数理解析研究所講究録
第 1555 巻 2007 年 104-109

104

A Note on the Stability Spectrum of Generic
Structures
法政大学・経営学部 池田宏一郎 (Koichiro IKEDA) *Faculty of Business Administration, Hosei University

Abstract「超安定であるが 安定でないような K-generic 構造が存在するか」 という問題がある (Baldwin [1]). このノートでは, クラス $K$ が部分グラフに関して閉じているときは, そのような K-generic 構造が存在しないことを示した. なお, このノートは [4] の内容を整理改良したものである.
$\omega$

Let be a countable relational language and $K$ a class of finite L-structures
be a class of L-structures such that any finite
closed under subgraphs. Let
substructure belongs to K.
$L$$\overline{K}$. Then and
Deflnition 1 Let), if it satisfies the following:
symbol,$ABC\in\overline{K}$

$B$

$C$

are said to be flee over

$A$

(in

$\delta(A_{i})>\delta(A_{i+1})$

for$B\perp {}_{A}C$(i) $B\cap C\subset A$ ;(ii) $R^{ABC}=R^{AB}\cup R^{AC}$ for any $R\in L$ .

Remark 2 Let $ABCD\in\overline{K}$ . Then(i) If $A1_{B}C$ and $A\perp BCD$ , then.
(ii) If $BC\perp AD$ , then $B\perp CAD$ .
(iii) If $BC\perp AD$ , thenif and only if$A\perp {}_{B}CD$$B\perp {}_{A}C$$B\perp {}_{D}C$.

Deflnition 3 $\delta:Karrow R\geq 0$ is said to be a predimension, if(i) if or $B\in K$ , then $\delta(A)=\delta(B)$ ;(ii);(iii) for all $AB\in K,\delta(A/B)\leq\delta(A/A\cap B)$ ;(iv) there is no infinite chain
of $A_{i}\in K$ with$A$$\delta(\emptyset)=0$$A_{1}\subset A_{2}\subset\ldots$’Research partially supported by Grants-in-Aid for Scientific Research (no.16540123), Ministry of Education, Science and Culture.

105;(v) for any $AB\in K,$ $A\perp A\cap BB$ if and only if $\delta(A/B)=\delta(A/A\cap B)$ ;(vi) for any $ABCD\in K$ with $B\cap ACD=\emptyset,$ $\delta(B/AC)-\delta(B/A)\leq\delta(B/DAC)-$$\delta(B/DA)$ ,$\dot{j}\in\omega$where $\delta(X/Y)$ means $\delta(XY)-\delta(Y)$ .

, we define $A\leq B$ , if $\delta(X/A’)\geq 0$ for any
Definition 4 (i) For, define $c1_{B}(A)=\cap\{A’$ :finite $X\subset B-A$ and $A’\subset A$ . For$A’\leq B\}.$ By the definition of a predimension, there exists such a $c1_{B}(A)$ , andmoreover if is finite, then so is $c1_{B}(A)$ .
(ii) Fix. For finite $A\subset M$ , define $d_{M}(A)=\delta(c1_{M}(A))$ . For finite$M,$ $d_{M}(A/B)=d_{M}(AB)-d_{M}(B)$ . For infinite $B,$ $d_{M}(A/B)= \inf\{d_{M}(A/B’)$ :$B’\subset B$
finite}. For (possibly) infinite $A,$ $B,$ $C\subset M,$ $d_{M}(B/C)=d_{M}(B/A)$
means $d_{M}(B’/C)=d_{M}(B’/A)$ for any finite $B’\subset B$ .
(iii) A countable L-structure $M$ is said to be $(K, \leq)$ -generic, if $A\in K$ for any
finite $A\subset M$ ; If $A\leq B\in K$ , then there is $B’\cong_{A}B$ with $B’\leq M$ .
$A\subset B\in\overline{K}$$A\subset B\in\overline{K}$$A\subset$$A$$B\subset$$M\in\overline{K}$Let$\mathcal{M}$

Fact 5be a big model. The following facts can be found in [2], [5] and [6].

Let $B,$$C\leq \mathcal{M}$(i) $d(B/C)=d(B/A)$ ;(ii)
and$B\perp {}_{A}C$

Proof

and $A=B\cap C$ . Then the following are equivalent:$BC\leq \mathcal{M}$$(i)\Rightarrow(ii)$.

