x]$ with $f(b)=0$. By Proposition 3.1.2., p.212, we have: a ring extension $A\!\leq\!B$ is finite iff it is integral and of finite type. If $A,B$ are both affine $K$-algebras (i.e. $A\!=\!K[x_1,\ldots,x_m]/I$ and $B\!=\!K[y_1,\ldots,y_n]/J$, or equivalently, $K\!\leq\!A$ and $K\!\leq\!B$ is of finite type) and if $A\!\leq\!B$, then this extension is automatically of finite type. Hence an extension of affine $K$-agebras is finite iff it is integral.

Theorem(Cohen, Seidenberg - 1946): If $A\leq B$ is an integral extension of commutative unital rings, then the following holds (propositions 3.1.10.a), 3.3.3, 3.1.10.b), 3.2.9, and induction):
a) (lying over) $\forall \mathfrak{p}\!\in\!\mathrm{Spec}(A)\; \exists\mathfrak{q}\!\in\!\mathrm{Spec}(B)\!: \mathfrak{p}=\mathfrak{q}\cap A$.
b) (incomparable) $\forall \mathfrak{q}_0,\ldots,\mathfrak{q}_n\!\in\!\mathrm{Spec}(B):\; \mathfrak{q}_0\subsetneq\ldots\subsetneq\mathfrak{q}_n \;\Rightarrow\; \mathfrak{q}_0\!\cap\!A\subsetneq\ldots\subsetneq\mathfrak{q}_n\!\cap\!A$.
c) (going up) Let $\mathfrak{p}_0\!\subseteq\!\ldots\!\subseteq\!\mathfrak{p}_n$ be prime ideals of $A$ and $q_0\!\subseteq\!\ldots\!\subseteq\!\mathfrak{q}_k$ $(k\!<\!n)$ prime ideals of $B$ with $\mathfrak{p}_i\!=\!\mathfrak{q}_i\!\cap\!A$. Then $\exists$ prime ideals $\mathfrak{q}_{k+1}\!\subseteq\!\ldots\!\subseteq\!\mathfrak{q}_n$ of $B$ with $\mathfrak{p}_i\!=\!\mathfrak{q}_i\!\cap\!A$.
d) (going down) Let $\mathfrak{p}_0\!\subseteq\!\ldots\!\subseteq\!\mathfrak{p}_n$ be prime ideals of $A$ and $q_{k+1}\!\subseteq\!\ldots\!\subseteq\!\mathfrak{q}_n$ $(k\!<\!n)$ prime ideals of $B$ with $\mathfrak{p}_i\!=\!\mathfrak{q}_i\!\cap\!A$. If $A,B$ are domains and $A$ is integrally closed, then $\exists$ prime ideals $\mathfrak{q}_0\!\subseteq\!\ldots\!\subseteq\!\mathfrak{q}_k$ of $B$ with $\mathfrak{p}_i\!=\!\mathfrak{q}_i\!\cap\!A$.


