\documentstyle[12pt]{article} \title{Supersimple fields and division rings} \author{A. Pillay\thanks{Partially supported by NSF grant DMS 96-96268}\\University of Illinois and MSRI \and T. Scanlon\thanks{Supported by an NSF Postdoctoral Fellowship}\\MSRI \and F.O. Wagner\thanks{Supported by DFG grant Wa899/2-1}\\University of Oxford and MSRI} \newtheorem{Theorem}{Theorem}[section] \newtheorem{Proposition}[Theorem]{Proposition} \newtheorem{Lemma}[Theorem]{Lemma} \newtheorem{Fact}[Theorem]{Fact} \newtheorem{Definition}[Theorem]{Definition} \newtheorem{Remark}[Theorem]{Remark} \newtheorem{Corollary}[Theorem]{Corollary} \begin{document} \maketitle \begin{abstract} It is proved that any supersimple field has trivial Brauer group, and more generally that any supersimple division ring is commutative. As prerequisites we prove several results about generic types in groups and fields whose theory is simple. \end{abstract} \section{Introduction} Simple theories were introduced by Shelah in \cite{Shelah}. In \cite{Kim}, Kim, continuing Shelah's work, showed how the theory of forking transfers almost completely from the stable context to the more general simple context. In \cite{Kim-Pillay} the ``Independence Theorem" was proved for simple theories, giving a satisfactory analogue of the theory of stationarity from stability theory. In \cite{Pillay}, Poizat's theory (\cite{Poizat}) of generic types and stabilizers in stable groups was generalized to the case of groups definable in simple theories. Further generalizations appear in \cite{Wagner}. Stability-theoretic algebra studied, among other things, the algebraic consequences of imposing stability-theoretic conditions on a group, ring or field. Among the main results in the area was the Macintyre-Cherlin-Shelah theorem (\cite{Macintyre}, \cite{Cherlin-Shelah}) saying that any infinite field whose theory is superstable must be algebraically closed. Supersimple fields (namely fields with supersimple theory) on the other hand form a broader class. In \cite{Hrushovski} it is shown that any perfect pseudo-algebraically closed ($PAC$) field with ``small" absolute Galois group is supersimple; in fact of $SU$-rank $1$ in the language of rings. (Ultraproducts of finite fields furnish examples.) The work in this paper is partly motivated by the conjecture that these are the only cases. It follows from \cite{Pillay-Poizat} that any supersimple field is perfect and has small absolute Galois group. In the present paper we show that a supersimple field $K$ has trivial Brauer group. As supersimplicity is preserved under finite extensions, this is, by \cite{Serre}, X.Prop.11, equivalent to the norm map $N : L^{*} \rightarrow K^{*}$ being surjective for any finite Galois extension $L$ of $K$. Triviality of the Brauer group also implies that any rational variety over $K$ has a $K$-rational point, yielding in a sense a first approximation to the conjecture that any supersimple field must be $PAC$. We assume acquaintance with the notions and machinery from \cite{Kim-Pillay} and \cite{Pillay}. In section 2, some additional results are obtained about generic types and connected components of groups in simple theories. In section 3, we show that if $K$ is a field in a simple theory then the notions of additive and multiplicative generic coincide. We also make some observations about the interaction between the additive and multiplicative connected components. These results are applied in section 4 to show triviality of the Brauer group in the supersimple case. Some additional algebraic arguments yield in section 5 commutativity of any supersimple division ring. The first author would like to thank Zoe Chatzidakis for many fruitful discussions around the time when he first began