. First we show that

$B,\overline{c}\in C,\overline{e}\in c1(\overline{b}\overline{c})-BC$$d(\overline{b}/A)<\gamma/2$

Then

and

with

$BC\leq \mathcal{M}$. If not, then there are. Take$\delta(\overline{e}/\overline{b}\overline{c})=-\gamma<0$$d(\overline{c}/\overline{a})-d(\overline{c}/A)<\gamma/2$

. Let

with
and と

$\overline{a}\leq A$$\overline{b}’=c1(\overline{b}\overline{a})$$\overline{b}\in$$d(\overline{b}/\overline{a})-$$=c1(\overline{c}\overline{a})$.

$d(\overline{b}’\overline{c}’/\overline{a})=d(-\ -/\overline{a})\geq d(\overline{b}\overline{c}/A)=d(\overline{b}/A\overline{c})+d(\overline{c}/A)=d(\overline{b}/A)+d(\overline{c}/A)>$$d(\overline{b}/\overline{a})+d(\overline{c}/\overline{a})-\gamma=\delta(\overline{b}’/\overline{a})+\delta(\overline{c}’/\overline{a})-\gamma\geq\delta(\overline{b}’\overline{c}’/\overline{a})-\gamma$

hand, we have

. On the other. A$d(\overline{b}’-d/\overline{a})\leq\delta(\overline{e}\overline{b}’\overline{c}’/\overline{a})\leq\delta(\overline{b}’\overline{c}’/\overline{a})+\delta(\overline{e}/\overline{b}’\partial)=\delta(\overline{b}’-d/\overline{a})-\gamma$

contradiction. Next we show that $B1_{A}C$ . If not, then there arewith. Take
. Let
where
with
. By remark,
and
. Let
and
. Thenwe have. A contradiction.
. By (ii),. If not, then there are
with
and と
we ct takesuch that. A. Thenwhere$\overline{b}\in B,\overline{c}\in C$$\delta(\overline{b}/\overline{c})<\delta(\overline{b}/\overline{a})$$\overline{a}=\overline{b}\cap\overline{c}$$d(\overline{b}/\overline{a}^{l})-d(\overline{b}/A)<\gamma$$\overline{a}\subset\overline{a}’$$\gamma=\delta(\overline{b}/\overline{a})-\delta(\overline{b}/\overline{c})$$‘=c1(\overline{a}^{l}\overline{b})$$\delta(\overline{b}’/\overline{a}’)-\delta(\overline{b}’/\overline{c}\overline{a}’)\geq\delta(\overline{b}/\overline{a})-\delta(\overline{b}/\overline{c})=\gamma$$\overline{a}’\leq A$$d-=c1(\overline{a}’\overline{c})$$\delta(\overline{b}’/\overline{c}\overline{a}’)\geq d(\overline{b}’/\overline{c}A)=$$d(\overline{b}’/A)=d(\overline{b}/A)>d(\overline{b}/\overline{a}^{l})-\gamma=\delta(\overline{b}’/\overline{a}’)-\gamma\geq\delta(\overline{b}’/\overline{c}\overline{a}’)$$\overline{b}\in B,\overline{c}\in C$$(ii)\Rightarrow(i)$$\overline{b}’,\overline{c}’$$\overline{a}’=\overline{b}’\cap\overline{c}’$$d(\overline{b}/\overline{c})<d(\overline{b}/A)$$\overline{b}\subset\overline{b}’\leq-B,\overline{c}\subset\overline{c}’\leq C,\overline{b}’\perp\overline{a}’\overline{c}’$$-b’$$\leq \mathcal{M}$$d(\overline{b}/\overline{c})=\delta(b’/\overline{c}’)=\delta(\overline{b}’/\overline{a}’)\geq d(b/\overline{a}’)\geq d(b/A)$

contradiction.

