数理解析研究所講究録 1283 巻 2002 年 55-60

55

A Remark on Generic Pseudoplanes
豊田工業高等専門学校

Toyota National College of Technology
池田宏一郎

Koichiro IKEDA
Abstract
We prove that if -generic saturated pseudoplane is strictly stable,
then the algebraic closure of afinite set is finite.
$\delta$

1Generic structures
Let be afinite relational language and If a class of finite -structures closed
under isomorphism and substructures. For any , @ with $A\subset B$ let $A\leq B$
be areflexive and transitive relation which is invariant under isomorphism. Inwhat follows , $K$ satisfies the following set of axioms.
$L$$L$$A$$B$$I\acute{\backslash }$(A1) $A\subset B\subset C\in K$ and $A\leq C$ implies $A\leq B$ ;for any $A\in K$ ;(A1)(A3) $A\leq B\in K$ and $X\subset B$ implies $A\cap X\leq X$ ;such that, for each $i<\omega$ ,(A4) There are no infinite chains$A_{i}\in K$ ,and any proper non-empty subset $X$ of $A_{i+1}-A_{i}$ satisfies

Axiom 1.1$\emptyset\leq A$$A_{1}\subset A_{2}\subset\ldots$$A_{\dot{1}}$$A:\leq A_{:}X$$\not\leq A_{i+1}$.

For an infinite -structure $M$ satisfying $A\in K$ for any finite $A\subset M$ , define$A\leq M$ if $A\leq B$ for all finitewith $A\subset B\subset M$ .
$L$

$B$

, for aNote 1.2 Let $M$ satisfy $A\in K$ for all finite $A\subset M$ . Byof with $B^{*}\leq M$ . Suchfinite $B\subset M$ there is aunique smallest superset).
ais called the closure of in M. (in symbol$(\mathrm{A}1)-(\mathrm{A}4)$$B^{*}$$B$$B^{*}$$B$$\mathrm{c}1_{M}(B)$. Astructure $M$ is said to besatisfy
Definition 1.3 Let$(K, \leq)$ -generic, if(i) If is afinite substructure of $M$ then $A\in K$ .
$(I\mathrm{f},$$A$$\leq)$$(\mathrm{A}1)-(\mathrm{A}4)$

56(ii) If A$f(B)\ovalbox{\tt\small REJECT}$

M and A BE K then there is an -embedding f B
M. (An -embedding is an embedding fixing A pointwise.)$\ovalbox{\tt\small REJECT}$$A$$\ovalbox{\tt\small REJECT}$$\ovalbox{\tt\small REJECT}+M$$\ovalbox{\tt\small REJECT}$

with

$A$

Whenever we consider a $(K, \leq)$ -generic structure,is supposed tosatisfy the above conditions. However, even if $(K, \leq)$ satisfies(A4), then a $(K, \leq)$ -generic structure does not necessarily exist.
$(I\mathrm{f},$$\leq)$$(\mathrm{A}1)-(\mathrm{A}4)$$(\mathrm{A}1)-$

Definition 1.4 $(K, \leq)$ is said to have the amalgamation prvyperty if for any$A\leq B\in K$ and $A\leq C\in K$ there is $D\in K$ such that $f(B)\leq D$ and $g(C)\leq D$
for some -embeddings : $Barrow D$ and $g:Carrow D$ .
$A$

$f$

Fact 1.5([1],[2],[5]) If (K,has the amalgamation property, then there
exists aunique (K, -generic structurt.
$\leq)$$\leq)$

2Theorem and Proof
Let L be alanguage of bipartite graphs: L {P, Q, R} where P,Q are unarypredicates and R. Let be areal number. Then$=$$\subset P\cross Q$$\alpha$\bullet For afinite L structure A, Sa(A) $=|P^{A}|+|Q^{A}|-\alpha|R^{A}|$ .
\bullet

$K_{\alpha}=$\bullet For ANote 2,1{ A : A is afinite L structure,$\subset B\in K_{\alpha}$$\forall B\subset A[\delta_{\alpha}(B)\geq 0]$, A $\leq B$ is defined by

It is easily checked that$\delta_{\alpha}(XA)\geq \mathrm{S}\mathrm{Q}(\mathrm{A})$$(K_{\alpha}, \leq)$

satifies}.

for any X $\subset B-A$ .