considering the issues dealt with in this paper. In particular the result that if $K$ is a supersimple field of characteristic $0$, then any rational {\em curve} over $K$ has a $K$-rational point, was obtained together with her. The result on the triviality of the Brauer group was obtained by all three authors, and the results in section 5 on division rings were obtained by Wagner. \section{More on generic types and stabilizers} Let us fix a saturated model $\bar{M}$ of a simple theory, and an infinite group $G$ which is type-definable in $\bar{M}$ over $\emptyset$ say. All complete types $p(x)$ we consider will be types of elements of $G$ (namely we assume $``p(x) \rightarrow x \in G"$). We will often wish to work with Lascar strong types (so that we can use the Independence Theorem). This is most easily accomplished by working with types over models. So $M$ will denote a small elementary substructure of $\bar{M}$, which may vary. $A,B$ as usual denote small subsets of $\bar{M}$. We briefly recall (from \cite{Pillay}) the notions of generic type, stabilizer, and connected component, \noindent {\bf Generics.} Let $a\in G$. Then $tp(a/A)$ is (left-) generic (for $G$) if whenever $b\in G$ is independent from $a$ over $A$ then $b.a$ is independent from $A\cup \{b\}$ over $\emptyset$. (We also say that $a$ is a generic element of $G$ over $A$.) We have associated notions of a (type)-definable subset of $G$ being generic in $G$. Moreover the notions of left-generic and right-generic coincide. \noindent {\bf Stabilizers.} For $p(x)\in S(M)$, $St(p)$ is defined to be $\{b\in G$: for some realization $a$ of $p$ which is independent from $b$ over $M$, $b.a$ realises $p$ (and is independent from $b$ over $M$\}.$St(p)$ is type-definable over $M$. Let $Stab(p)$ be the subgroup of $G$ generated by $St(p)$. Then it turns out that $Stab(p)$ is type-definable over $M$, and moreover that $St(p)$ is ``large" in $Stab(p)$ in the sense that any generic element of $Stab(p)$ over $M$ is in $St(p)$. \begin{Remark} One can also define $St(p,q)$ for $p(x), q(x) \in S(M)$ in the obvious way: $St(p,q)$ = $\{b\in G$: for some $a$ realising $p$ such that $a$ is independent from $b$ over $M$, $b.a$ realises $q$ and is independent from $b$ over $M$\}. $St(p,q)$ is again type-definable over $M$ and one sees easily, using the Independence Theorem, that $St(p,q)$ is, if nonempty a large subset of a translate of $Stab(p)$. \end{Remark} {\bf Connected components.} For any set $A$ of parameters, $G^{0}_{A}$, the connected component of $G$ over $A$, is by definition the smallest type-definable over $A$ subgroup of $G$ of bounded index. A key fact is that $p(x)\in S(M)$ is a generic type of $G$ if and only if $Stab(p)$ contains $G^{0}_{M}$. Also clearly any generic type of $G^{0}_{A}$ is a generic type of $G$. In contrast to the stable situation, the connected component of $G$ over $A$ may vary with $A$. On the other hand $G^{0}_{A}$ is always a normal subgroup of $G$. If $G$ happens to be stable then we know that $G^{0}$ has a unique generic type, whose stabilizer is $G^{0}$. The next lemma gives an analogue of this in the simple case. \begin{Proposition} Let $p(x), q(x), r(x) \in S(M)$ be generic types of $G^{0}_{M}$. Then there are realizations $a,b,c$ of $p,q,r$ respectively, which are pairwise $M$-independent, and with $a.b = c$. \end{Proposition} {\em Proof.