Fact 6 Let $B,$
and $A=B\cap C$ be algebraically closed. Then thefollowing are equivalent:(i) $tp(B/C)$ does not fork over ;
(ii)
and
.
$C\leq \mathcal{M}$$A$$B\perp {}_{A}C$$BC\leq \mathcal{M}$

106
Proof

$(i)\Rightarrow(ii)$. Suppose that

$B\downarrow AC$. Take a sufficiently saturated model.
and. Then we have
with
Claim 1: $d(B/N)=d(B/NC)$ .
with
Proof: If $d(B/N)>d(B/NC)$ , then there are. Take countable. By the saturation of $N$ , we can
with
pick
, we have. Sinceandwith. ThenU. Henceand$d(b/N)$ . A contradiction.
Claim 2: $d_{\backslash }(B/A)=d(B/N)$ .
Proof: Let $B^{*}=ac1(B)$ . We can takewith $d(B^{*}/N)=d(B^{*}/A_{1})$ where$A\subset A_{1}\subset N$ and $|A_{1}|=|B|+\aleph_{0}$ .
acl??? By the saturation of $N$ there is. Note thatwith $tp(A_{2}/A)=tp(A_{1}/A)$ andby. Let $B_{1}^{*}=c1(A_{1}B^{*})$ and $B_{2}^{*}=c1(A_{2}B^{*})$ . Then
. By fact
6, we havesince $d(B^{*}/N)=d(B^{*}/A_{1})=d(B"/A_{2})$ . Hence$B^{*}N=B_{1}^{*}N\cap B_{2}^{*}N\leq \mathcal{M}$ .
On the other hand, we have $B^{*}\perp AN$ . (Proof:
Suppose that $B^{*}4_{A}N$ . Note thatand $B^{*}1_{A_{2}}N$ sinoe $d(B”/N)=$$d(B”/A_{1})=d(B^{*}/A_{2})$ . So we have $B^{*}J_{A}A_{1}$ and $B^{*}1_{A}A_{2}$ . Since,$A_{1}\cap
A_{2}=A$
we have. A contradiction. ) Hence $d(B/N)=d(B/A)$ .
By claim 1,2, we have $d(B/A)=d(B/NC)$ , and hence $d(B./A)=d(B/C)$ .
. Take $B’$ such that $tp(B’/C)$ does not fork over and $tp(B/A)=$$tp(B’/A)$ . By
, we haveand. So we have $tp(BC/A)=$$tp(B’C/A)$ , and henoe $tp(B/C)$ does not fork over .
$N\supset A$$BC\downarrow_{A}N$$B\downarrow_{N}C$$B\downarrow_{A}N$$d(\overline{b}/N)>$$\overline{b}\in B,\overline{c}\in NC$$A_{0}\subset N$$d(\overline{b}/\overline{c})$$\overline{b}\overline{c}\downarrow A_{0}N$$stp(\overline{c}/A_{0})=stp(\overline{c}’/A_{0})$$\overline{c}’\in N$$\overline{b}\downarrow A_{0}\overline{c}$$tp(b\overline{c}/A_{0})=tp(b\overline{c}’/A_{0})$$\overline{b}\downarrow A_{0}$$\overline{b}\downarrow N^{\overline{C}}$$\overline{b}\overline{c}\downarrow A_{0}N$$d(b/\overline{c})=d(b/\overline{c}’)\geq$$A_{1}$$A_{1}$$A_{2}\subset N$$A_{1}\downarrow AA_{2}$$A_{1}\downarrow B^{*}A_{2}$$B\downarrow_{A}N$$B_{1}^{*}\cap B_{2}^{*}=B^{*}$$B_{1}^{*}N,B_{2}^{*}N\leq \mathcal{M}$$B^{*}\perp A_{1}N$$A_{1}\downarrow AA_{2}$$A$$(ii)\Rightarrow(i)$$B’\perp {}_{A}C$$(i)\Rightarrow(ii)$$B’C\leq \mathcal{M}$$A$

For each $A\leq B\in K,$ $B$ is said to be minimal, if $C=A$ or$A\leq C\leq B$.

Lemma 7 Let $A\leq B\in K$ with
$tp(B/A)$ is algebraic, thenfor any$B\leq \mathcal{M}$$B\perp {}_{A}C$. Let$C\leq \mathcal{M}$

$B$

$B$

for any$C$be minimal overwith $B\cap C=A$

with

$A$

.

