On the proof of elimination of imaginaries in algebraically closed valued fields

Let $v_{1},\ldots,v_{n}$ be non-zero vectors with $\operatorname{val}(v_{n})$ in different orbits of $\Gamma(K)$. We will show that the $v_{i}$ are linearly independent. If not, let $w_{1},\ldots,w_{m}$ be a minimal subset which is linearly dependent. Then $\sum_{i}x_{i}w_{i}=0$ for some $x_{i}\in K^{\times}$. But by assumption,

$$\operatorname{val}(x_{i}w_{i})=\operatorname{val}(x_{i})+\operatorname{val}(w_{i})\neq\operatorname{val}(x_{j})+\operatorname{val}(w_{j})=\operatorname{val}(x_{j}w_{j})$$

for any $i\neq j$. By the ultrametric inequality in $V$, $\sum_{i}x_{i}w_{i}$ cannot be zero, a contradiction. ∎

We need to show that

$$\operatorname{val}(v+\alpha a)=\min(\operatorname{val}(v),\operatorname{val}(\alpha a))$$

(1)

for $v\in V$ and $\alpha\in K$. Replacing $v$ and $\alpha$ with $\alpha^{-1}v$ and $\alpha^{-1}\alpha$ changes both sides of (1) by the same amount, so we may assume that $\alpha=0$ or $\alpha=1$.

The $\alpha=0$ case is trivial. Suppose that $\alpha=1$; we want to show $\operatorname{val}(v+a)=\min(\operatorname{val}(v),\operatorname{val}(a))$. If $\operatorname{val}(v)\neq\operatorname{val}(a)$, then $\operatorname{val}(v+a)=\min(\operatorname{val}(v),\operatorname{val}(a))$ by the ultrametric inequality. In the case where $\operatorname{val}(v)=\operatorname{val}(a)$, the ultrametric inequality only implies

$$\operatorname{val}(v+a)\geq\min(\operatorname{val}(v),\operatorname{val}(a))=\operatorname{val}(a).$$

(2)

But $\operatorname{val}(v+a)\leq\operatorname{val}(a)$, by the assumption on $a$. So equality holds in (2). ∎

Proceed by induction on $\dim_{K}V$. The one-dimensional case is easy. Let $V^{\prime}$ be a codimension 1 subspace. By induction, $V^{\prime}$ is isomorphic to $K^{\oplus(n-1)}$, so $V^{\prime}$ is spherically complete. Choose some $a_{0}\in V\setminus V^{\prime}$ and let $a$ be an element of $a_{0}+V^{\prime}$ maximally close to 0. By Lemma 1.4, $K\cdot a$ is perpendicular to $V^{\prime}$. Thus $V\cong V^{\prime}\oplus K\cong K^{\oplus(n-1)}\oplus K=K^{\oplus n}$. ∎

We are trying to prove a conjunction of first-order sentences, so we may replace $K$ with an elementarily equivalent model. Therefore, we may assume $K$ is spherically complete.

Replacing $V$ with the $K$-span of $N$, we may assume that $V$ is the $K$-span of $N$. Similarly, if $W$ denotes the largest $K$-vector space contained in $N$, then by quotienting out $W$, we may assume that $N$ contains no nontrivial $K$-vector spaces. Now $\bigcup_{\alpha\in K^{\times}}\alpha N=V$ and $\bigcap_{\alpha\in K^{\times}}\alpha N=0$.

For any nonzero $v\in V$, let

$$\operatorname{val}(v)=\sup\{\operatorname{val}(\alpha):v\in\alpha N\}=\inf\{\operatorname{val}(\alpha):v\notin\alpha N\}.$$

This is well-defined by o-minimality of $\Gamma$, and one easily checks that

$$\operatorname{val}(\beta v)=\operatorname{val}(\beta)+\operatorname{val}(v).$$

(3)

$$\operatorname{val}(v)>0\implies v\in N$$

(4)

$$\operatorname{val}(v)<0\implies v\notin N$$

(5)

for all $\beta\in K$, $v\in V$. We claim that $\operatorname{val}:V\to\Gamma$ makes $V$ into a valued $K$-vector space. Given (3), we merely need to check the ultrametric inequality

$$\operatorname{val}(v+w)\geq\min(\operatorname{val}(v),\operatorname{val}(w)).$$

If this failed, then multiplying everything by an appropriate scalar, we would get

$$\operatorname{val}(v+w)<0<\min(\operatorname{val}(v),\operatorname{val}(w)).$$

But then $v,w\in N$ and $v+w\notin N$, contradicting the fact that $N$ is a module.

So $\operatorname{val}:V\to\Gamma$ makes $V$ into a valued $K$-vector space. By Theorem 1.5, we can assume that $V$ is $K^{\oplus n}$. Then (4-5) mean the following for $\vec{x}\in K^{n}$:

• •

  If $\operatorname{val}(x_{i})>0$ for every $i$, then $\vec{x}\in N$. In other words, $\mathfrak{M}^{n}\subseteq N$.