} This can be seen using Remark 2.1, but we give a direct proof. Let $b, c$ be independent (over $M$) realizations of $q,r$ respectively. Let $a'$ = $c.b^{-1}$. Then $a'.b = c$, $a'$ is generic in $G^{0}_{M}$ over $M$ and $\{a',b,c\}$ is pairwise $M$-independent. We can similarly find $d$ generic in $G^{0}_{M}$ over $M$, such that $a = a'.d$ realises $p$ and $\{d,a',a\}$ is pairwise $M$-independent. By the facts above on connected components and generics, $Stab(q)$ contains $G^{0}_{M}$ and hence any generic element of $G^{0}_{M}$ over $M$ is in $St(q)$. In particular $d^{-1} \in St(q)$. By the Independence Theorem we can assume that $\{d,a',b\}$ is pairwise $M$-independent and that $d^{-1}.b = b'$ realises $q$. Then $(a'.d).b' = c'$ realises $r$. That is $a.b' = c$, and easily $\{a,b',c\}$ is pairwise $M$-independent. \section{Simple fields} In this section $F$ will be an infinite field which is type-definable over $\emptyset$ in the saturated model $\bar{M}$ of a simple theory $T$. Types $p(x)$, $q(x)$ will be types of elements of $F$. We are going to apply Proposition 2.2 in order to understand the interaction between generic additive subgroups and generic multiplicative subgroups in simple fields. (In the stable case the situation is clear: a stable field is connected both additively and multiplicatively.) We begin by pointing out that in simple fields additive and multiplicative generics coincide. \begin{Proposition} Let $F$ be an infinite field which is type-definable in $\bar{M}$ over $\emptyset$. Then a type $p(x)\in S(A)$ is an additive generic type of $F$ iff it is a multiplicative generic type of $F$. \end{Proposition} {\em Proof.} We begin with \newline \noindent {\em Claim 1.} If $p(x) \in S(A)$ is additively generic then for any nonzero $a\in F$ and realisation $c$ of $p$ independent from $a$ over $A$, $tp(a.c/A\cup\{a\})$ is an additive generic of $F$. \newline \noindent {\em Proof of Claim 1.} Let $a\in F$ and $M$ contain $A\cup\{a\}$. Let $p'$ be a nonforking extension of $p$ over $M$, realised by $b$. Then $p'$ remains an additive generic of $F$. So the additive stabilizer of $p'$, $Stab^{+}(p')$ is a type-definable over $M$ subgroup of $F^{+}$ of bounded index. It follows that $Stab^{+}(a.p')$ = $a.Stab^{+}(p')$ also has bounded index. By the remarks above on connected components $a.p'$ is an additive generic type of $F$. In particular $a.p'$ does not fork over $\emptyset$. \newline \noindent From Claim 1, using the fact that $tp(a.c/A\cup\{a\})$ there, being an additive generic, does not fork over $\emptyset$, we see that: \newline \noindent {\em Claim 2.} Any additive generic type is a multiplicative generic type. \newline \noindent {\em Claim 3.} Any multiplicative generic type of $F$ is an additive generic type of $F$. \newline \noindent {\em Proof of Claim 3.} Let $q(x)\in S(A)$ be a multiplicative generic type. We may replace $q$ any time by a nonforking extension. By Claim 2 (and the existence of generic types) there is an additive generic type $p(x) \in S(A)$ which is also multiplicatively generic. It follows that there is $c \in F$, $M$ containing $A\cup\{c\}$ and nonforking extensions $q',p'$ of $q,p$ over $M$ such that $c.q'$ = $p'$. So $c.q'$ is an additive generic type. By Claim 1, $q'$ is also an additive generic type. \begin{Lemma} Let $T$ be a multiplicative subgroup of $F$ of bounded index, type-definable over $A$. Then every nonzero coset $a.T$ of $T$ meets $(F^{+})^{0}_{A}$ in a generic set. \end{Lemma} {\em Proof.} Note that if $B\supset A$, then $(F^{+})^{0}_{B}$ is contained in $(F^{+})^{0}_{A}$. So we may assume that $A$ is a model $M$ which contains a representative of each coset $c.T$ of $T$ in $F^{*}$. Note that \vspace{2mm} \noindent (*) For each nonzero $c \in F^{M}$, $c.(F^{+})^{0}_{M}$ = $(F^{+})^{0}_{M}$ (as multiplication by $c$ is an additive endomorphism of $F$ which interchanges $M$-definable sets). \vspace{2mm} \noindent Let $p(x)$ be a generic type of $G^{0}_{M}$ over $M$. Let $S$ be a coset of $T$ in $F^{*}$. Then there is $c\in F^{M}$ such that $c.p$ is the type of an element of $S$. By Claim 1 in the proof of Proposition 3.1, together with (*), $c.p$ is a generic type of $(F^{+})^{0}_{M}$. This proves the lemma. \begin{Remark} In fact, with the assumptions of Lemma 3.2, every coset of $(F^{+})^{0}_{A}$ in $F^{+}$ meets every coset of $T$ in $F^{*}$ in a generic set. \end{Remark} {\em Proof.