$(\mathrm{A}1)-(\mathrm{A}3)$

Definition 2.2 We say that abipartite graph $M$ isgeneric for some and.

$\delta$.

-generic, if $M$ is $(K, \leq)-$$K\subset K_{\alpha}$$\alpha$Our goal is to show the following theorem.
Theorem Let $M$ be a -generic saturated pseudoplane. If $M$ is strictly stable,
then the algebraic closure of any finite set is finite.
$\delta$

To prove this theorem, we need some preparations.
In what follows, we assume that $K$ (:satisfies the amalgamation property, and that $M$ is a $(K, \leq)$ -generic saturated pseudoplane.
$K_{\alpha}$

Note 2.3([1],[5])If$\alpha$is apositive rational number, then Th(M) is$\omega$

stable,57
Definition 2.4 Let AB be afinite bipartite graph. Then(i) Apair $(B, A)$ is said to be nomal, if $A\leq AB\in K$ and $A\cap B=\emptyset$ .
(ii) Anormal pair $(B, A)$ is said to be small, if there are no normal pairs $(D, C)$such that $A\subset C$ , $B\subset D$ and.
(iii) Anormal pair $(B, A)$ is said to be minimal, if there are no non-empty
proper subsets $C$ of with $AC\leq AB$ .
$\mathrm{S}(\mathrm{D}/\mathrm{C})<\mathrm{S}(\mathrm{D}/\mathrm{C})$$B$

To simplify our notation, we denote $R(x, y)\vee R(y, x)$ by $S(x, y)$ . For any
elements , , of abipartite graph we say apair ( , ab) is special, if$S(e, b)$ holds.
$e$

$a$

$b$

$e$

$\mathrm{S}(\mathrm{d}, a)\wedge$Note 2.5 Suppose that a \"A, -generic bipartite graph is apsuedoplane.
Letbe afinite bipartite graph with no loops, i.e., for each $n>2$ there do
not exist dintinct , , , $a_{n}\in A$ with ( , a),, $S(a_{n-1}, a_{n})$ and$S(a_{n}, a_{1})$ . Then we can see that $A\in K$. (The proof is by induction.)$($$\leq)$$A$$a_{1}$

Lemma 2.6$\alpha\leq 1$$a_{2}$$S$$\ldots$$a_{1}$$S(a_{2}, a_{3}),\ldots$.

Proof Suppose by way of contradiction that $\alpha>1$ . By genericity there is$a\in M$ with $a\leq M$ . Then there are no element $b\in M$ with $S(b, a)$ . (If not,then $\delta(b/a)=1-\alpha<0$ . This contradicts $a\leq M.$ ) But this contradicts thedefinition of pseudoplanes.

Lemma 2.7$\frac{1}{3}<\alpha$.

Proof Suppose by way contradiction that
. Let abed be an L-structurewith the relations $S(d, a)$ , $S(d, b)$ ,2.5,By.
we have,. By
we have $\delta(d/abc)\geq 0$ , and so $a6c\leq a6cd$ . By amalgamation property, we caninductively construct asequencesuch that (i) $S(e:, a)$ , $S(e_{i}, b)$ ,$i<\omega$for each, and (ii)for each $i<\omega$ . In particular we have$S(e_{i}, a)\wedge S(e:, b)$ for each $i<\omega$ . This contradicts the definition of psuedoplane.
$\alpha\leq\frac{1}{3}$$abed\in I\mathrm{f}$$\mathrm{S}(\mathrm{d}, c)$$\neg S(e_{i}, c)$$\{e_{i}\}_{i<\omega}$$abcde_{1}\ldots e_{\dot{l}}$$\in I\mathrm{f}$Lemma 2.8 Let
be an irrational number with
$n\geq 2$ . Then aspecial pair is not small.
$\alpha$Proof Let
$S(a_{n}, d)$$\alpha\leq\frac{1}{3}$$\frac{n-1}{2n-1}<\alpha\leq\frac{n}{2n+1}$, wherebe afinite -structure with the relations $S(a_{1}, c)$ ,and $\{S(a_{i}, a_{i+1})\}_{i=1,\ldots,n-1}$ . Let $A=\{a_{i}\}_{i=1,\ldots,n}$ and. By 2.5, we have $ABcd\in K$ .

