Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Definition : A Topological space $\mathcal{D}$ is called noetherian if it satisfies the descending chain condition for closed subsets. We define the dimension of $\mathcal{D}$ to be the supremum of all integers n such that there exists a chain $C_{0}\subset C_{1}\subset\dots\subset C_{n}$ of distinct irreducible closed subsets of $\mathcal{D}$. This dimension is called Krull dimension. ($Kdim\mathcal{(D)}=n$).

Definition : Let $\varphi(\bar{x})$ be a formula with parameters in the monster model. We define the Morley rank of $\varphi(\bar{x})$ by indution as follow:

$MR(\varphi)\geq0$ if $\varphi(M)$ is nonempty;

$MR(\varphi)\geq\beta+1$ if there is an infinite family $(\varphi_{i}(\bar{x}))_{i\in\mathbb{N}}$ of formulas (in the same variables $\bar{x}$) which imply $\varphi$, are pairwise inconsistent and such that $MR(\varphi_{i})\geq\beta$ for all $i$;

$MR(\varphi)\geq \lambda$ (for a limit ordinal $\lambda$) if $MR(\varphi)\geq\beta$ for all $\beta<\lambda.$

(AF) Addition formula: For any irreducible $S\subseteq_{cl}U\subseteq_{op}D^{n}$ and a projection map $\pi:M^{n}\longrightarrow M^{m}$

$dim(S)=dim(\pi(S))+min\{ dim(\pi^{-1}(a)\bigcap S): a\in \pi(S)\}$

(FC) Fiber Condition : For any irreducible $S\subseteq_{cl}U\subseteq_{op}D^{n}$ and a projection map $\pi:M^{n}\longrightarrow M^{m}$ there exists $V\subseteq_{op}\pi(S)$ (relatively open) such that

$min\{ dim(\pi^{-1}(a)\bigcap S): a\in \pi(S)\}=dim(\pi^{-1}(a)\bigcap S)$, for any $a\in V$.

Question(1) : ŔźDoes Krull dimension have the (AF) and (FC) properties?

Question(2) : Does Morley Rank have the (AF) and (FC) properties?

Question(3) : let D be a Zariski geometry. Are Krull dimension and Morley rank equal in D ?

share|improve this question
    
It's "Does X have property Y?" rather than your "Has X property Y?". The latter would be OK German grammar though :) –  Dima Pasechnik Sep 4 '13 at 0:40
    
Dear Dima, Thank you very much for your correction. –  M.Mirabi Sep 7 '13 at 7:16

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.