} We may assume that $A$ is a model $M$ which contains representatives of all cosets of $T$ in $F^{*}$. To prove the remark, it is enough (using Lemma 3.2) to show that if $S$ is a coset of $(F^{+})^{0}_{M}$ in $F^{+}$ then $S\cap T$ is generic. Let $b\in S$ be nonzero. Let $M'$ be a model containing $M \cup\{b\}$. By Lemma 3.2, we can find generic $d$ in $(F^{+})^{0}_{M'}$ over $M'$ such that $d \in b^{-1}.T$. Let $c = d-1$. Then $c$ is generic in $(F^{+})^{0}_{M'}$ over $M'$ too, and as in the proof of Lemma 3.2, $b.c$ is generic in $(F^{+})^{0}_{M'}$ over $M'$. So $b.d$ = $b.c + b$ is generic in $S$ over $M$. On the other hand $b.d \in T$. Thus $S \cap T$ is generic, as required. \begin{Proposition} Let $T$ be any multiplicative subgroup of $F$ of bounded index. Let $S_{1}, S_{2}$ be cosets of $T$ in $F^{*}$. Then $S_{1} + S_{2}$ (the set of $s_{1} + s_{2}$ for $s_{1} \in S_{1}$ and $s_{2} \in S_{2}$) contains $F^{*}$. In fact for any nonzero $d\in F$ there are $a,b \in S_{1}, S_{2}$ respectively, each generic over $d$ such that $a+b = d$. \end{Proposition} {\em Proof.} Again we work over a reasonably saturated model $M$ (so in particular $T, S_{1}, S_{2}$ are defined over $M$). Let $d\in F^{M}$ be nonzero, and let $S_{3}$ be $d.T$. By Lemma 3.2, let $p(x)$, $q(x)$, $r(x)$ be generic types of $(F^{+})^{0}_{M}$ such that $p(x) \rightarrow x\in S_{1}$, $q(x) \rightarrow x\in S_{2}$ and $r(x) \rightarrow x\in S_{3}$. By Proposition 2.2, there are realisations $a,b,c$ of $p,q,r$ respectively, such that $a+b = c$, where moreover $\{a,b,c\}$ is pairwise independent over $M$. Multiplying on the right by $c^{-1}.d \in T$, we obtain $a' \in S_{1}$, $b'\in S_{2}$ with $a' + b' = d$. Clearly each of $a', b'$ are generic over $M$. This completes the proof. \vspace{5mm} \noindent [Here is a sketch of another proof of 3.4, avoiding use of Proposition 2.2 and containing some possibly useful ideas: Work again over a model $M$. Let $X$ = $St(S_{1},S_{2})$ which by definition equals $\{a\in F:$ for each generic type $p(x)\in S(M)$ of $S_{1}$ there is $c$ realising $p$ independent from $a$ such that $a.c \in S_{2}$\}. $X$ is type-definable over $M$ and moreover (see 2.2) there is a type-definable set $X'$, a union of cosets of $(F^{+})^{0}_{M}$ in $F^{+}$ such that $X \subset X'$ and also every generic element of $X'$ over $M$ is in $X$. Note that $X$ is invariant under multiplication by $T$. Let now $H$ be the set-theoretic stabilizer of $X'$. $H$ is a type-definable over $M$ subgroup of $F^{+}$ containing $(F^{+})^{0}_{M}$. $H$ is also invariant under multiplication by $T$, but we know that $(F^{+})^{0}_{M}$ intersects each coset of $T$ in $F^{*}$. Thus $H$ = $F^{+}$. It follows that $X'$ = $F$, and thus $X$ contains all generic types over $M$. As $X$ is also invariant under $T$ it follows that $X$ contains $F^{*}$.] \vspace{5mm} \noindent The characteristic $0$ case of the following was observed earlier together with Zoe Chatzidakis. \begin{Corollary} If $F$ is supersimple, then any conic defined over $F$ has an $F$-rational point. \end{Corollary} {\em Proof.