Computing the "Lying Over", "Going Up", "Going Down" Ideals: Let $\{x_1,\ldots,x_m\}\!\subseteq\!\{y_1,\ldots,y_n\}$ be variables, $I\!\unlhd\!K[x_1,\ldots,x_m]\! =\!K[\mathbf{x}]$, $A\!=\!K[\mathbf{x}]/I$, $J\!\unlhd\!K[y_1,\ldots,y_n]\! =\!K[\mathbf{y}]$, $B\!=\!K[\mathbf{y}]/J$, and $A\!\leq\!B$ via the identification $f(\mathbf{x})\!+\!I\!\mapsto\!f(\mathbf{x})\!+\!J$ (this map is injective iff $J\!\cap\!K[\mathbf{x}]\!\subseteq\!I$, which we assume). In general, we have (more or less by definition)
$$\begin{align*}
\mathrm{Spec}(A/\mathfrak{a})&=\{\mathfrak{p}/\mathfrak{a};\,\mathfrak{p}\!\in\!\mathrm{Spec}(A),\mathfrak{a}\!\subseteq\!\mathfrak{p}\}\text{ and }\\ 
\mathrm{Max}(A/\mathfrak{a})&=\{\mathfrak{m}/\mathfrak{a};\,\mathfrak{m}\!\in\!\mathrm{Max}(A),\mathfrak{a}\!\subseteq\!\mathfrak{m}\}\\
\end{align*}.$$
Recall that if $A\!\leq\!B$ and $\mathfrak{a}\!\unlhd\!A$, then $\mathfrak{a}B\!:=\!\langle\mathfrak{a}\rangle_B\!=\!\{\sum_{i=1}^n\!s_ia_i;\, n\!\in\!\mathbb{N}_0, s_i\!\in\!B, a_i\!\in\!\mathfrak{a}\}$, the ideal of $B$ generated by $\mathfrak{a}$. We investigate the situation where we have $\mathfrak{p}_0\!\subseteq\!\mathfrak{p}_1\!\subseteq\!\mathfrak{p}_2,\,$ $\mathfrak{p}_i\!=\!\langle P_i\rangle/I\!\in\!\mathrm{Spec}(A),\,$ $P_i\!\subseteq\!K[\mathbf{x}],\,$ $\mathfrak{q}_0\!\subseteq\!\mathfrak{q}_1\!\subseteq\!\mathfrak{q}_2,\,$ $\mathfrak{q}_i\!=\!\langle Q_i\rangle/J\!\in\!\mathrm{Spec}(B),\,$ $Q_i\!\subseteq\!K[\mathbf{y}]$, and $\mathfrak{q}_i\!\cap\!A\!=\!\mathfrak{p}_i$, for $i\!=\!0,1,2$.
 $(\ast)$
Claim: If $A\!\leq\!B$ is integral, $\dim(B)\!<\!\infty$, $\mathfrak{p}\!\in\!\mathrm{Spec}(A)$, $\mathfrak{q}\!\in\!\mathrm{Spec}(B)$, $\mathfrak{p}B\!\subseteq\!\mathfrak{q}$, then $$\,\dim(B/\mathfrak{p}B)=\dim(B/\mathfrak{q})\,\Longleftrightarrow\,\mathfrak{q}\cap A=\mathfrak{p}\,\Longrightarrow\,\mathfrak{q}\in\mathrm{minAss}(\mathfrak{p}B).$$
Proof: $(\Rightarrow)$ Suppose $\mathfrak{q}\!\cap\!A\!\supsetneq\!\mathfrak{p}$. If $\mathfrak{q}\!=\!\mathfrak{q}_0\!\subsetneq\!\ldots\!\subsetneq\!\mathfrak{q}_n$ is the longest chain of primes in $B/\mathfrak{q}$, i.e. $n\!=\!\dim(B/\mathfrak{q})$, then after intersecting with $A$, by "incomparable" and "lying over" we get (together with $\mathfrak{p}$) a chain of primes in $A$ of length $n\!+\!1$. By "lying over" and "going up" we can lift this to a chain of primes $\tilde{\mathfrak{q}}\!\subsetneq\!\tilde{\mathfrak{q}}_0\!\subsetneq\!\ldots\!\subsetneq\!\tilde{\mathfrak{q}}_n$ in $B$. Since $\tilde{\mathfrak{q}}\!\cap\!A\!=\!\mathfrak{p}$, we have $\mathfrak{p}B\!\subseteq\!\tilde{\mathfrak{q}}$. This implies $\dim(B/\mathfrak{p}B)\!\geq\!n\!+\!1$, which contradicts the assumption.
$(\Leftarrow)$ Since $\mathfrak{p}B\!\subseteq\!\mathfrak{q}$, we automatically have $\dim(B/\mathfrak{p}B)\!\geq\!\dim(B/\mathfrak{q})$. Let $\mathfrak{q}\!\cap\!A\!=\!\mathfrak{p}$ and let $\mathfrak{p}B\!\subseteq\!\mathfrak{q}_0\!\subsetneq\!\ldots\!\subsetneq\!\mathfrak{q}_n$ be the longest chain of primes in $B/\mathfrak{p}B$, i.e. $n\!=\!\dim(B/\mathfrak{p}B)$. After intersecting with $A$, by "incomparable" and "lying over" we get $\mathfrak{p}\!\subseteq\!\mathfrak{p}B\!\cap\!A\!\subseteq\!\mathfrak{q}_0\!\cap\!A\!\subsetneq\!\ldots\!\subsetneq\!\mathfrak{q}_n\!\cap\!A$, where $\mathfrak{q}_i\!\cap\!A$ and $\mathfrak{p}$ are prime. Since $\mathfrak{q}$ is prime and lies over $\mathfrak{p}$, by "going up" we can lift this chain to  $\mathfrak{q}\!\subseteq\!\tilde{\mathfrak{q}}_0\!\subsetneq\!\ldots\!\subsetneq\!\tilde{\mathfrak{q}}_n$, with $\tilde{\mathfrak{q}_i}\!\in\!\mathrm{Spec}(B)$, i.e. $n\!\leq\!\dim(B/\mathfrak{q})$.
$(\Rightarrow)$ Let $\mathfrak{q}\!\cap\!A\!=\!\mathfrak{p}$, $\mathfrak{p}B\!\subseteq\!\mathfrak{q}'\!\subseteq\!\mathfrak{q}$, $\mathfrak{q}'\!\in\!\mathrm{Spec}(B)$. After intersecting with $A$, we get $\mathfrak{p}\subseteq\mathfrak{p}B\!\cap\!A \subseteq \mathfrak{q}'\!\cap\!A  \subseteq \mathfrak{p}$, i.e. $\mathfrak{q}'\!\cap\!A\!=\!\mathfrak{p}$, so by "incomparable", $\mathfrak{q}'\!=\!\mathfrak{q}$. $\blacksquare$