$a_{1}b_{1}a_{2}b_{2}\ldots a_{n}b_{n}cd$$L$, $\{S(a_{i}, b:)\}_{i=1,\ldots,n}$$B=\{b_{\dot{1}}\}:=1,\ldots,n$Claim 1: $Bed\leq ABcd$ .
Proof: Take any $X\subset A$ . It is easily seen that if $X\neq A$ then $\delta(X/Bcd)\geq$
then$\delta(X/Bcd)=n-(2n+1)\alpha-(2n+1)\frac{\mathrm{v}\mathrm{e}\delta n}{2n+1}=0$.
.
$\mathrm{H}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e}Bcd\leq ABcd|X|-2|X|\alpha.\mathrm{S}\mathrm{o},\mathrm{b}\mathrm{y}\alpha\leq\frac{n}{2n+1,\geq n’}\leq\frac{1}{2}\mathrm{w}\mathrm{e}\mathrm{h}\mathrm{a}(X/Bcd)\geq 0.\mathrm{I}\mathrm{f}X=A$58
Claim $45(a_{1}/Bcd)>\mathit{6}(A/Bcd)$ .
Proof: 6
By claim 1,2, special pairthis lemma.
$2\ovalbox{\tt\small REJECT}$$(A/Bcd)-\mathit{6}(a_{\mathit{1}}/Bcd)\ovalbox{\tt\small REJECT}$ $(\mathrm{n}-1)-(2\mathrm{n}-1)0<(\mathrm{n}-1)-(2\mathrm{n}-1)\ovalbox{\tt\small REJECT}\ovalbox{\tt\small REJECT}\ovalbox{\tt\small REJECT}$$(a_{1},6_{1}\mathrm{c})$Note 2.9

Let X$=$Lemma 2.10 Letpair is not small.

{a$\alpha$: a, b$-b\alpha$Q.

is not smalL This completes the proof ofa $-b\alpha>0\}$ . Then $\inf X=0$ .

$<\omega,$be an irrational number with$\alpha>\frac{1}{2}$. Then any minimal. By 2.9, there are
Proof Let $(B, A)$ be aminimal pair with
, $q<\omega$ such that $m\leq p$ , $n\leq q$ and $0<p-q\alpha<m-\mathrm{n}\mathrm{a}$ . To show that $(B,A)$
is not small, it is enough to see that there is anormal pair $(D,C)$ such that
$A\subset C$ , $B\subset D$ and. Pick aelement $b_{0}\in B$ . Let $k=p-m$,, , $S(b_{k-1},b_{k})$ . Let
and take , , , with the relations$S(a:,b:)$with the relations$l=q-n$ and take , a2, ,for $1\leq:\leq l-k$ . Let$C=Aa_{1}a_{2}\ldots a_{l-k}$ and. By 2.5, $CD\in K$ . On the other hand,$\delta(D/C)=\delta(B/A)+k$ $-(k+l-k)\alpha=(m+k)$ $-(n+l)\alpha=p-q\alpha$ . Also we cansee that $C\leq CD$ . (It can be shown as follows: Take any $X\subset D-C$ and let$X_{C}=X\cap C$ and $X_{D}=X\cap(D-C)$ . Then $\delta(X/C)=\delta(X_{B}/C)+\delta(X_{D}/CX_{B})=$$\delta(X_{B}/A)+\delta(X_{D}/CX_{B})$ . Note that $B\geq A$ and. Hence $\delta(X/C)\geq 0.)$ It
follows that $(D, C)$ is normal.
$\delta(B/A)=m-\mathrm{n}\mathrm{a}$$p$$\delta(D/C)=p-\mathrm{q}\mathrm{a}$$b_{1}$$S(b_{0},b_{1})$$b_{k}$$b_{2}$$\ldots$$a_{1}$$\ldots$$S(b_{1},b_{2})$$\ldots$$a\iota-k$$D=Bb_{1}b_{\mathit{2}}\ldots b_{k}$$\alpha>\frac{1}{2}$Lemma 2.11 If$\alpha$is irrational, then any minimal pair is not small.