} A conic defined over $F$ can be put into the form $x^{2}+ay^{2} = b$ for some nonzero $a,b \in F$. As we note in the next section, supersimplicity of $F$ implies that the squares form a definable subgroup of $F^{*}$ of finite index. By 3.4, the equation has a solution in $F$. \section{Triviality of the Brauer group for supersimple fields} We will prove that if $F$ is an infinite field possibly with extra structure, whose theory is supersimple, then $F$ has trivial Brauer group. We will see in the course of the proof that this is a first order property, so we will assume $F$ to be saturated. We will work with the formalism of the previous section, namely we assume $F$ to be definable over $\emptyset$ in a big model $\bar{M}$ of a supersimple theory. At this point it is worth mention the relevant facts about the $SU$-rank, most of which will be used in the next section. Each complete type has ordinal valued $SU$-rank, and the generic types of a (type-)definable group $G$ are precisely the types of maximal $SU$-rank in $G$ (which exist). The following is mentioned explicitly in \cite{Wagner} \begin{Fact} (i) for any elements $a,b$ and set $A$ of parameters, $SU(a/A\cup\{b\}) + SU(b/A) \leq SU(a,b/A) \leq SU(a/A\cup \{b\}) \oplus SU(b/A)$. \newline \noindent (ii) Let $G$ be a type-definable group and $H$ a type-definable subgroup, then $SU(H) + SU(G/H) \leq SU(G) \leq SU(H) \oplus SU(G/H)$ \newline \noindent (iii) Suppose $D$ is a type-definable division ring, then $D$ is definable and has monomial $SU$-rank. \end{Fact} The relevant consequences for this section are: \begin{Fact} (i) for each n, the group of nth. powers $(F^{*})^{n}$ has finite index in $F^{*}$, \newline \noindent (ii) $F$ is perfect. \end{Fact} {\em Proof.} (i) Let $a \in F$ be generic over $\emptyset$. Let $b = a^{n}$. Then $a \in acl(b)$, so $SU(a/b) = 0$. Thus by Fact 4.1. (i) $SU(a) = SU(b)$, whereby $b$ is generic in $F$ too. Thus $(F^{*})^{n}$ is a generic subgroup of $F^{*}$ and has finite index (as it is definable). \newline \noindent (ii) Supposing $F$ to have characteristic $p > 0$, note that as in (i) the additive group of $F^{p}$ has finite index in $F$. But $F/F^{p}$ is a vector space over the field $F^{p}$, a contradiction (the latter being infinite) unless $F^{p} = F$. \noindent Note that any finite extension of $F$ is also definable in $\bar{M}$, so Fact 4.2 applies to all finite extensions of $F$. We will now recall relevant notions and facts concerning the Brauer group, the norm map and Galois cohomology. In fact one can extract a general result which says that for the Brauer group of every finite extension of a field $F$ to be trivial it is enough that for any finite extension K of F and Kummer extension $L$ of $K$, $N_{L/K}$ is surjective. (Here it is assumed that every finite extension of $F$ is perfect.) Proposition 3.4 will allow us to conclude triviality of the Brauer group for supersimple $F$. Rather than simply state this reduction, we will include an explanation of it as part of the proof, which will entail giving some definitions. We first discuss the Brauer group. The reader can look at \cite{Jacobson} and \cite{Serre} for further details. We will assume that $F$ and all its finite extensions are perfect. By a central simple algebra over $F$ we mean a finite dimensional $F$-algebra $A$ whose centre is $F$ and which has no nontrivial two-sided ideals. If $A$ and $B$ are two such objects then so is the tensor product $A \otimes _{F} B$. $A$ and $B$ are called equivalent (or similar) if for some $m,n$ the matrix algebras $M_{m}(A)$, $M_{n}(B)$ are isomorphic (as $F$-algebras). The tensor product operation respects this equivalence relation and turns the set of classes into an abelian group. This group is called the {\em Brauer group of $F$}, $Br(F)$, and is an important invariant of the field $F$. Any central simple $F$-algebra $A$ will be a matrix algebra over a certain finite dimensional division algebra $D$ with centre $F$. Moreover the equivalence class of $A$ is determined by and determines the isomorphism type of $D$. The trivial element of $Br(F)$ then corresponds to $F$ itself. On the other