To sum up, consider the situation $(\ast)$, i.e. $\mathbf{x},\mathbf{y},I,J$ are given and $K[\mathbf{x}]/I\!\leq\!K[\mathbf{y}]/J$ is finite. We have established the following recipes:


*

*(lying over) Suppose $P_1$ is given. How do we compute $Q_1$? We compute $\mathrm{minAss}(\langle P_1,J\rangle_{K[\mathbf{y}]})$ and then any $Q$ in this (finite) set of ideals, for which $\dim(K[\mathbf{y}]/\langle P_1,J\rangle)\!=\!\dim(K[\mathbf{y}]/\langle Q\rangle)$ holds, suffices as $Q_1$, by the claim above. These are all the possible $Q_1$.

*(going up) Suppose $P_1,P_2,Q_1$ are given. How do we compute $Q_2$?
We compute $\mathrm{minAss}(\langle P_2,J\rangle_{K[\mathbf{y}]})$
and then any $Q$ in this set, for which $\dim(K[\mathbf{y}]/\langle
   P_2,J\rangle)\!=\!\dim(K[\mathbf{y}]/\langle Q\rangle)$ and $\langle
   Q_1\rangle\!\subseteq\!\langle Q\rangle$ holds, suffices as $Q_2$, by
the claim. These are all the possible $Q_2$.

*(going down) Suppose $P_0,P_1,Q_1$ are given. How do we compute $Q_0$? We compute $\mathrm{minAss}(\langle P_0,\!J\rangle_{K[\mathbf{y}]})$ and then any $Q$ in this set, for which $\dim(K[\mathbf{y}]/\langle P_0,J\rangle)\!=\!\dim(K[\mathbf{y}]/\langle Q\rangle)$ and $\langle Q\rangle\!\subseteq\!\langle Q_1\rangle$ holds, suffices as $Q_0$, by the claim. These are all the possible $Q_0$. $\blacklozenge$ 

A: 1) As Zhen Yie write, a ring $R$ of finite type over $K$ is a finitely-generated $K$-algebra.
2) This is correct, as a direct consequence of the homomorphism theorem.
3.1) This looks correct.
3.2) $\mathfrak{q} \in \mathrm{minAss(\mathfrak{p}B})$ does not neccesarily imply $\dim(B/\mathfrak{p}B) = \dim(B/\mathfrak{q})$. In general, this property holds only if $B/\mathfrak{p}B$ is a catenary ring, which is not always the case. For example, consider the finite ring extension $A := C[x,y] \hookrightarrow B:= C[x,y,z]/\langle x(z-x), y(z-x)\rangle$, and $P = 0$. Then $\dim(B/PB) = 2$ via the chain $\langle x \rangle \subseteq \langle x, y \rangle$, but $\langle z-x \rangle \in \mathrm{minAss}(B/PB)$, where $\dim(B/\langle z \rangle) = 1$.