• •

  If $\operatorname{val}(x_{i})<0$ for some $i$, then $\vec{x}\notin N$. In other words, $N\subseteq\mathcal{O}^{n}$.

So $N$ is sandwiched between $\mathcal{O}^{n}$ and $\mathfrak{M}^{n}$. But the possibilities for $N$ then correspond to the submodules of $\mathcal{O}^{n}/\mathfrak{M}^{n}$, i.e., the $k$-subspaces of $k^{n}$. These are easy to deal with.

Specifically, note that $N/\mathfrak{M}^{n}$ is a $k$-subspace of $\mathcal{O}^{n}/\mathfrak{M}^{n}=k^{n}$. Let $\gamma$ be an element of $GL_{n}(k)$ sending $N/\mathfrak{M}^{n}\subseteq k^{n}$ to $k^{\ell}\times 0^{n-\ell}\subseteq k^{n}$ for $\ell=\dim_{k}N/\mathfrak{M}^{n}$. Then $\gamma$ can be lifted to some $\gamma^{\prime}\in GL_{n}(\mathcal{O})$, because $\mathcal{O}$ is a local ring. If $N^{\prime}=\gamma^{\prime}(N)$, then $N^{\prime}/\mathfrak{M}^{n}$ is $k^{\ell}\times 0^{n-\ell}=(\mathcal{O}^{\ell}\times\mathfrak{M}^{n-\ell})/\mathfrak{M}^{n}$. So $N^{\prime}=\mathcal{O}^{\ell}\times\mathfrak{M}^{n-\ell}\subseteq K^{n}$. But $N^{\prime}$ and $N$ are isomorphic. ∎

(a)

If $p$ and $q$ are generically stable, and $r$ is arbitrary, then $p\otimes q\otimes r=p\otimes r\otimes q=r\otimes p\otimes q$.

(b)

Suppose $p$ is generically stable, $f$ is a definable function, and $q$ is arbitrary. Then $p(x)\otimes q(y)=q(y)\otimes p(x)$. Pushing both sides forwards along $(f\times id)$ and applying Remark 2.1, we get that $f_{*}p(x^{\prime})\otimes q(y)=q(y)\otimes f_{*}p(x^{\prime})$.

(c)

Let $q$ be another invariant type; we will show that $p(x)\otimes q(y)=q(y)\otimes p(x)$. Let $B$ be a set over which $p,q,f$ are defined. Let $(b,a)$ realize $q\otimes p|B$. By Remark 2.1, $(b,f(a))\models q\otimes f_{*}p|B$. Since $f_{*}p$ is generically stable, $(f(a),b)\models f_{*}p\otimes q|B$. By (6), $(a,b)\models p\otimes q|B$. So $p\otimes q$ and $q\otimes p$ agree when restricted to the arbitrary set $B$.

(d)

Let $B$ be a sufficiently big set. Suppose that $(f(a),g(b))\models f_{*}p\otimes g_{*}q|B$. We need to show that $(a,b)\models p\otimes q|B$. By (6), $(a,g(b))\models p\otimes g_{*}q|B$. By generic stability of $p$, $(g(b),a)\models g_{*}q\otimes p|B$. By (6) again, $(b,a)\models q\otimes p|B$. By generic stability again, $(a,b)\models p\otimes q|B$.

(e)

Suppose $p(x)$ commutes with itself, but $p(x)\otimes q(y)\neq q(y)\otimes p(x)$. Choose some formula $\phi(x;y;c)$ which is in $p(x)\otimes q(y)$ and not in $q(y)\otimes p(x)$. We will prove that $\phi(x;y,z)$ has the independence property. Let $n$ be arbitrary. Let $a_{1},\ldots,a_{n},b,a_{n+1},\ldots,a_{2n}$ realize $p^{\otimes n}\otimes q\otimes p^{\otimes n}$ restricted to $c$. Then $\models\phi(a_{i};b;c)\iff i\leq n$, by choice of $\phi(x;y;c)$. The fact that $p$ commutes with itself implies that all permutations of $(a_{1},\ldots,a_{2n})$ have the same type over $c$. Therefore, for each permutation $\pi$ of $\{1,\ldots,2n\}$, we can find a $b_{\pi}$ such that $\phi(a_{i};b_{\pi};c)$ holds iff $\pi(i)\leq n$. It follows that for any $S\subseteq\{1,\ldots,n\}$, we can find a $b_{S}$ such that $\phi(a_{i};b_{S};c)$ holds if and only if $i\in S$. As $n$ was arbitrary, $T$ has the independence property, a contradiction.