hand, for any central simple algebra $A$ over $F$ there i s some finite extension $K$ of $F$ such that $A \otimes K$ is isomorphic to a matrix algebra over $K$, hence represents the trivial element of $Br(K)$. ($K$ is called a splitting field for $A$.) If $K$ is a finite extension of $F$ then tensoring with $K$ determines a homomorphism from $Br(F)$ into $Br(K)$. The kernel of this homomorphism is denoted $Br(K/F)$. From the previous paragraph, $Br(F)$ is the union of all $Br(K/F)$ as $K$ runs over finite extensions of $F$, in fact, over finite Galois extensions of $F$. For $K$ a finite Galois extension of $F$ with Galois group $G$, there is a classical isomorphism of $Br(K/F)$ with the Galois cohomology group $H^{2}(G,K^{*})$: given a $2$-cocycle $f:G \times G \rightarrow K^{*}$, define a $F$-algebra structure on the $K$-vector space $A$ with basis $\{u_{s}:s\in G\}$ by: $u_{s}\cdot u_{t} = f(s,t)u_{st}$ and for $e\in K$, $u_{s}e = (se)u_{s}$. $A$ then becomes a central simple $F$-algebra $A_{f}$, and the map $f \rightarrow A_{f}$ determines an isomorphism between $H^{2}(G,K^{*})$ and $Br(K/F)$. \vspace{3mm} \noindent We have the following fact (see \cite{Serre}, X.6): \begin{Fact} Let $F 0$, $b^{n} \notin Z(D)$ \newline \noindent (ii) for all $a \in D\setminus Z(D)$, $a^{D}-a \supseteq Z(D)$. \end{Lemma} {\em Proof.} (i) Note that as $char(D) = 0$, $D$ has no additive subgroups of finite index, and thus by Lemma 5.3 (ii), $\{x^{b}-x:x\in D\} = D$. Thus there are nonzero $c\in Z(D)$ and $a\notin Z(D)$ such that $a^{b}-a = c$. Then $b$ is in the normaliser in $D$ of $C_{D}(a)$ and $b \notin C_{D}(a)$. If $b^{n}\in Z(D)$ then the division ring generated by $C_{D}(a)$ and $b$ is noncommutative and finite-dimensional over the field $C_{D}(a)$, contradicting Corollary 4.7 \newline \noindent (ii) As $a\notin Z(D)$, $(a+Z(D))\cap Z(D) = \emptyset$. Lemma 5.3 (i) says there are only finitely many noncentral conjugacy classes in $D$ and so one of them, say $b^{D}$, must intersect $a+Z(D)$ in a generic subset of $a+Z(D)$ (namely a subset of $a+Z(D)$ of maximal $SU$-rank). Without loss of generality, $b \in a+Z(D)$. Thus $(b^{D}\cap (a+Z(D)) - b)$ which equals $(b^{D}-b)\cap Z(D)$ is a generic subset of $Z(D)$ which is easily seen to be an additive subgroup, and thus an additive subgroup of finite index. It follows, as we are in characteristic $0$ that $b^{D}-b$ contains $Z(D)$. In particular, note $a\in b+Z(D) \subset b^{D}$, so also $a^{D}-a$ contains $Z(D)$. \vspace{5mm} \noindent With the above lemmas we can finish the proof. We divide into cases. \vspace{5mm} \noindent {\bf Case I. $char(D) = p > 0$.} \newline \noindent By Lemma 5.3(ii) and compactness there is a uniform bound, say $m$, on the index in $D^{+}$ of the subgroup $\{x^{a}-x:x\in D\}$ for $a\in D\setminus Z(D)$. We will show that every element of $D^{*}/Z(D)^{*}$ has finite order, which by Kaplansky's Theorem (see 5.15 in \cite{Lam}) implies that $D$ is commutative. If $C_{D}(a)$ has cardinality at most $m$ then clearly $a$ has exponent at most $m$ in $D$. Otherwise we can (by choice of $m$) find nonzero $a' \in C_{D}(a)$ of the form $b^{a} - b$ for some $b\in D$. Note that $a$ and $b$ do not commute. But $a^{p}$ and $b$ do commute ($b^{a} = b + a'$ implies $b^{a^{p}} = b + pa' = b$). So $a^{p}$ is in $C_{D}(a)\cap C_{D}(b)$ which by Lemma 5.4 equals $Z(D)$. So $a$ has order $p$ in $D^{*}/Z(D)^{*}$. This finishes the proof of Theorem 5.1 in the positive characteristic case. \vspace{5mm} \noindent {\bf Case II. $char(D) = 0$.} \newline \noindent Suppose to begin with that every element of $Z(D)$ has a square root in $Z(D)$. Thus, we obtain infinitely many roots of unity $\{\omega _{i}:i < \omega\}$ in $Z(D)$. Let $a\notin Z(D)$. Then for some $i