# Central division and quaternion algebras

I would like to know if there are some central-simple algebras $D_1$, $D_2$ and $D_3$ over a field $k$ satisfying the following properties :

• $ind(D_1)=exp(D_1)=4$ ($ind$ is the Schur index and $exp$ the exponent);
• $D_2$ and $D_3$ are two non-isomorphic quaternion algebras;
• $ind(D_1^{\otimes2} \otimes D_2)=ind(D_1^{\otimes 2} \otimes D_3)=4$ (recall that $ind(D_1^{\otimes 2})$ is always $2$ in this setting).

To give a little motivation, i'm interested in the existence of "square roots" (in the Brauer group of a field) of quaternion algebras.

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There aren't any conditions on $D_2$, and $D_4$ seems to appear out of nowhere. –  S. Carnahan Aug 11 '11 at 8:41
oh my god sorry, i meant D_2 but wrote D_4... I corrected the typo, thank you. –  Louis Aug 11 '11 at 13:50
Could you explain what you mean by $D_1^2D_2$? Are you taking the tensor product of $D_1$, $D_1$, and $D_2$? –  André Henriques Aug 11 '11 at 13:59
exactly, i also correct this. Thank you. –  Louis Aug 11 '11 at 14:01
If $D$ is a quaternion algebra, then its local Hasse invariant is $1/2$ at a finite even number of places and zero elsewhere, so you can get a square root of the Brauer class by changing the nonzero invts to $1/4$ at half the places and $3/4$ at the other half. –  David Hansen Aug 11 '11 at 22:45

You can find such division algebras over $\mathbb{Q}(x_1,x_2,x_3)$ or $k(x_1,x_2,x_3,x_4)$ where $k$ is any field using the results of "Nakayama, T. Über die direkte Zerlegung einer Divisionsalgebra. Japanese J. of Mathematics 12 (1935), 65–70". A simplified proof is given in the book "Associative algebras" by Pierce, Corollary c, p. 381.
In the notation of Pierce, let $r=3$, $n_1 = 4$, $n_2 = 2$, $n_3=2$ and set $D_i$ to be the cyclic algebras denoted as $B_i$ by Pierce. The only thing one needs to observe is that $B_1^{\otimes 2}$ is equivalent to the cyclic algebra associated to the quadratic subfield of $E_1$ (and the element $x_1$) so we may apply the corollary to $D_1^{\otimes 2} \otimes D_2$ and $D_1^{\otimes 2} \otimes D_3$.