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Let $K$ be a field and let $Q$ be a quaternion algebra over $K$. Then it is well-known that the class $[Q]$ of $Q$ in $Br(K)$ has order $2$. One can show this by constructing an explicit isomorphism $Q \otimes_K Q \cong M_2(K)$. My question is about the converse.

Does there exist a field $K$ and a division algebra $D$ over $K$ such that the class $[D]$ of $D$ in $Br(K)$ has order $2$, but such that $D$ is not isomorphic to a quaternion algebra over $K$?

If such a $K$ and $D$ exist, then it would also be nice to see an explicit example.

As a non-example, I believe that it follows from local and global class field theory that if $K$ is a local or global field, then every element of order $2$ in $Br(K)$ may indeed be represented by a quaternion algebra.

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It is a theorem of Merkurjev that the subgroup of elements of order $2$ is generated by quaternion algebras, if the field $K$ in question is of characteristic not $2$. In fact, the theorem is much more precise, and tells you that the group is $K_2(K)/2K_2(F)$. Thus such elements in the Brauer group are tensor products of quaternion algebras. –  Mariano Suárez-Alvarez Oct 23 '12 at 16:22
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(There is a similar result for the subgroup of elements of order $n$, using cyclic algebras, by Merrkujev and Suslin) –  Mariano Suárez-Alvarez Oct 23 '12 at 16:23
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See mathoverflow.net/questions/28237/… –  Felipe Voloch Oct 23 '12 at 16:28
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That is not a division algebra, though! –  Mariano Suárez-Alvarez Oct 23 '12 at 16:36
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@Felipe: Thanks for the link! There is a remark in there which says that there exist examples where K is a complex function field of transcendence degree three, which answers my question. –  Daniel Loughran Oct 23 '12 at 17:19
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4 Answers 4

up vote 2 down vote accepted

I have a vague memory that examples of $2$-torsion classes which are not exactly quaternion algebras can be found in one of the papers on Merkurjev's Theorem in "Applications of algebraic K-theory to algebraic geometry and number theory, Part 2", Contemporary Math. 1986 (either the one of Merkurjev, or the one of Wadsworth), but I can not check.

On the other hand, there are examples and references in this paper of A. Kresch:

arxiv.org/pdf/math/0009115

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Yes. Such an example is provided by a biquaternion algebra over $K$. It is a well known result due to Albert (see Lam, Introduction to Quadratic Forms over Fields) that the tensor product of two quaternion algebras is a division algebra iff they do not have a common quadratic splitting field.

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And to expand on this, if $F$ is any field of characteristic not 2, and one considers the purely transcendental extension $F(x, y, z, w)$, then over this field, the product of the two quaternion algebras: $(x,y) \otimes__{F(x,y,z,w)} (z,w)$ is a division algebra, by Albert's criteria above, and hence has index 4. It has period two because each quaternion algebra does. –  Danny Krashen Oct 24 '12 at 12:58
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I think what you're after is, in modern parlance, an algebra of period 2 but index strictly greater than 2. Googling "period-index problem" gives plenty of references and examples.

I had a look on Colliot-Thélène's web page and found the following article: "Exposant et indice d'algèbres centrales simples non ramifiées" (avec un appendice par Ofer Gabber), L'Enseignement Mathématique 48 (2002) 127–146. In that article, there is a nice introduction with a big list of references. Apparently the first example of the type you're after was given by Brauer himself: R. Brauer, "Untersuchungen über die arithmetischen Eigenschaften von Gruppen linearer Substitutionen", Zweite Mitteilung, Math. Zeitschrift 31 (1929) 733–747. The article by Brauer is available online, but only one page at a time from a rather clunky web site, so I haven't gone into it to track down the example.

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Thanks for the help Martin! –  Daniel Loughran Oct 24 '12 at 8:36
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By the Merkurjev-Suslin theorem, $K^M_2(K) \to \mathrm{Br}(K)[2]$ is surjective in characteristic $\neq 2$, and $K^M_2(K)$ is generated by symbols $\{a,b\}$, $a,b\in K^\times$. For characteristic $2$, see Gille-Szamuely, Chapter 9.

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Thanks for the answer, but I just want to clarify how it relates to my question. This theorem says that such a divison algebra $D$ is similar to a tensor product of quaternion algebras $Q_1\ldots Q_n$. But my question is asking whether there exists a field $K$ and a $D$ where one may not choose such a representation with $n=1$. –  Daniel Loughran Oct 23 '12 at 16:42
    
I suppose not every element of $K^M_2(K)/2$ is equivalent to a pure symbol $\{a,b\}$, so I guess the answer is "no". –  Timo Keller Oct 23 '12 at 16:59
    
What I am getting at is do you know examples of fields for which this can this happen? –  Daniel Loughran Oct 23 '12 at 17:04
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