, then any minimal pair. If
Proof By 2.6 and 2.7, we have
, then there is $n<\omega$ with, andis not small by 2.10. If
therefore any minimal pair is not small, by 2.8.
$\alpha\in(\frac{1}{3},1]$$\alpha>\frac{1}{2}$$\alpha\in(\frac{n-1}{2n-1},$ $\frac{n}{2n+1}]$$\alpha<\frac{1}{2}$Lemma 2.12 Let $A\leq AB\leq M$ . Let $(B, A)$ be aminimal pair. If
is algebraic, then $(B, A)$ is small.

$\mathrm{t}\mathrm{p}(B/A)$

Proof Suppose by way of contradiction that $(B,A)$ is not small. Then there
is anormal pair $(D, C)$ such that $A\subset C$, $B\subset D$ and $\delta(D/C)<\delta(B/C)$ . By
minimality of $(B, A)$ we can assume that $(D, C)$ is minimal.
with the following conditions:
Claim 1: There is asequence$i<\omega$;(i)for any, $CB_{0}\ldots B:-1D\leq CB_{0}\ldots B:D\in K$ for any $i<\omega$ ;(\"u)(\"ui) $D$ , , , , are pairwise disjoint.
Proof of Claim: We prove by induction. Suppose $(B:):\leq n$ has constructed.
By (ii), we have CBO...Bn $\leq CB_{0}\ldots B_{n}D\in K$ , and therefore CBO...Bn$CB_{0}\ldots B_{n}B\in K$ .
So, by amalgamation property, we can take $B_{n+1}$ so that$(B_{\dot{|}}):<\mathrm{I}d$$B:\cong cB_{\mathrm{O}}\ldots B_{j-1}B$$CB_{0}\cdots B:$$B_{0}$$B_{1}$$B_{2}$$\ldots$$\leq$

59$B_{n}B_{n+1}\leq CB_{0}\ldots$ $B_{n}B_{n+1}D\in K$ .
andBi B.
Thus $B_{n+1}$ satisfies (i) and (ii). For (iii) it is enough to show that $B_{n+1}\cap$. Suppose that $D’=B_{n+1}\cap D\neq\emptyset$ . We have had $CB_{0}\ldots B_{n}B_{n+1}$$CB_{0}\ldots B_{n}B_{n+1}D$ , so $CD’\leq CD$ . Note that $D’\neq D$ . This contradicts minimality of $(D, C)$ . Hence $B_{n+1}\cap D=\emptyset$ . (End of Proof of Claim 1)
Claim 2: AB, $ABj\leq AB_{0}\ldots B_{i}B$ for $j\leq i<\omega$
Proof: We prove by induction on . By (ii) of claim 1, $AB_{0}\ldots B_{i}B\leq AB_{0}\ldots B_{i+1}B$ .
By induction hypothesis, we have AB, $AB_{j}\leq AB_{0}\ldots B_{i}B$ for $j\leq i$ . Hence
AB, $AB_{j}\leq AB_{0}\ldots B_{i+1}B$ for $j\leq i$ . So, it is enough to show that $AB_{i+1}\leq$$AB_{0}\ldots B_{i+1}B$ .
By induction hypothesis, we have $AB\leq AB_{0}\ldots B_{i}B$ . By (i)of claim 1, we have $AB_{i+1}\leq AB_{0}\ldots B_{i+1}$ . By (ii) of claim 1, $AB_{0}\ldots B_{i+1}\leq$$AB_{0}\ldots B:+1B$ . Hence we have $AB:+1\leq AB_{0}\ldots B\dot{.}+1B$ . (End of Proof of Claim2)
is non-algebraic. Fix any $n<\omega$ . By claim 2,
We show thatand $AB\leq ABB_{0}^{*}\ldots B_{n}^{*}\leq M$ .
such that
there arefor all $i\leq n$ . Therefore we have
Again, by claim 2,’s are pairwise disjoint. Hence. By (iii) of claim 1,
is not algebraic.