∎

Naming the parameters from $C$, we may assume $C=\emptyset$. Let $\phi(x_{2},x_{1};y)$ be a formula; we must produce a $\phi$-definition (over $\emptyset$) for $\operatorname{tp}(a_{2}a_{1}/B)$. Since $\operatorname{tp}(a_{2}/Ba_{1})$ is $a_{1}$-definable, the $\phi(x_{2};x_{1},y)$-type of $a_{2}$ over $Ba_{1}$ has a definition $\psi(y,a_{1})$. In particular, for every tuple $b$ from $B$,

$$\models\phi(a_{2},a_{1},b)\leftrightarrow\psi(b,a_{1}).$$

Meanwhile, since $\operatorname{tp}(a_{1}/B)$ is 0-definable, there is a formula $\chi(y)$ such that for every $b$ in $B$,

$$\models\psi(b,a_{1})\leftrightarrow\chi(b).$$

Thus, for every $b$ in $B$,

$$\models\phi(a_{2},a_{1},b)\leftrightarrow\chi(b),$$

so $\chi(y)$ is the $\phi(x_{2},x_{1};y)$-definition of $\operatorname{tp}(a_{2}a_{1}/B)$. ∎

(1$\implies$2)

The restriction of a global $C$-definable type to $BCD$ is $C$-definable.

(2$\implies$3)

Given $D$, (2) implies that there is $A^{\prime}\equiv_{BC}A$ such that $\operatorname{tp}(A^{\prime}/BCD)$ is $C$-definable. Choose $D^{\prime}$ such that $A^{\prime}D\equiv_{BC}AD^{\prime}$. Then $\operatorname{tp}(A/BCD^{\prime})$ is $C$-definable and $D^{\prime}\equiv_{BC}D$.

(3$\implies$4)

Applying (3) to $D$ a small model containing $BC$, we get a small model $D^{\prime}$ containing $BC$ such that $\operatorname{tp}(A/BCD^{\prime})=\operatorname{tp}(A/D^{\prime})$ is $C$-definable.

(4$\implies$1)

$C$-definable types over models have unique $C$-definable extensions to elementary extensions. This is true even for $*$-types.

∎

We use condition (3) of Lemma-Definition 2.8. Let $D$ be a small set of parameters. Since $a_{1}\operatorname*{\raise 0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}^{\textrm{def}}_{C}B$, there is $D^{\prime}\equiv_{BC}D$ such that $\operatorname{tp}(a_{1}/BCD^{\prime})$ is $C$-definable. As $a_{2}\operatorname*{\raise 0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}^{\textrm{def}}_{Ca_{1}}B$ there is $D^{\prime\prime}\equiv_{Ca_{1}B}D^{\prime}$ such that $\operatorname{tp}(a_{2}/BCa_{1}D^{\prime\prime})$ is $Ca_{1}$-definable. Note that $\operatorname{tp}(a_{1}/BCD^{\prime\prime})$ is $C$-definable. By Lemma 2.7, it follows that $\operatorname{tp}(a_{2}a_{1}/BCD^{\prime\prime})$ is $C$-definable. Since $D^{\prime\prime}\equiv_{BC}D^{\prime}\equiv_{BC}D$, we have verified $a_{2}a_{1}\operatorname*{\raise 0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}^{\textrm{def}}_{C}B$ using condition (3). ∎

Replacing $C$ with $\operatorname{acl}(C)$, we may assume that $C=\operatorname{acl}(C)$. By condition (4) of Lemma-Definition 2.8, there is a small model $M$ containing $BC$ such that $\operatorname{tp}(A/M)$ is $C$-definable. We need to show that $\operatorname{tp}(\operatorname{acl}(AC)/M)$ is $C$-definable. This is equivalent to showing that for each $\operatorname{acl}(AC)$-definable set $X$, there is some $C$-definable set $X^{\prime}$ such that $X\cap M=X^{\prime}\cap M$.

Given such an $X$, let $X_{1},\ldots,X_{n}$ be the conjugates of $X$ over $AC$. Let $E$ be the equivalence relation

$$xEy\leftrightarrow\bigwedge_{i=1}^{n}(x\in X_{i}\leftrightarrow y\in X_{i})$$

Then $E$ is $AC$-definable. Since $\operatorname{tp}(AC/M)$ is $C$-definable, the restriction $E^{\prime}=E\cap M$ of $E$ to $M$ is $C$-definable. Since $E$ has finitely many equivalence classes, so does $E^{\prime}$, and hence each equivalence class of $E^{\prime}$ is $C$-definable, as $\operatorname{acl}(C)=C$. But $X\cap M$ is a union of finitely many $E^{\prime}$-equivalence classes, so $X\cap M$ is $C$-definable. ∎

By Lemma 2.11, we know that $\operatorname{acl}(a_{1}C)\operatorname*{\raise 0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}^{\textrm{def}}_{\operatorname{acl}(C)}B$. We are given $a_{2}\operatorname*{\raise 0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}^{\textrm{def}}_{\operatorname{acl}(Ca_{1})}B$. Combining these using Lemma 2.9, we conclude that $a_{2}\operatorname{acl}(Ca_{1})\operatorname*{\raise 0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}^{\textrm{def}}_{\operatorname{acl}(C)}B$. This easily implies $a_{2}a_{1}\operatorname*{\raise 0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}^{\textrm{def}}_{\operatorname{acl}(C)}B$, as desired. ∎