. If

Proof Suppose that $\delta(B/C)<\delta(B/A)$ for some $C\leq BC\in K$ with $B\cap C=A$ .
Claim: There is a setof copies of $B$ over with the following conditions:(i) $C\leq CB_{j}\leq CB_{0}B_{1}\cdots B_{i}\in K$ for each $j\leq i<\omega$ ;(ii) $B_{i}\cap B_{j}=A$ for each $j<i<\omega$ ;(iii)
are free over for each $i<w$ .
Proof: Suppose thathas been defined. By our assumption, we have
$C\leq CB\in K$ , and by (i) we have $C\leq CB_{0}B_{1}\ldots B_{n}\in K$ . By amalgamation, we
can take a copyof $B$ over $C$ such that.
$B_{i}\neq
B^{*}$By (iii) and $\delta(B^{*}/C)<\delta(B^{*}/A)$ , we havefor all $i\leq n$ . Since$B$$K$$B^{*}\cap
B_{i}=A$,is minimal over
we have. Since is closed under Lsubgraphs, there is $B_{n+1ABoB_{1}\ldots B_{n}}\cong B^{*}$ such that
and
$B_{n+1},$
are free over . So (ii) and (iii) hold. It is not difficult to check that
$CB_{j}\leq CB_{0}B_{1}\cdots B_{n+1}\in K$ for each $j\leq n+1$ . So (i) holds. (End of Proof of$\{B_{i}\}_{i<\omega}$$B_{i},$

$C$

$A$

$A$

$\{B_{i}\}_{i\leq \mathfrak{n}}$$B^{*}$$CB_{0}\ldots B_{\mathfrak{n}},$$CB”\leq CB_{0}\ldots B_{\mathfrak{n}}B^{*}\in K$$A$$CB_{0}B_{1}\ldots B_{n}B_{\mathfrak{n}+1}\in K$$C$

Claim)$A$

107
By claim, we have $AB_{j}\leq AB_{0}\ldots B_{i}\in K$ for each $j\leq i<w$ . We can assume. Thus we have $tp(B_{j}/A)=tp(B/A)$ for each $j\leq i$ . By (ii)thatof claim,are pairwise distinct. Hence $tp(B/A)$ is not algebraic.
$AB_{0}\ldots B_{i}\leq \mathcal{M}$$B_{j}’ s$Lemma 8 Let $A\leq B\in K$ with
$tp(B/A)$ is algebraic, thenfor any

Let $B$ be minimal overwith $B\cap C=A$ .