$B_{n+1}\cong_{CB_{0}}B_{n}B$$CB_{0}\ldots$$CB_{0}\ldots$$D=\emptyset$$\leq$$i$$\mathrm{t}\mathrm{p}(B/A)$$B_{0}^{*}\ldots B_{n}^{*}\cong_{AB}B_{0}\ldots B_{n}$$B_{i}^{*}’ \mathrm{s}$$AB_{i}^{*}\leq ABB_{0}^{*}\ldots B_{n}^{*}\leq M$$\mathrm{t}\mathrm{p}(B_{i}^{*}/A)=\mathrm{t}\mathrm{p}(B/A)$$B^{*}.\cdot$$\mathrm{t}\mathrm{p}(B/A)$

Lemma 2.13
of $M$ .

If$\alpha$is irrational, then$\mathrm{a}\mathrm{c}1(X)=\mathrm{c}1(X)$for any finite subset $X$is clear. We
Proof Take any finite subset $X$ of $M$ . Then
$A=\mathrm{c}1(X)$and. Let. If not, there isshow$B=\mathrm{c}1(aX)$ . Take amaximal chain
with $A=B0\leq B_{1}\leq\ldots\leq B_{n}=B$ .
Then, for each $i<\omega$ , $(B_{i+1}-B_{i}, B:)$ is minimal and $A\leq ABj\leq M$ . By 2.11,
is not algebraic. In particular we havethey are not small, and so. Acontradiction.
$\mathrm{c}1(X)\subset \mathrm{a}\mathrm{c}1(X)$$\mathrm{a}\mathrm{c}1(X)$$a\in \mathrm{a}\mathrm{c}1(X)-\mathrm{c}1(X)$$\subset \mathrm{c}1(X)$$\{B_{*}.\}:<\omega$$\mathrm{t}\mathrm{p}(B_{i+1}/B_{i})$$B\not\subset \mathrm{a}\mathrm{c}1(A)=\mathrm{a}\mathrm{c}1(X)$Proof of Theorem Let $M$ be a $(K, \leq)$ -generic saturated pseudoplane for. Suppose that $M$ is strictly stable. By 2.3, is irrational. By
someis finite by (A4) offor any finite $X\subset M$ . Note that
2.13,
is finite.
Axiom 1.1. Hence$K\subset K_{\alpha}$$\alpha$$\mathrm{c}1(X)$$\mathrm{a}\mathrm{c}1(X)=\mathrm{c}1(X)$$\mathrm{a}c1(X)$

Question Are -generic pseudoplanes -categorical?
$\delta$$\omega$

Reference[1] J. T. Baldwin and N. Shi, Stable generic structures Annals of Pure andApplied Logic 79 (1996) 1-35
[2] J. Goode, Hrushovski’s geometries, In Helmut Wolter Bernd Dahn, editor,
Proceedings of 7th Easter Conference on Model Theory (1989) 106-118[3] E. Hrushovski, Astable -categorical pseudoplane, preprint, 19 8$\mathrm{N}_{0}$60[4] E. Hrushovski, Anew strongly minimal set, Annals of Pure and AppliedLogic, 46 (1990) 235-264[5] F. O. Wagner, Relational structures and dimensions, Kaye, Richai (ed.)
et al., Automorphisms of first-0rder structures. Oxford: Clarendon Press.
153-180 (1994)

Department of Mathematics
Toyota National College of Technology
2-1, Eiseicho,Toyota, 471-8525, JAPANikeda@toyota-ct.ac.j$\mathrm{p}$