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?