$B\leq \mathcal{M}$$BC\leq \mathcal{M}$

$A$

. If

$C\leq \mathcal{M}$withfor some
Proof Suppose by way of contradiction that
$B\cap C=A$ . Then there is finite
such that $\delta(X/BC)<0$ .
of copies of with the following conditions:
Claim 1: There is a set(i) $B_{i}\cong cB_{0}\ldots B:-1B$ for each $i<w$ ;$CB_{0}\ldots B_{i-1}BX\leq CB_{0}\ldots B_{i}BX\in K$ for each $i<w$ ;(ii)(iii) $XB\cap B_{i}=B_{j}\cap B_{i}=A$ for each $j<i<w$ .
Proof: Suppose thathas been defined. By (ii), $CB_{0}\ldots B_{n}\leq CB_{0}\ldots B_{n}BX\in$$K$ , and so we have $CB_{0}\ldots B_{n}\leq CB_{0}\ldots B_{n}B\in$ K.
By amalgamation, we can$B_{n+1}$
such thatof overtake a copy
$CB_{0}\ldots B_{n}B_{n+1}BX\in$ K. Hence (i) and (ii) hold. On the other hand,$B_{i}=A$ for each $i\leq n$ , since. So, to see that (iii) holds, it isenough to show that $B_{n+1}\cap XB=A$ . Let $B’=B_{n+1}\cap XB$ . First, suppose. Then we have $B_{n+1}\subset BX$ , and so $CB_{n+1}\not\leq CBX$ , sincethat$C\leq \mathcal{M}$$BC\not\leq \mathcal{M}$$X\subset \mathcal{M}-\mathcal{B}C$$B$$\{B_{i}\}_{i<w}$$CB_{0}\ldots B_{i},$$\{B_{i}\}_{i\leq n}$$B$$CB_{0}\ldots B_{n}BX,CB_{0}\ldots B_{\mathfrak{n}}B_{n+1}\leq$$CB_{0}\ldots B_{n}$$B_{\mathfrak{n}+1}\cap$$B_{n+1CB_{0}\ldots B_{n}}\underline{\simeq}B$$B’=B_{\mathfrak{n}+1}$$\delta(XB/CB_{n+1})=\delta(XB/C)-\delta(B_{n+1}/C)=\delta(XB/C)-\delta(B/C)=\delta(X/BC)<$$0$. This contradicts our choice of$B_{n+1}$. Hence we have$B’\neq B_{\mathfrak{n}+1}$. We have to, we havesee that $B’=A$ . This can be shown as follows: By our choice of, and so $B’\leq B_{n+1}$ . Since is minimal and$B’\neq B_{n+1}$ , we have $B’=A$ . (End of Proof of Claim 1)
Claim 2: $B,$ $B_{j}\leq B_{0}\ldots B_{i}B(\in K)$ for $j\leq i<\omega$.
Proof: We prove by induction on . By (ii) of claim 1,$B,$
for $j\leq i$ . Hence
By induction hypothesis, we have $B,$
$B_{i+1}\leq
B_{0}\ldots B_{t+1}B$ . By infor $j\leq i$ . So, it is enough to show that
. From (i) of claim 1, it follows
duction hypothesis again, we have
. Henoe we. By (ii) of claim 1,
that
have
B. (End of Proof of Claim 2)$tp(B/A)$
We show thatis non-algebraic. By claim 2, we can assume that$B,$
for each $i,j$ with $j\leq i<\omega$ . So we have $tp(B_{j}/A)=$$tp(B/A)$ for each $j<w$ . By (iii) of claim 1,
are pairwise distinct. Hence$B_{n+1}$$B$$CB_{0}\ldots B_{n}BX\leq CB_{0}\ldots B_{n}B_{\mathfrak{n}+1}BX$$B_{0}\ldots B_{i}B\leq B_{0}\ldots B_{i+1}B$$i$$B_{j}\leq$$B_{j}\leq B_{0}\ldots B_{i}B$$B_{0}\ldots B_{i+1}B$$B\leq B_{0}\ldots B_{i}B$$B_{0}\ldots B_{i+1}\leq B_{0}\ldots B_{i+1}B$$B_{i+1}\leq B_{0}\ldots B_{i+1}$$B_{t+1}\leq B_{0}\ldots B_{i+1}$$B_{j}\leq BB_{0}\ldots B_{i}\leq \mathcal{M}$$B_{j}’ s$$tp(B/A)$ is not algebraic.

Proposition 9$ac1(A)\cup B\leq \mathcal{M}$.

Let$A\leq B\leq \mathcal{M}$and $A=ac1(A)\cap B$ . Then $ac1(A)\perp AB$ andand
Proof We can assume that $A,B$ are finite. We wil show that
. Take $A=A_{0}\leq A_{1}\leq\ldots\leq$for any finite $A^{*}\leq ac1(A)$ with
$A_{n}=A^{*}$ withminimal overfor each $i<n$ . Then it is enough to show$A^{*}\perp AB$$A\subset A^{*}$$A^{*}B\leq \mathcal{M}$$A_{i+1}$$A_{i}$

108
for each $i\leq n$ . (Proof: We prove by induction
on . Clearly $A_{i}\leq A_{i+1},$ $A;+1\cap A_{i}B=A_{i}$ and $tp(A_{i+1}/A_{i})$ is algebraic. By
by. So we have $A_{i+1A}\perp: B$ andinducition hypothesis,, and henoe $A_{i+1}1_{A_{0}}B$ . )lemma. By induction hypothesis,that$A_{iA_{0}}\perp B$

and$A_{i}B\leq \mathcal{M}$$i$$A_{i+1}B\leq \mathcal{M}$$A_{i}B\leq \mathcal{M}$$A_{iA_{0}}\perp B$

and $A=B\cap C$ . Then the following are equivaTheorem 10 Let $B,$
lent:(i) $tp(B/C)$ does not fork over $A$ ;(ii) $B1_{A}C$. and $BC\cup ac1(A)\leq \mathcal{M}$ .
$C\leq \mathcal{M}$

