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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:

enter image description here

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?

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3  
If $R \to S$ is a ring homomorphism, then $S$ is of finite type over $R$ if $S$ is finitely-generated as a $R$-algebra. When $R$ is a field, then this is the same thing as a finitely-presented $R$-algebra. A similar definition can be made for schemes. This is not to be confused with a finite $R$-algebra, which is an $R$-algebra which is finitely-generated as an $R$-module. –  Zhen Lin Mar 14 '12 at 22:41
    
@ZhenLin: "When $R$ is a field, then this is the same thing as a finitely-presented $R$-algebra." Umm, more generally, when $R$ is commutative unital Noetherian, 'finitely generated R-algebra' is the same as 'finitely presented $R$-algebra', by the Hilbert basis theorem, yes? –  Leon Mar 17 '12 at 2:42
    
Yes. We just need to use the fact that any polynomial ring in finitely many variables over a noetherian ring is again noetherian. –  Zhen Lin Mar 17 '12 at 9:15

2 Answers 2

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 $\dim(B/\langle z \rangle) = 1$.

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I just checked my Commutative Algebra lecture notes. In fact, for the general Lying-Over theorem, it is sufficient that we have an integral ring extension $A \to B$. –  Johannes Kloos Mar 17 '12 at 7:23
    
Wiki says: $A\leq B$ is a finite / of finite type / integral extension when $B$ is a finitely generated $R$-module / $B$ is a finitely generated $R$-algebra / $\forall b\!\in\!B$ $\exists$ monic $f\in A[x]$ with $f(b)=0$. A ring extension is finite iff it is integral and of finite type. Q1: Isn't $A=K[\mathbf{x}]/I\leq B=K[\mathbf{y}]/J$ always an extension of finite type, because already $K\leq K[\mathbf{y}]/J$ is an extension of finite type? Thus "finite" and "integral" extensions of affine $K$-algebras are equivalent notions, right? –  Leon Mar 18 '12 at 7:01
    
Q2: We always have $\dim(R/\mathfrak{a})\!\leq\!\dim(R)$, right? Q3: Regarding (3.2), Kemper's A Course in Commutative Algebra, p.117 thm.8.22, says that all affine $K$-algebras are catenary. I do not understand you counterexample. If we have $A=\mathbb{Q}[y,z]\leq B=K[x,y,z]/J$, $J=\langle x(x-z),y(x-z)\rangle$, and $P=0\in\mathrm{Spec}(A)$, then $PB=0$, $\mathrm{minAss}_B(PB)=\mathrm{minAss}_{K[x,y,z]}(J)/J=\{\langle x,y\rangle/J,\langle x-z\rangle/J\}$, and $\dim(B/PB)=\dim(K[x,y,z]/J)=2=\dim(K[x,y,z]/\langle J,x-z\rangle)$, but $\dim(K[x,y,z]/\langle J,x,y\rangle=1$, according –  Leon Mar 18 '12 at 7:23
    
to SINGULAR. Perhaps that's what you meant. But what is then wrong with the following "proof". Let $Q\in\mathrm{minAss}(PB)$. Since $PB\subseteq Q$, automatically $\dim(B/PB)\geq\dim(B/Q)$. Let $PB\subseteq P_0\subseteq\ldots\subseteq P_n$ be the longest chain of primes in $B/PB$. Then $Q\subseteq P_0\subseteq\ldots\subseteq P_n$ is also a chain of primes in $B/Q$, hence $\dim(B/PB)=\dim(B/Q)$. I still do not see what catenary rings (i.e. those where any two maximal chains of primes between two prime ideals have the same length) have to do with anything, since $PB$ isn't in general prime. –  Leon Mar 18 '12 at 7:31
1  
Thanks for noticing. It's strange - I actually constructed it from a geometric example ($x$-$y$ plane and non-perpendicular, non-parallel line cutting through it). Its projection on the $x$-$y$ plane has finite fibers, so I expected the ring extension to be finite as well. –  Johannes Kloos Apr 1 '12 at 19:23
up vote 1 down vote accepted

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 $\mathfrak{p}_i\!=\!\mathfrak{q}_i\!\cap\!A$.

enter image description here


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 $i\!=\!0,1,2$.

enter image description here $(\ast)$

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 $Q_0$. $\blacklozenge$
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