Proof By proposition 9, $B\cup ac1(A),$ $C\cup ac1(A)\leq \mathcal{M}$ . So, by fact 7, (i)is equivalent to $B\perp Cac1(A)$ and $BC\cup ac1(A)\leq \mathcal{M}$ . Therefore, proving that(i) and (ii) are equivalent, it is enough to show that $B\perp ac1(A)C$ if and only if. We can assume that $A,$ $B,$ $C$ is finite. Take any finite $A^{*}\leq ac1(A)$ with
$BC\cap ac1(A)\subset A^{*}$ . Then we will show that
if and only if. Let$B’=B\cap A^{*}$ , $C’=C\cap A^{*}$ .
Sinoe $tp(A^{*}/B’C’)$ is algebraic, we have $A”\perp_{B’C’}BC$ . So, fromit$B\perp
B’C’C$. On the other hand, sinoe $tp(B’/C’)$ andfollows thatare
. Henoe we have.
algebraic, we haveand$B\perpB^{l}C’C$$B1_{A}C$$tp(A^{*}/B’C’)$, we have
By. On the other hand, sinoeis$A^{*}\perp
B’C’BC$
algebraic, we have
. Henoe.
$B\perp {}_{A}C$$B\perp A^{*}C$$B\perp {}_{A}C$$B\perp {}_{A}C$$(\Rightarrow)$$t^{\backslash }p(C’/A)$$B’\perp c\prime C$$B\perp {}_{A}C’$$B\perp {}_{A}C$$(\Leftarrow)$$B\perp {}_{A}C$

Corollary 11 Let be a countable relational language and $K$ a class df finte
L-structures that is derived from a predimension . Then there is no K-genericstructure that is superstable but not w-stable.
$L$$\delta$

Proof Suppose that a theory of a K-generic structure is superstable. Takeany countable model $N$ of .
Claim: For any $p\in S(N)$ there is finite $A\subset N$ such that does not fork over$A$
and $p|A$ is stationary.
Proof: Take a realization of . By superstability, there is finite $X\subset N$ suchthat does not fork over $X$ . Let $B=c1(X\overline{b})$ and $A=B\cap N$ . Clearly does
not fork over . We show thatis stationary. Take any
such that
anddoes not fork over . Let $B=c1(\overline{b}’A)$ . Then$tp(B/N)$ and $tp(B’/N)$ do not fork over . Sinoe, we have$B\cong AB’$ . Note that $B\cap N=B’\cap N=A$ . By theorem, $B\perp AN,$ $B’\perp AN$ and$BN,$. In particular, $BN\cong B’N$ . It follows that $tp(BN)=tp(B’N)$
and hence $tp(b/N)=tp(b’/N)$ . (End of Proof of Claim)
By claim, we have $|S(N)|\leq\aleph_{0}\cdot|S(T)|=\aleph_{0}$ . Henoe is w-stable.
$T$

$T$

$p$

$\overline{b}$

$p$

$p$

$p$

$A$

$tp(\overline{b}’/A)=tp(\overline{b}/A)$$tp(\overline{b}/A)$$\overline{b}’$$tp(\overline{b}/N)$

$A$

$A$

$tp(\overline{b}/A)=tp(\overline{b}’/A)$$B’N\leq \mathcal{M}$$T$

Reference

109[1] J. T. Baldwin, Problems on pathological structures, In Helmut Wolter
Martin Weese, editor, Proceedings of 10th Easter Conferenoe in Model Theory(1993) 1-9[2] J. T. Baldwin and N. Shi, Stable generic structures, Annals of Pure andApplied Logic 79 (1996) 1-35
[3] K. Ikeda, A remark on the stability of saturated generic graphs, to appear
in Journal of the Mathematical Society of Japan, vol.57, no.4 (2005)[4] K. Ikeda, Forking in generic structures, Kokyuroku 1469(2006), 86-91[5] V. Verbovskiy and I, Yoneda, CM-triviality and relational structures,
Annals of ure and Applied Logic 122 (2003), 175-194[6] F. O. Wagner, Relational structures and dimensions, Kaye, Richard (ed.)
et al., Automorphisms of first-order structures. Oxford: Clarendon Press.
153-180 (1994)$\grave{P}$Faculty of Business Administration
Hosei University
2-17-1, Fujimi, Chiyoda
Tokyo, 102-8160
JAPAN
ikeda@hosei.ac.jp